Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix multiplied by any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Let me think about how to approach this.

First, I remember that for a matrix A, the operator norm (which is the maximum value of ||Av|| / ||v|| for all non-zero vectors v) is the largest singular value of A. So, if I can find the largest singular value of this matrix, that should be the smallest such C. But maybe I need to verify this.

Alternatively, maybe I can think about it in terms of eigenvalues? Wait, but the operator norm isn't directly the eigenvalues unless the matrix is symmetric. This matrix isn't symmetric, though. The transpose of the matrix is [[2, 0], [3, -2]], which isn't the same as the original matrix. So maybe singular values are the way to go.

Singular values are the square roots of the eigenvalues of A*A, right? So let me compute A*A first. Let me write down matrix A:

A = [[2, 3],
     [0, -2]]

So A transpose is [[2, 0], [3, -2]]. Then A*A is:

First row first column: (2)(2) + (3)(3) = 4 + 9 = 13

First row second column: (2)(0) + (3)(-2) = 0 - 6 = -6

Second row first column: (0)(2) + (-2)(3) = 0 - 6 = -6

Second row second column: (0)(0) + (-2)(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]]. Now, to find the singular values, I need to find the eigenvalues of this matrix. Then take their square roots.

So let me find the eigenvalues of [[13, -6], [-6, 4]]. The characteristic equation is det(A*A - λI) = 0.

So determinant of [[13 - λ, -6], [-6, 4 - λ]] is (13 - λ)(4 - λ) - (-6)(-6) = 0.

Calculating that:

(13 - λ)(4 - λ) - 36 = 0

First expand (13 - λ)(4 - λ):

13*4 - 13λ - 4λ + λ² = 52 - 17λ + λ²

Then subtract 36: 52 - 17λ + λ² - 36 = 16 - 17λ + λ² = 0

So the equation is λ² -17λ +16 = 0

Wait, let me check that again. Wait, 52 - 36 is 16, yes. So the quadratic equation is λ² -17λ +16 = 0.

Wait, solving this quadratic equation: λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2.

So sqrt(225) is 15. Therefore, the eigenvalues are (17 + 15)/2 = 32/2 = 16, and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A*A are 16 and 1. Therefore, the singular values of A are sqrt(16) = 4 and sqrt(1) = 1. Therefore, the largest singular value is 4, so the operator norm is 4. Therefore, the smallest C is 4?

Wait, but let me check if this is correct. Let me think again.

Alternatively, maybe I can approach this problem by considering the norm of Av for any vector v. Let me write v as [x, y]^T. Then Av is [2x + 3y, -2y]^T. Then the norm squared of Av is (2x + 3y)^2 + (-2y)^2. Let me compute that:

(2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y².

So the norm squared of Av is 4x² + 12xy + 13y². The norm squared of v is x² + y². Therefore, the ratio squared is (4x² + 12xy + 13y²)/(x² + y²). Therefore, the maximum value of this ratio is the square of the operator norm, so the square root of that maximum is the operator norm. Therefore, if I can find the maximum of (4x² + 12xy + 13y²)/(x² + y²), then take its square root, that would be C.

Alternatively, since this is a homogeneous function, we can set x² + y² = 1, and maximize 4x² + 12xy + 13y². So, let me consider that.

Let me denote that expression as f(x, y) = 4x² + 12xy + 13y², with the constraint x² + y² = 1.

To maximize f(x, y) under the constraint x² + y² = 1. This is a standard problem in optimization, which can be solved using Lagrange multipliers.

Let me set up the Lagrangian: L = 4x² + 12xy + 13y² - λ(x² + y² - 1).

Taking partial derivatives:

∂L/∂x = 8x + 12y - 2λx = 0

∂L/∂y = 12x + 26y - 2λy = 0

∂L/∂λ = -(x² + y² - 1) = 0

So the equations are:

1) 8x + 12y = 2λx

2) 12x + 26y = 2λy

3) x² + y² = 1

Let me rearrange equations 1 and 2:

From equation 1: 8x + 12y = 2λx ⇒ 8x - 2λx + 12y = 0 ⇒ x(8 - 2λ) + 12y = 0 ⇒ x(4 - λ) + 6y = 0

Similarly, equation 2: 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0 ⇒ 12x + y(13 - λ) = 0

So now we have:

(4 - λ)x + 6y = 0

12x + (13 - λ)y = 0

This is a homogeneous system, so for non-trivial solutions, the determinant of the coefficients must be zero.

So determinant:

| (4 - λ)    6        |

| 12     (13 - λ) |  = 0

Calculating determinant:

(4 - λ)(13 - λ) - 6*12 = 0

Which is the same as before: (4 - λ)(13 - λ) - 72 = 0

Expanding (4 - λ)(13 - λ):

4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Then subtract 72: 52 - 17λ + λ² - 72 = λ² -17λ -20 = 0?

Wait, wait, earlier when I calculated the eigenvalues of A*A, I had λ² -17λ +16 = 0. But here, I have λ² -17λ -20 = 0? Wait, that seems conflicting. Wait, maybe I made a mistake here.

Wait, let me check again. The determinant is (4 - λ)(13 - λ) - 6*12. Let me compute that again:

(4 - λ)(13 - λ) = 4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Then subtract 6*12 = 72. So 52 - 17λ + λ² - 72 = λ² -17λ -20. Wait, 52 -72 is -20. So the equation is λ² -17λ -20 = 0. But earlier, when I calculated the eigenvalues of A*A, I had λ² -17λ +16 = 0. Wait, so there's a discrepancy here. Which one is correct?

Wait, let me check again the calculation of A*A. Original matrix A is [[2, 3], [0, -2]]. Then A transpose is [[2, 0], [3, -2]]. Then A*A is:

First row first column: 2*2 + 3*3 = 4 + 9 = 13

First row second column: 2*0 + 3*(-2) = 0 -6 = -6

Second row first column: 0*2 + (-2)*3 = 0 -6 = -6

Second row second column: 0*0 + (-2)*(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]], which is correct. Then the characteristic equation is det(A*A - λI) = (13 - λ)(4 - λ) - (-6)(-6) = (13 - λ)(4 - λ) - 36. Which is 52 -17λ + λ² - 36 = λ² -17λ +16 = 0. So that's correct. But in the Lagrange multiplier method, I ended up with λ² -17λ -20 = 0. So why is there a discrepancy?

Wait, maybe I made a mistake in the Lagrange multiplier equations. Let me check again.

Original equations:

From equation 1: 8x + 12y = 2λx ⇒ 8x - 2λx + 12y = 0 ⇒ x(8 - 2λ) + 12y = 0 ⇒ divide both sides by 2: x(4 - λ) + 6y = 0. That's correct.

Equation 2: 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0 ⇒ divide by 2: 6x + y(13 - λ) = 0. Wait, earlier I wrote 12x + (13 - λ)y = 0. But if I divide by 2, it's 6x + (13 - λ)y/2? Wait no, original equation is 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0. If I divide by 2, it's 6x + y(13 - λ) = 0. So that's correct. So the equations are:

(4 - λ)x + 6y = 0

6x + (13 - λ)y = 0

Wait, previously I had written 12x + (13 - λ)y = 0, but that was a mistake. It should be 6x + (13 - λ)y = 0. Therefore, the determinant is:

| (4 - λ)    6        |

| 6     (13 - λ) |  = 0

So determinant is (4 - λ)(13 - λ) - 6*6 = (4 - λ)(13 - λ) - 36. Which is same as before, which is 52 -17λ + λ² - 36 = λ² -17λ +16 = 0. Which matches the previous result. So earlier, I had mistakenly written 12x + ... instead of 6x + ..., which led to an error. So the correct equation is λ² -17λ +16 = 0, which gives λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2, so 16 and 1. Therefore, the eigenvalues of A*A are 16 and 1, so the singular values are 4 and 1. Therefore, the maximum singular value is 4, so the operator norm is 4. Therefore, the minimal C is 4. Therefore, the answer is 4. But let me check with another approach to be sure.

Alternatively, since the matrix is 2x2, maybe I can find the maximum value of ||Av|| by considering the maximum stretching factor. For that, maybe I can look at the eigenvalues of A? Wait, but A isn't symmetric, so its eigenvalues aren't necessarily related directly to the operator norm. However, the operator norm is the maximum singular value, which is the square root of the maximum eigenvalue of A*A, which we found to be 4. So that seems consistent.

Alternatively, let me test with specific vectors. For example, take v = [1, 0]^T. Then Av = [2, 0]^T, whose norm is 2. The norm of v is 1, so ratio is 2. If I take v = [0, 1]^T, then Av = [3, -2]^T, whose norm is sqrt(9 + 4) = sqrt(13) ≈ 3.605. Which is less than 4. What about another vector? Let me try to find a vector that gives a higher ratio.

Suppose I take a vector in the direction of the singular vector corresponding to the largest singular value. Since the largest singular value is 4, then there exists a vector v such that ||Av|| = 4||v||. Let me try to find such a vector.

Alternatively, since we found that the maximum of (4x² + 12xy + 13y²)/(x² + y²) is 16, because the maximum eigenvalue of A*A is 16. Therefore, the square root is 4. Therefore, the maximum ratio is 4. Therefore, the minimal C is 4.

But let me check with another vector. Suppose I take a vector that is an eigenvector of A*A corresponding to eigenvalue 16. Let me find that eigenvector. The eigenvalues of A*A are 16 and 1. Let me find the eigenvector for λ = 16.

From the equations:

(4 - λ)x + 6y = 0 ⇒ (4 - 16)x + 6y = 0 ⇒ -12x + 6y = 0 ⇒ -2x + y = 0 ⇒ y = 2x.

Similarly, the second equation: 6x + (13 - λ)y = 0 ⇒ 6x + (13 - 16)y = 0 ⇒ 6x - 3y = 0 ⇒ 2x - y = 0 ⇒ y = 2x. So same equation. Therefore, the eigenvectors are scalar multiples of [1, 2]^T. Let me check if this is correct.

Let me take v = [1, 2]^T. Then Av = [2*1 + 3*2, 0*1 + (-2)*2]^T = [2 + 6, -4]^T = [8, -4]^T. The norm of Av is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). The norm of v is sqrt(1² + 2²) = sqrt(5). Therefore, the ratio is 4*sqrt(5)/sqrt(5) = 4. Perfect! So that gives exactly 4. Therefore, this vector achieves the ratio 4, so C cannot be smaller than 4. And since the operator norm is 4, that's the minimal C.

Alternatively, let me check another vector. Suppose I take v = [1, 1]^T. Then Av = [2 + 3, -2]^T = [5, -2]^T. The norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(29)/sqrt(2) ≈ 5.385 / 1.414 ≈ 3.807, which is less than 4. So that's consistent.

Another test: take v = [1, 2]^T as before, which gives ratio 4. If I take v = [2, 4]^T, which is a scalar multiple, the ratio remains the same. So that's good.

Alternatively, let me check the other singular value. The other eigenvalue of A*A is 1, so the corresponding singular value is 1. Let me check with the eigenvector. For λ = 1, solving:

(4 - 1)x + 6y = 0 ⇒ 3x + 6y = 0 ⇒ x + 2y = 0 ⇒ x = -2y.

Second equation: 6x + (13 - 1)y = 0 ⇒ 6x + 12y = 0 ⇒ x + 2y = 0, same as before. So eigenvectors are scalar multiples of [-2, 1]^T. Let me check with v = [-2, 1]^T. Then Av = [2*(-2) + 3*1, 0*(-2) + (-2)*1]^T = [-4 + 3, -2]^T = [-1, -2]^T. The norm is sqrt(1 + 4) = sqrt(5). The norm of v is sqrt(4 + 1) = sqrt(5). Therefore, ratio is 1. So that's consistent with the singular value 1. Therefore, the operator norm is indeed 4.

Therefore, the minimal C is 4. Therefore, the answer is \boxed{4}.

But just to make sure, let me think if there's another approach. For example, using the fact that for any vector v, ||Av|| ≤ ||A|| ||v||, where ||A|| is the operator norm. But we need to find the minimal C such that this holds. Since we found that the operator norm is 4, and there exists a vector for which equality holds, then 4 is indeed the minimal such C.

Alternatively, maybe I can think in terms of the matrix's effect on vectors. The matrix A is upper triangular, so its eigenvalues are the diagonal entries, 2 and -2. However, the operator norm isn't directly the maximum of the absolute values of the eigenvalues. For example, if the matrix were diagonal, then the operator norm would be the maximum absolute value of the diagonal entries. But since this matrix isn't diagonal, but upper triangular, the operator norm can be larger. For instance, the entry 3 in the upper right can contribute to a larger norm when multiplied by a vector. So in this case, the operator norm is indeed larger than 2 and 2, which is 4, as we found.

Alternatively, maybe I can use the formula for the operator norm of a 2x2 matrix. For a general 2x2 matrix [[a, b], [c, d]], the operator norm is sqrt( (max eigenvalue of A*A) ). But we already did that.

Alternatively, maybe I can use the fact that for any vector v, ||Av||^2 = (Av) ⋅ (Av) = v^T A^T A v. Therefore, the maximum value of ||Av||^2 / ||v||^2 is the maximum eigenvalue of A^T A, which is 16, so sqrt(16) = 4. Therefore, again, the same result.

Therefore, I think I can be confident that the minimal C is 4.

**Final Answer**
\boxed{4}

To find the smallest positive real number $ C $ such that the inequality

$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$

holds for all two-dimensional vectors $ \bold{v} $, we need to determine the **operator norm** of the matrix $ A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} $.

---

### Step 1: Understanding the Operator Norm

The **operator norm** of a matrix $ A $ is defined as:

$$
\|A\| = \sup_{\|\bold{v}\| = 1} \|A\bold{v}\|
$$

This is the **maximum stretching factor** of the matrix $ A $ when applied to unit vectors. It is also equal to the **largest singular value** of $ A $, which is the square root of the largest eigenvalue of $ A^T A $.

---

### Step 2: Compute $ A^T A $

Given:

$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}, \quad A^T = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix}
$$

Then:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 4 + 0 & 6 - 6 \\ 6 - 6 & 9 + 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 13 \end{pmatrix}
$$

Wait — that's not correct. Let's recompute:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}
$$

Compute each entry:

- (1,1): $ 2 \cdot 2 + 0 \cdot 0 = 4 $
- (1,2): $ 2 \cdot 3 + 0 \cdot (-2) = 6 $
- (2,1): $ 3 \cdot 2 + (-2) \cdot 0 = 6 $
- (2,2): $ 3 \cdot 3 + (-2) \cdot (-2) = 9 + 4 = 13 $

So:

$$
A^T A = \begin{pmatrix} 4 & 6 \\ 6 & 13 \end{pmatrix}
$$

---

### Step 3: Find the Eigenvalues of $ A^T A $

We solve the characteristic equation:

$$
\det(A^T A - \lambda I) = \det\left( \begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} \right) = (4 - \lambda)(13 - \lambda) - 36
$$

$$
= 52 - 17\lambda + \lambda^2 - 36 = \lambda^2 - 17\lambda + 16
$$

Solving:

$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
$$

$$
\lambda = 16 \quad \text{or} \quad \lambda = 1
$$

---

### Step 4: Compute the Singular Values

The singular values of $ A $ are the square roots of the eigenvalues of $ A^T A $:

$$
\sigma_1 = \sqrt{16} = 4, \quad \sigma_2 = \sqrt{1} = 1
$$

So the **largest singular value** is $ \sigma_1 = 4 $, which is the **operator norm** of $ A $.

---

### Step 5: Conclusion

The smallest positive real number $ C $ such that

$$
\left\| A \bold{v} \right\| \le C \|\bold{v}\|
$$

for all vectors $ \bold{v} $ is the **operator norm** of $ A $, which is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix multiplied by any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Let me think about how to approach this.

First, I remember that for a matrix A, the operator norm (which is the maximum value of ||Av|| / ||v|| for all non-zero vectors v) is the largest singular value of A. So, if I can find the largest singular value of this matrix, that should be the smallest such C. But maybe I need to verify this.

Alternatively, maybe I can think about it in terms of eigenvalues? Wait, but the operator norm isn't directly the eigenvalues unless the matrix is symmetric. This matrix isn't symmetric, though. The transpose of the matrix is [[2, 0], [3, -2]], which isn't the same as the original matrix. So maybe singular values are the way to go.

Singular values are the square roots of the eigenvalues of A*A, right? So let me compute A*A first. Let me write down matrix A:

A = [[2, 3],
     [0, -2]]

So A transpose is [[2, 0], [3, -2]]. Then A*A is:

First row first column: (2)(2) + (3)(3) = 4 + 9 = 13

First row second column: (2)(0) + (3)(-2) = 0 - 6 = -6

Second row first column: (0)(2) + (-2)(3) = 0 - 6 = -6

Second row second column: (0)(0) + (-2)(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]]. Now, to find the singular values, I need to find the eigenvalues of this matrix. Then take their square roots.

So let me find the eigenvalues of [[13, -6], [-6, 4]]. The characteristic equation is det(A*A - λI) = 0.

So determinant of [[13 - λ, -6], [-6, 4 - λ]] is (13 - λ)(4 - λ) - (-6)(-6) = 0.

Calculating that:

(13 - λ)(4 - λ) - 36 = 0

First expand (13 - λ)(4 - λ):

13*4 - 13λ - 4λ + λ² = 52 - 17λ + λ²

Then subtract 36: 52 - 17λ + λ² - 36 = 16 - 17λ + λ² = 0

So the equation is λ² -17λ +16 = 0

Wait, let me check that again. Wait, 52 - 36 is 16, yes. So the quadratic equation is λ² -17λ +16 = 0.

Wait, solving this quadratic equation: λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2.

So sqrt(225) is 15. Therefore, the eigenvalues are (17 + 15)/2 = 32/2 = 16, and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A*A are 16 and 1. Therefore, the singular values of A are sqrt(16) = 4 and sqrt(1) = 1. Therefore, the largest singular value is 4, so the operator norm is 4. Therefore, the smallest C is 4?

Wait, but let me check if this is correct. Let me think again.

Alternatively, maybe I can approach this problem by considering the norm of Av for any vector v. Let me write v as [x, y]^T. Then Av is [2x + 3y, -2y]^T. Then the norm squared of Av is (2x + 3y)^2 + (-2y)^2. Let me compute that:

(2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y².

So the norm squared of Av is 4x² + 12xy + 13y². The norm squared of v is x² + y². Therefore, the ratio squared is (4x² + 12xy + 13y²)/(x² + y²). Therefore, the maximum value of this ratio is the square of the operator norm, so the square root of that maximum is the operator norm. Therefore, if I can find the maximum of (4x² + 12xy + 13y²)/(x² + y²), then take its square root, that would be C.

Alternatively, since this is a homogeneous function, we can set x² + y² = 1, and maximize 4x² + 12xy + 13y². So, let me consider that.

Let me denote that expression as f(x, y) = 4x² + 12xy + 13y², with the constraint x² + y² = 1.

To maximize f(x, y) under the constraint x² + y² = 1. This is a standard problem in optimization, which can be solved using Lagrange multipliers.

Let me set up the Lagrangian: L = 4x² + 12xy + 13y² - λ(x² + y² - 1).

Taking partial derivatives:

∂L/∂x = 8x + 12y - 2λx = 0

∂L/∂y = 12x + 26y - 2λy = 0

∂L/∂λ = -(x² + y² - 1) = 0

So the equations are:

1) 8x + 12y = 2λx

2) 12x + 26y = 2λy

3) x² + y² = 1

Let me rearrange equations 1 and 2:

From equation 1: 8x + 12y = 2λx ⇒ 8x - 2λx + 12y = 0 ⇒ x(8 - 2λ) + 12y = 0 ⇒ x(4 - λ) + 6y = 0

Similarly, equation 2: 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0 ⇒ 12x + y(13 - λ) = 0

So now we have:

(4 - λ)x + 6y = 0

12x + (13 - λ)y = 0

This is a homogeneous system, so for non-trivial solutions, the determinant of the coefficients must be zero.

So determinant:

| (4 - λ)    6        |

| 12     (13 - λ) |  = 0

Calculating determinant:

(4 - λ)(13 - λ) - 6*12 = 0

Which is the same as before: (4 - λ)(13 - λ) - 72 = 0

Expanding (4 - λ)(13 - λ):

4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Then subtract 72: 52 - 17λ + λ² - 72 = λ² -17λ -20 = 0?

Wait, wait, earlier when I calculated the eigenvalues of A*A, I had λ² -17λ +16 = 0. But here, I have λ² -17λ -20 = 0? Wait, that seems conflicting. Wait, maybe I made a mistake here.

Wait, let me check again. The determinant is (4 - λ)(13 - λ) - 6*12. Let me compute that again:

(4 - λ)(13 - λ) = 4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Then subtract 6*12 = 72. So 52 - 17λ + λ² - 72 = λ² -17λ -20. Wait, 52 -72 is -20. So the equation is λ² -17λ -20 = 0. But earlier, when I calculated the eigenvalues of A*A, I had λ² -17λ +16 = 0. Wait, so there's a discrepancy here. Which one is correct?

Wait, let me check again the calculation of A*A. Original matrix A is [[2, 3], [0, -2]]. Then A transpose is [[2, 0], [3, -2]]. Then A*A is:

First row first column: 2*2 + 3*3 = 4 + 9 = 13

First row second column: 2*0 + 3*(-2) = 0 -6 = -6

Second row first column: 0*2 + (-2)*3 = 0 -6 = -6

Second row second column: 0*0 + (-2)*(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]], which is correct. Then the characteristic equation is det(A*A - λI) = (13 - λ)(4 - λ) - (-6)(-6) = (13 - λ)(4 - λ) - 36. Which is 52 -17λ + λ² - 36 = λ² -17λ +16 = 0. So that's correct. But in the Lagrange multiplier method, I ended up with λ² -17λ -20 = 0. So why is there a discrepancy?

Wait, maybe I made a mistake in the Lagrange multiplier equations. Let me check again.

Original equations:

From equation 1: 8x + 12y = 2λx ⇒ 8x - 2λx + 12y = 0 ⇒ x(8 - 2λ) + 12y = 0 ⇒ divide both sides by 2: x(4 - λ) + 6y = 0. That's correct.

Equation 2: 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0 ⇒ divide by 2: 6x + y(13 - λ) = 0. Wait, earlier I wrote 12x + (13 - λ)y = 0. But if I divide by 2, it's 6x + (13 - λ)y/2? Wait no, original equation is 12x + 26y = 2λy ⇒ 12x + 26y - 2λy = 0 ⇒ 12x + y(26 - 2λ) = 0. If I divide by 2, it's 6x + y(13 - λ) = 0. So that's correct. So the equations are:

(4 - λ)x + 6y = 0

6x + (13 - λ)y = 0

Wait, previously I had written 12x + (13 - λ)y = 0, but that was a mistake. It should be 6x + (13 - λ)y = 0. Therefore, the determinant is:

| (4 - λ)    6        |

| 6     (13 - λ) |  = 0

So determinant is (4 - λ)(13 - λ) - 6*6 = (4 - λ)(13 - λ) - 36. Which is same as before, which is 52 -17λ + λ² - 36 = λ² -17λ +16 = 0. Which matches the previous result. So earlier, I had mistakenly written 12x + ... instead of 6x + ..., which led to an error. So the correct equation is λ² -17λ +16 = 0, which gives λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2, so 16 and 1. Therefore, the eigenvalues of A*A are 16 and 1, so the singular values are 4 and 1. Therefore, the maximum singular value is 4, so the operator norm is 4. Therefore, the minimal C is 4. Therefore, the answer is 4. But let me check with another approach to be sure.

Alternatively, since the matrix is 2x2, maybe I can find the maximum value of ||Av|| by considering the maximum stretching factor. For that, maybe I can look at the eigenvalues of A? Wait, but A isn't symmetric, so its eigenvalues aren't necessarily related directly to the operator norm. However, the operator norm is the maximum singular value, which is the square root of the maximum eigenvalue of A*A, which we found to be 4. So that seems consistent.

Alternatively, let me test with specific vectors. For example, take v = [1, 0]^T. Then Av = [2, 0]^T, whose norm is 2. The norm of v is 1, so ratio is 2. If I take v = [0, 1]^T, then Av = [3, -2]^T, whose norm is sqrt(9 + 4) = sqrt(13) ≈ 3.605. Which is less than 4. What about another vector? Let me try to find a vector that gives a higher ratio.

Suppose I take a vector in the direction of the singular vector corresponding to the largest singular value. Since the largest singular value is 4, then there exists a vector v such that ||Av|| = 4||v||. Let me try to find such a vector.

Alternatively, since we found that the maximum of (4x² + 12xy + 13y²)/(x² + y²) is 16, because the maximum eigenvalue of A*A is 16. Therefore, the square root is 4. Therefore, the maximum ratio is 4. Therefore, the minimal C is 4.

But let me check with another vector. Suppose I take a vector that is an eigenvector of A*A corresponding to eigenvalue 16. Let me find that eigenvector. The eigenvalues of A*A are 16 and 1. Let me find the eigenvector for λ = 16.

From the equations:

(4 - λ)x + 6y = 0 ⇒ (4 - 16)x + 6y = 0 ⇒ -12x + 6y = 0 ⇒ -2x + y = 0 ⇒ y = 2x.

Similarly, the second equation: 6x + (13 - λ)y = 0 ⇒ 6x + (13 - 16)y = 0 ⇒ 6x - 3y = 0 ⇒ 2x - y = 0 ⇒ y = 2x. So same equation. Therefore, the eigenvectors are scalar multiples of [1, 2]^T. Let me check if this is correct.

Let me take v = [1, 2]^T. Then Av = [2*1 + 3*2, 0*1 + (-2)*2]^T = [2 + 6, -4]^T = [8, -4]^T. The norm of Av is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). The norm of v is sqrt(1² + 2²) = sqrt(5). Therefore, the ratio is 4*sqrt(5)/sqrt(5) = 4. Perfect! So that gives exactly 4. Therefore, this vector achieves the ratio 4, so C cannot be smaller than 4. And since the operator norm is 4, that's the minimal C.

Alternatively, let me check another vector. Suppose I take v = [1, 1]^T. Then Av = [2 + 3, -2]^T = [5, -2]^T. The norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(29)/sqrt(2) ≈ 5.385 / 1.414 ≈ 3.807, which is less than 4. So that's consistent.

Another test: take v = [1, 2]^T as before, which gives ratio 4. If I take v = [2, 4]^T, which is a scalar multiple, the ratio remains the same. So that's good.

Alternatively, let me check the other singular value. The other eigenvalue of A*A is 1, so the corresponding singular value is 1. Let me check with the eigenvector. For λ = 1, solving:

(4 - 1)x + 6y = 0 ⇒ 3x + 6y = 0 ⇒ x + 2y = 0 ⇒ x = -2y.

Second equation: 6x + (13 - 1)y = 0 ⇒ 6x + 12y = 0 ⇒ x + 2y = 0, same as before. So eigenvectors are scalar multiples of [-2, 1]^T. Let me check with v = [-2, 1]^T. Then Av = [2*(-2) + 3*1, 0*(-2) + (-2)*1]^T = [-4 + 3, -2]^T = [-1, -2]^T. The norm is sqrt(1 + 4) = sqrt(5). The norm of v is sqrt(4 + 1) = sqrt(5). Therefore, ratio is 1. So that's consistent with the singular value 1. Therefore, the operator norm is indeed 4.

Therefore, the minimal C is 4. Therefore, the answer is \boxed{4}.

But just to make sure, let me think if there's another approach. For example, using the fact that for any vector v, ||Av|| ≤ ||A|| ||v||, where ||A|| is the operator norm. But we need to find the minimal C such that this holds. Since we found that the operator norm is 4, and there exists a vector for which equality holds, then 4 is indeed the minimal such C.

Alternatively, maybe I can think in terms of the matrix's effect on vectors. The matrix A is upper triangular, so its eigenvalues are the diagonal entries, 2 and -2. However, the operator norm isn't directly the maximum of the absolute values of the eigenvalues. For example, if the matrix were diagonal, then the operator norm would be the maximum absolute value of the diagonal entries. But since this matrix isn't diagonal, but upper triangular, the operator norm can be larger. For instance, the entry 3 in the upper right can contribute to a larger norm when multiplied by a vector. So in this case, the operator norm is indeed larger than 2 and 2, which is 4, as we found.

Alternatively, maybe I can use the formula for the operator norm of a 2x2 matrix. For a general 2x2 matrix [[a, b], [c, d]], the operator norm is sqrt( (max eigenvalue of A*A) ). But we already did that.

Alternatively, maybe I can use the fact that for any vector v, ||Av||^2 = (Av) ⋅ (Av) = v^T A^T A v. Therefore, the maximum value of ||Av||^2 / ||v||^2 is the maximum eigenvalue of A^T A, which is 16, so sqrt(16) = 4. Therefore, again, the same result.

Therefore, I think I can be confident that the minimal C is 4.

**Final Answer**
\boxed{4}

To find the smallest positive real number $ C $ such that the inequality

$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$

holds for all two-dimensional vectors $ \bold{v} $, we need to determine the **operator norm** of the matrix $ A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} $.

---

### Step 1: Understanding the Operator Norm

The **operator norm** of a matrix $ A $ is defined as:

$$
\|A\| = \sup_{\|\bold{v}\| = 1} \|A\bold{v}\|
$$

This is the **maximum stretching factor** of the matrix $ A $ when applied to unit vectors. It is also equal to the **largest singular value** of $ A $, which is the square root of the largest eigenvalue of $ A^T A $.

---

### Step 2: Compute $ A^T A $

Given:

$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}, \quad A^T = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix}
$$

Then:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 4 + 0 & 6 - 6 \\ 6 - 6 & 9 + 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 13 \end{pmatrix}
$$

Wait — that's not correct. Let's recompute:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}
$$

Compute each entry:

- (1,1): $ 2 \cdot 2 + 0 \cdot 0 = 4 $
- (1,2): $ 2 \cdot 3 + 0 \cdot (-2) = 6 $
- (2,1): $ 3 \cdot 2 + (-2) \cdot 0 = 6 $
- (2,2): $ 3 \cdot 3 + (-2) \cdot (-2) = 9 + 4 = 13 $

So:

$$
A^T A = \begin{pmatrix} 4 & 6 \\ 6 & 13 \end{pmatrix}
$$

---

### Step 3: Find the Eigenvalues of $ A^T A $

We solve the characteristic equation:

$$
\det(A^T A - \lambda I) = \det\left( \begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} \right) = (4 - \lambda)(13 - \lambda) - 36
$$

$$
= 52 - 17\lambda + \lambda^2 - 36 = \lambda^2 - 17\lambda + 16
$$

Solving:

$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
$$

$$
\lambda = 16 \quad \text{or} \quad \lambda = 1
$$

---

### Step 4: Compute the Singular Values

The singular values of $ A $ are the square roots of the eigenvalues of $ A^T A $:

$$
\sigma_1 = \sqrt{16} = 4, \quad \sigma_2 = \sqrt{1} = 1
$$

So the **largest singular value** is $ \sigma_1 = 4 $, which is the **operator norm** of $ A $.

---

### Step 5: Conclusion

The smallest positive real number $ C $ such that

$$
\left\| A \bold{v} \right\| \le C \|\bold{v}\|
$$

for all vectors $ \bold{v} $ is the **operator norm** of $ A $, which is:

$$
\boxed{4}
$$

Solution:
Let $\bold{v} = \begin{pmatrix} x \\ y \end{pmatrix}$.  Then
\[\|\bold{v}\| = \left\| \begin{pmatrix} x \\ y \end{pmatrix} \right\| = \sqrt{x^2 + y^2},\]and
\begin{align*}
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| &= \left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right\| \\
&= \left\| \begin{pmatrix} 2x + 3y \\ -2y \end{pmatrix} \right\| \\
&= \sqrt{(2x + 3y)^2 + (-2y)^2} \\
&= \sqrt{4x^2 + 12xy + 13y^2},
\end{align*}so the given inequality becomes
\[\sqrt{4x^2 + 12xy + 13y^2} \le C \sqrt{x^2 + y^2},\]or
\[\sqrt{\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}} \le C.\]Thus, we can think of $C$ as the maximum value of the expression in the left-hand side.

Maximizing the expression in the left-hand side is equivalent to maximizing its square, namely
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}.\]Let $k$ be a possible value of this expression, which means the equation
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2} = k\]has a solution in $x$ and $y$.  We can re-write this equation as
\[(4 - k) x^2 + 12xy + (13 - k) y^2 = 0.\]For this quadratic expression to have a solution in $x$ and $y$, its discriminant must be nonnegative.  In other words,
\[12^2 - 4 (4 - k)(13 - k) \ge 0,\]or $4k^2 - 68k + 64 \le 0$.  This inequality factors as $4(k - 1)(k - 16) \le 0$.  The largest value of $k$ that satisfies this inequality is 16, so the value of $C$ we seek is $\sqrt{16} = \boxed{4}$.  Note that equality occurs for
\[\bold{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]


Explanation:
Okay, so I need to factor the polynomial $ x^8 + 3x^4 - 4 $ into monic polynomials with integer coefficients that can't be factored further over the integers. Then, I have to compute the sum of each of these polynomials evaluated at 1. Let me start by trying to factor the given polynomial.

First, I notice that the polynomial is $ x^8 + 3x^4 - 4 $. It looks like a quadratic in terms of $ x^4 $. Maybe I can make a substitution to simplify it. Let me set $ y = x^4 $. Then the polynomial becomes $ y^2 + 3y - 4 $. Let me try to factor this quadratic.

Looking for two numbers that multiply to -4 and add up to 3. Let me think... 4 and -1? Yes, 4 * (-1) = -4 and 4 + (-1) = 3. So, the quadratic factors as $ (y + 4)(y - 1) $. Therefore, substituting back $ y = x^4 $, we get:

$ (x^4 + 4)(x^4 - 1) $.

So, the original polynomial factors into $ (x^4 + 4)(x^4 - 1) $. Now, I need to check if these factors can be factored further over the integers.

Starting with $ x^4 - 1 $. This is a difference of squares: $ x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) $. Then, $ x^2 - 1 $ is also a difference of squares: $ (x - 1)(x + 1) $. So, putting it all together:

$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $.

Now, check if $ x^4 + 4 $ can be factored further. Hmm, I remember that $ x^4 + 4 $ is a special case. It's a biquadratic polynomial. Maybe it can be factored into quadratics? Let me try to factor it.

I recall that $ x^4 + 4 $ can be factored using the Sophie Germain identity. The identity is $ a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) $. Let me check if that applies here. If I set $ a = x $ and $ b = 1 $, then:

$ x^4 + 4(1)^4 = x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Yes! So, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Let me verify this multiplication:

Multiply $ (x^2 + 2x + 2)(x^2 - 2x + 2) $:

First, multiply $ x^2 $ by each term in the second polynomial: $ x^2 * x^2 = x^4 $, $ x^2 * (-2x) = -2x^3 $, $ x^2 * 2 = 2x^2 $.

Then, multiply 2x by each term: $ 2x * x^2 = 2x^3 $, $ 2x * (-2x) = -4x^2 $, $ 2x * 2 = 4x $.

Then, multiply 2 by each term: $ 2 * x^2 = 2x^2 $, $ 2 * (-2x) = -4x $, $ 2 * 2 = 4 $.

Now, add all these terms together:

- $ x^4 $
- $ -2x^3 + 2x^3 = 0 $
- $ 2x^2 - 4x^2 + 2x^2 = 0 $
- $ 4x - 4x = 0 $
- $ +4 $

So, the result is $ x^4 + 4 $. Perfect! So, that factorization is correct.

Therefore, the original polynomial factors as:

$ (x^2 + 2x + 2)(x^2 - 2x + 2)(x - 1)(x + 1)(x^2 + 1) $.

Now, I need to check if any of these factors can be factored further over the integers. Let's go through each factor:

1. $ x - 1 $: This is linear, so it can't be factored further.
2. $ x + 1 $: Similarly, linear, can't be factored further.
3. $ x^2 + 1 $: This is a quadratic. Over integers, it doesn't factor because the discriminant is $ 0^2 - 4*1*1 = -4 $, which is negative. So, it's irreducible over integers.
4. $ x^2 + 2x + 2 $: Let's check discriminant: $ (2)^2 - 4*1*2 = 4 - 8 = -4 $. Negative discriminant, so irreducible over integers.
5. $ x^2 - 2x + 2 $: Discriminant is $ (-2)^2 - 4*1*2 = 4 - 8 = -4 $. Also negative, so irreducible over integers.

Therefore, all factors are irreducible over integers. So, the complete factorization is into these five polynomials: $ (x - 1) $, $ (x + 1) $, $ (x^2 + 1) $, $ (x^2 + 2x + 2) $, and $ (x^2 - 2x + 2) $.

Wait, but the problem says "each non-constant polynomial $ p_i(x) $ is monic with integer coefficients, and cannot be factored further over the integers." So, all these factors are monic, have integer coefficients, and are irreducible. Therefore, the factorization is correct.

Now, the question asks for the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $. So, I need to evaluate each of these polynomials at x=1 and add them up.

Let me compute each one:

1. $ p_1(x) = x - 1 $: $ p_1(1) = 1 - 1 = 0 $
2. $ p_2(x) = x + 1 $: $ p_2(1) = 1 + 1 = 2 $
3. $ p_3(x) = x^2 + 1 $: $ p_3(1) = 1^2 + 1 = 1 + 1 = 2 $
4. $ p_4(x) = x^2 + 2x + 2 $: $ p_4(1) = 1^2 + 2*1 + 2 = 1 + 2 + 2 = 5 $
5. $ p_5(x) = x^2 - 2x + 2 $: $ p_5(1) = 1^2 - 2*1 + 2 = 1 - 2 + 2 = 1 $

Adding them up: 0 + 2 + 2 + 5 + 1 = 10.

Wait, let me check again:

First polynomial: 0

Second: 2

Third: 2

Fourth: 5

Fifth: 1

Total: 0 + 2 is 2, plus 2 is 4, plus 5 is 9, plus 1 is 10. So, 10?

But let me check if I have the correct factors. Let me verify the factorization again.

Original polynomial: $ x^8 + 3x^4 - 4 $. Let me compute the product of the factors:

First, multiply $ (x - 1)(x + 1) = x^2 - 1 $.

Then, multiply by $ x^2 + 1 $: $ (x^2 - 1)(x^2 + 1) = x^4 - 1 $.

Then, multiply by $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Earlier, we saw that this product is $ x^4 + 4 $.

Therefore, multiplying $ x^4 - 1 $ and $ x^4 + 4 $ gives $ (x^4 - 1)(x^4 + 4) = x^8 + 4x^4 - x^4 - 4 = x^8 + 3x^4 - 4 $, which matches the original polynomial. So, the factorization is correct.

Therefore, the factors are indeed those five polynomials. Then, evaluating each at 1 gives 0, 2, 2, 5, 1. Sum is 10. So, the answer is 10?

Wait, but let me check again the evaluation of each polynomial at 1:

For $ x - 1 $: 1 - 1 = 0. Correct.

For $ x + 1 $: 1 + 1 = 2. Correct.

For $ x^2 + 1 $: 1 + 1 = 2. Correct.

For $ x^2 + 2x + 2 $: 1 + 2 + 2 = 5. Correct.

For $ x^2 - 2x + 2 $: 1 - 2 + 2 = 1. Correct.

Adding them up: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Hmm. But let me think again. Wait, the problem says "each non-constant polynomial p_i(x) is monic with integer coefficients, and cannot be factored further over the integers." So, the factors are as above. But maybe there's another way to factor the original polynomial? Let me check if there are other possible factorizations?

Alternatively, maybe I can factor the original polynomial in another way? Let me think. For example, maybe there are more factors? Let me check.

But given that we have already factored it into irreducible polynomials, and the factorization is unique up to ordering, so the answer should be 10. However, let me check if maybe some of the factors can be combined or if I missed something.

Wait, but let me think again. Suppose I didn't use the Sophie Germain identity. Let me try to factor $ x^4 + 4 $ another way. Suppose I try to factor it into quadratics with integer coefficients. Let me assume that $ x^4 + 4 = (x^2 + ax + b)(x^2 + cx + d) $. Then expanding:

$ x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd $.

Set equal to $ x^4 + 0x^3 + 0x^2 + 0x + 4 $. Therefore, we have the system:

1. $ a + c = 0 $
2. $ ac + b + d = 0 $
3. $ ad + bc = 0 $
4. $ bd = 4 $

From equation 1: $ c = -a $.

Substitute into equation 3: $ ad + b(-a) = a(d - b) = 0 $. So, either a = 0 or d = b.

Case 1: a = 0. Then c = 0. Then equation 2 becomes 0 + b + d = 0 => b + d = 0. Equation 4: bd = 4. So, solving b + d = 0 and bd = 4. Then, d = -b, so b*(-b) = -b² = 4 => b² = -4. No real solutions, so no integer solutions here.

Case 2: d = b. Then equation 4: b*d = b² = 4. Therefore, b = ±2. Then, equation 2: ac + b + d = (-a²) + 2b = 0. Since c = -a, ac = -a². So:

- a² + 2b = 0 => a² = 2b.

Since b is ±2:

If b = 2: a² = 4 => a = ±2.

If b = -2: a² = -4, which is impossible.

So, for b = 2, a = ±2. Let's check:

If a = 2, then c = -2. Then, equation 3 is already satisfied because d = b = 2. So, the factors would be:

$ (x^2 + 2x + 2)(x^2 - 2x + 2) $, which is exactly the factorization we had earlier. Similarly, if a = -2, then c = 2, and the factors would be $ (x^2 - 2x + 2)(x^2 + 2x + 2) $, same thing. So, that's the only factorization. Therefore, our previous factorization is correct.

Therefore, the factorization into irreducible polynomials is correct, and the sum is 10. However, let me check once more the problem statement to make sure I didn't misread anything.

The problem says: "Compute $ p_1(1) + p_2(1) + \dots + p_k(1) $." So, if there are five factors, then sum is 10. But let me check if maybe the problem counts factors with multiplicity? But no, since each factor is distinct and irreducible. Wait, but maybe I made a mistake in the number of factors?

Original polynomial: degree 8. The factors are:

- (x - 1): degree 1

- (x + 1): degree 1

- (x^2 + 1): degree 2

- (x^2 + 2x + 2): degree 2

- (x^2 - 2x + 2): degree 2

Total degrees: 1 + 1 + 2 + 2 + 2 = 8. Correct. So, five factors. Therefore, the sum is 0 + 2 + 2 + 5 + 1 = 10. So, 10. But let me check again the evaluations:

For (x - 1): 1 - 1 = 0. Correct.

For (x + 1): 1 + 1 = 2. Correct.

For (x² + 1): 1 + 1 = 2. Correct.

For (x² + 2x + 2): 1 + 2 + 2 = 5. Correct.

For (x² - 2x + 2): 1 - 2 + 2 = 1. Correct.

So, adding them up: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Therefore, the answer is 10? Wait, but I have a doubt because sometimes problems like this might have a trick. Let me check if there's another way to factor the original polynomial?

Alternatively, maybe I can factor the original polynomial as a product of quadratics? Let me try. Suppose I try to factor $ x^8 + 3x^4 - 4 $ as a product of quadratics and other terms. But since we already have the factorization into linear and quadratics, and we checked that they are irreducible, I think that's the complete factorization.

Alternatively, maybe there's a different factorization? Let me think. Suppose I try to factor $ x^8 + 3x^4 - 4 $ as a product of two quartic polynomials? But since we already factored it into a product of a quartic and another quartic, which then further factors into lower degrees. But since the problem says "cannot be factored further over the integers," so the answer is as above.

Alternatively, maybe I can check the value of the original polynomial at x=1 and compare with the product of the p_i(1). Let me compute the original polynomial at x=1: $ 1^8 + 3*1^4 - 4 = 1 + 3 - 4 = 0 $. On the other hand, the product of the p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. Which matches. So, that's consistent. Therefore, the sum of p_i(1) is 10. Therefore, I think the answer is 10.

But let me check again if there's a possibility that some factors are repeated or something. For example, maybe the polynomial can be factored into more factors? For example, if there are more linear factors? But the original polynomial evaluated at x=1 is 0, so x=1 is a root. Similarly, x=-1 is a root? Let me check:

Original polynomial at x=-1: (-1)^8 + 3*(-1)^4 - 4 = 1 + 3 - 4 = 0. Yes, so x=-1 is also a root. Therefore, (x - 1) and (x + 1) are factors, which we have. So, the factorization is correct.

Alternatively, maybe there are other roots? Let me check. Let me try to find all roots of the original polynomial. Let me set $ x^8 + 3x^4 - 4 = 0 $. Let me substitute y = x^4 again: y² + 3y - 4 = 0. Which factors as (y + 4)(y - 1) = 0. Therefore, y = -4 or y = 1. Therefore, x^4 = -4 or x^4 = 1. The solutions for x^4 = 1 are the 4th roots of unity: 1, -1, i, -i. The solutions for x^4 = -4 are complex numbers. Therefore, over real numbers, the roots are 1, -1, i, -i, and four complex roots from x^4 = -4. But over integers, we can only factor the polynomial into polynomials with integer coefficients, which we have done. Therefore, the factorization is correct.

Therefore, I think my answer is correct. The sum is 10. Therefore, the answer is \boxed{10}.

**Final Answer**
\boxed{10}

To solve the problem, we are given the polynomial:

$$
x^8 + 3x^4 - 4
$$

and asked to factor it into monic polynomials with integer coefficients that cannot be factored further over the integers. Then, we are to compute the sum of the values of these polynomials evaluated at $ x = 1 $.

---

### Step 1: Factor the Polynomial

We begin by recognizing that the polynomial is a **biquadratic** in terms of $ x^4 $. Let’s make the substitution:

$$
y = x^4
$$

Then the polynomial becomes:

$$
y^2 + 3y - 4
$$

This is a quadratic in $ y $, which we can factor:

$$
y^2 + 3y - 4 = (y + 4)(y - 1)
$$

Substituting back $ y = x^4 $, we get:

$$
x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1)
$$

---

### Step 2: Factor Each Component

Now we factor each of the two components:

- $ x^4 - 1 $ is a **difference of squares**:

$$
x^4 - 1 = (x^2 - 1)(x^2 + 1)
$$

- $ x^2 - 1 $ is also a **difference of squares**:

$$
x^2 - 1 = (x - 1)(x + 1)
$$

So:

$$
x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)
$$

Now, we factor $ x^4 + 4 $. This is a special case that can be factored using the **Sophie Germain identity**:

$$
x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)
$$

---

### Step 3: Final Factorization

Putting it all together, the full factorization of the original polynomial is:

$$
x^8 + 3x^4 - 4 = (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2)
$$

Each of these factors is **monic**, has **integer coefficients**, and is **irreducible over the integers**. Therefore, this is the complete factorization.

---

### Step 4: Evaluate Each Factor at $ x = 1 $

We now compute the value of each factor at $ x = 1 $:

- $ (x - 1) \Rightarrow 1 - 1 = 0 $
- $ (x + 1) \Rightarrow 1 + 1 = 2 $
- $ (x^2 + 1) \Rightarrow 1^2 + 1 = 2 $
- $ (x^2 + 2x + 2) \Rightarrow 1^2 + 2(1) + 2 = 1 + 2 + 2 = 5 $
- $ (x^2 - 2x + 2) \Rightarrow 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1 $

---

### Step 5: Compute the Sum

Now, summing the values:

$$
0 + 2 + 2 + 5 + 1 = 10
$$

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to factor the polynomial $ x^8 + 3x^4 - 4 $ into monic polynomials with integer coefficients that can't be factored further over the integers. Then, I have to compute the sum of each of these polynomials evaluated at 1. Let me start by trying to factor the given polynomial.

First, I notice that the polynomial is $ x^8 + 3x^4 - 4 $. It looks like a quadratic in terms of $ x^4 $. Maybe I can make a substitution to simplify it. Let me set $ y = x^4 $. Then the polynomial becomes $ y^2 + 3y - 4 $. Let me try to factor this quadratic.

Looking for two numbers that multiply to -4 and add up to 3. Let me think... 4 and -1? Yes, 4 * (-1) = -4 and 4 + (-1) = 3. So, the quadratic factors as $ (y + 4)(y - 1) $. Therefore, substituting back $ y = x^4 $, we get:

$ (x^4 + 4)(x^4 - 1) $.

So, the original polynomial factors into $ (x^4 + 4)(x^4 - 1) $. Now, I need to check if these factors can be factored further over the integers.

Starting with $ x^4 - 1 $. This is a difference of squares: $ x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) $. Then, $ x^2 - 1 $ is also a difference of squares: $ (x - 1)(x + 1) $. So, putting it all together:

$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $.

Now, check if $ x^4 + 4 $ can be factored further. Hmm, I remember that $ x^4 + 4 $ is a special case. It's a biquadratic polynomial. Maybe it can be factored into quadratics? Let me try to factor it.

I recall that $ x^4 + 4 $ can be factored using the Sophie Germain identity. The identity is $ a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) $. Let me check if that applies here. If I set $ a = x $ and $ b = 1 $, then:

$ x^4 + 4(1)^4 = x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Yes! So, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Let me verify this multiplication:

Multiply $ (x^2 + 2x + 2)(x^2 - 2x + 2) $:

First, multiply $ x^2 $ by each term in the second polynomial: $ x^2 * x^2 = x^4 $, $ x^2 * (-2x) = -2x^3 $, $ x^2 * 2 = 2x^2 $.

Then, multiply 2x by each term: $ 2x * x^2 = 2x^3 $, $ 2x * (-2x) = -4x^2 $, $ 2x * 2 = 4x $.

Then, multiply 2 by each term: $ 2 * x^2 = 2x^2 $, $ 2 * (-2x) = -4x $, $ 2 * 2 = 4 $.

Now, add all these terms together:

- $ x^4 $
- $ -2x^3 + 2x^3 = 0 $
- $ 2x^2 - 4x^2 + 2x^2 = 0 $
- $ 4x - 4x = 0 $
- $ +4 $

So, the result is $ x^4 + 4 $. Perfect! So, that factorization is correct.

Therefore, the original polynomial factors as:

$ (x^2 + 2x + 2)(x^2 - 2x + 2)(x - 1)(x + 1)(x^2 + 1) $.

Now, I need to check if any of these factors can be factored further over the integers. Let's go through each factor:

1. $ x - 1 $: This is linear, so it can't be factored further.
2. $ x + 1 $: Similarly, linear, can't be factored further.
3. $ x^2 + 1 $: This is a quadratic. Over integers, it doesn't factor because the discriminant is $ 0^2 - 4*1*1 = -4 $, which is negative. So, it's irreducible over integers.
4. $ x^2 + 2x + 2 $: Let's check discriminant: $ (2)^2 - 4*1*2 = 4 - 8 = -4 $. Negative discriminant, so irreducible over integers.
5. $ x^2 - 2x + 2 $: Discriminant is $ (-2)^2 - 4*1*2 = 4 - 8 = -4 $. Also negative, so irreducible over integers.

Therefore, all factors are irreducible over integers. So, the complete factorization is into these five polynomials: $ (x - 1) $, $ (x + 1) $, $ (x^2 + 1) $, $ (x^2 + 2x + 2) $, and $ (x^2 - 2x + 2) $.

Wait, but the problem says "each non-constant polynomial $ p_i(x) $ is monic with integer coefficients, and cannot be factored further over the integers." So, all these factors are monic, have integer coefficients, and are irreducible. Therefore, the factorization is correct.

Now, the question asks for the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $. So, I need to evaluate each of these polynomials at x=1 and add them up.

Let me compute each one:

1. $ p_1(x) = x - 1 $: $ p_1(1) = 1 - 1 = 0 $
2. $ p_2(x) = x + 1 $: $ p_2(1) = 1 + 1 = 2 $
3. $ p_3(x) = x^2 + 1 $: $ p_3(1) = 1^2 + 1 = 1 + 1 = 2 $
4. $ p_4(x) = x^2 + 2x + 2 $: $ p_4(1) = 1^2 + 2*1 + 2 = 1 + 2 + 2 = 5 $
5. $ p_5(x) = x^2 - 2x + 2 $: $ p_5(1) = 1^2 - 2*1 + 2 = 1 - 2 + 2 = 1 $

Adding them up: 0 + 2 + 2 + 5 + 1 = 10.

Wait, let me check again:

First polynomial: 0

Second: 2

Third: 2

Fourth: 5

Fifth: 1

Total: 0 + 2 is 2, plus 2 is 4, plus 5 is 9, plus 1 is 10. So, 10?

But let me check if I have the correct factors. Let me verify the factorization again.

Original polynomial: $ x^8 + 3x^4 - 4 $. Let me compute the product of the factors:

First, multiply $ (x - 1)(x + 1) = x^2 - 1 $.

Then, multiply by $ x^2 + 1 $: $ (x^2 - 1)(x^2 + 1) = x^4 - 1 $.

Then, multiply by $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Earlier, we saw that this product is $ x^4 + 4 $.

Therefore, multiplying $ x^4 - 1 $ and $ x^4 + 4 $ gives $ (x^4 - 1)(x^4 + 4) = x^8 + 4x^4 - x^4 - 4 = x^8 + 3x^4 - 4 $, which matches the original polynomial. So, the factorization is correct.

Therefore, the factors are indeed those five polynomials. Then, evaluating each at 1 gives 0, 2, 2, 5, 1. Sum is 10. So, the answer is 10?

Wait, but let me check again the evaluation of each polynomial at 1:

For $ x - 1 $: 1 - 1 = 0. Correct.

For $ x + 1 $: 1 + 1 = 2. Correct.

For $ x^2 + 1 $: 1 + 1 = 2. Correct.

For $ x^2 + 2x + 2 $: 1 + 2 + 2 = 5. Correct.

For $ x^2 - 2x + 2 $: 1 - 2 + 2 = 1. Correct.

Adding them up: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Hmm. But let me think again. Wait, the problem says "each non-constant polynomial p_i(x) is monic with integer coefficients, and cannot be factored further over the integers." So, the factors are as above. But maybe there's another way to factor the original polynomial? Let me check if there are other possible factorizations?

Alternatively, maybe I can factor the original polynomial in another way? Let me think. For example, maybe there are more factors? Let me check.

But given that we have already factored it into irreducible polynomials, and the factorization is unique up to ordering, so the answer should be 10. However, let me check if maybe some of the factors can be combined or if I missed something.

Wait, but let me think again. Suppose I didn't use the Sophie Germain identity. Let me try to factor $ x^4 + 4 $ another way. Suppose I try to factor it into quadratics with integer coefficients. Let me assume that $ x^4 + 4 = (x^2 + ax + b)(x^2 + cx + d) $. Then expanding:

$ x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd $.

Set equal to $ x^4 + 0x^3 + 0x^2 + 0x + 4 $. Therefore, we have the system:

1. $ a + c = 0 $
2. $ ac + b + d = 0 $
3. $ ad + bc = 0 $
4. $ bd = 4 $

From equation 1: $ c = -a $.

Substitute into equation 3: $ ad + b(-a) = a(d - b) = 0 $. So, either a = 0 or d = b.

Case 1: a = 0. Then c = 0. Then equation 2 becomes 0 + b + d = 0 => b + d = 0. Equation 4: bd = 4. So, solving b + d = 0 and bd = 4. Then, d = -b, so b*(-b) = -b² = 4 => b² = -4. No real solutions, so no integer solutions here.

Case 2: d = b. Then equation 4: b*d = b² = 4. Therefore, b = ±2. Then, equation 2: ac + b + d = (-a²) + 2b = 0. Since c = -a, ac = -a². So:

- a² + 2b = 0 => a² = 2b.

Since b is ±2:

If b = 2: a² = 4 => a = ±2.

If b = -2: a² = -4, which is impossible.

So, for b = 2, a = ±2. Let's check:

If a = 2, then c = -2. Then, equation 3 is already satisfied because d = b = 2. So, the factors would be:

$ (x^2 + 2x + 2)(x^2 - 2x + 2) $, which is exactly the factorization we had earlier. Similarly, if a = -2, then c = 2, and the factors would be $ (x^2 - 2x + 2)(x^2 + 2x + 2) $, same thing. So, that's the only factorization. Therefore, our previous factorization is correct.

Therefore, the factorization into irreducible polynomials is correct, and the sum is 10. However, let me check once more the problem statement to make sure I didn't misread anything.

The problem says: "Compute $ p_1(1) + p_2(1) + \dots + p_k(1) $." So, if there are five factors, then sum is 10. But let me check if maybe the problem counts factors with multiplicity? But no, since each factor is distinct and irreducible. Wait, but maybe I made a mistake in the number of factors?

Original polynomial: degree 8. The factors are:

- (x - 1): degree 1

- (x + 1): degree 1

- (x^2 + 1): degree 2

- (x^2 + 2x + 2): degree 2

- (x^2 - 2x + 2): degree 2

Total degrees: 1 + 1 + 2 + 2 + 2 = 8. Correct. So, five factors. Therefore, the sum is 0 + 2 + 2 + 5 + 1 = 10. So, 10. But let me check again the evaluations:

For (x - 1): 1 - 1 = 0. Correct.

For (x + 1): 1 + 1 = 2. Correct.

For (x² + 1): 1 + 1 = 2. Correct.

For (x² + 2x + 2): 1 + 2 + 2 = 5. Correct.

For (x² - 2x + 2): 1 - 2 + 2 = 1. Correct.

So, adding them up: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Therefore, the answer is 10? Wait, but I have a doubt because sometimes problems like this might have a trick. Let me check if there's another way to factor the original polynomial?

Alternatively, maybe I can factor the original polynomial as a product of quadratics? Let me try. Suppose I try to factor $ x^8 + 3x^4 - 4 $ as a product of quadratics and other terms. But since we already have the factorization into linear and quadratics, and we checked that they are irreducible, I think that's the complete factorization.

Alternatively, maybe there's a different factorization? Let me think. Suppose I try to factor $ x^8 + 3x^4 - 4 $ as a product of two quartic polynomials? But since we already factored it into a product of a quartic and another quartic, which then further factors into lower degrees. But since the problem says "cannot be factored further over the integers," so the answer is as above.

Alternatively, maybe I can check the value of the original polynomial at x=1 and compare with the product of the p_i(1). Let me compute the original polynomial at x=1: $ 1^8 + 3*1^4 - 4 = 1 + 3 - 4 = 0 $. On the other hand, the product of the p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. Which matches. So, that's consistent. Therefore, the sum of p_i(1) is 10. Therefore, I think the answer is 10.

But let me check again if there's a possibility that some factors are repeated or something. For example, maybe the polynomial can be factored into more factors? For example, if there are more linear factors? But the original polynomial evaluated at x=1 is 0, so x=1 is a root. Similarly, x=-1 is a root? Let me check:

Original polynomial at x=-1: (-1)^8 + 3*(-1)^4 - 4 = 1 + 3 - 4 = 0. Yes, so x=-1 is also a root. Therefore, (x - 1) and (x + 1) are factors, which we have. So, the factorization is correct.

Alternatively, maybe there are other roots? Let me check. Let me try to find all roots of the original polynomial. Let me set $ x^8 + 3x^4 - 4 = 0 $. Let me substitute y = x^4 again: y² + 3y - 4 = 0. Which factors as (y + 4)(y - 1) = 0. Therefore, y = -4 or y = 1. Therefore, x^4 = -4 or x^4 = 1. The solutions for x^4 = 1 are the 4th roots of unity: 1, -1, i, -i. The solutions for x^4 = -4 are complex numbers. Therefore, over real numbers, the roots are 1, -1, i, -i, and four complex roots from x^4 = -4. But over integers, we can only factor the polynomial into polynomials with integer coefficients, which we have done. Therefore, the factorization is correct.

Therefore, I think my answer is correct. The sum is 10. Therefore, the answer is \boxed{10}.

**Final Answer**
\boxed{10}

To solve the problem, we are given the polynomial:

$$
x^8 + 3x^4 - 4
$$

and asked to factor it into monic polynomials with integer coefficients that cannot be factored further over the integers. Then, we are to compute the sum of the values of these polynomials evaluated at $ x = 1 $.

---

### Step 1: Factor the Polynomial

We begin by recognizing that the polynomial is a **biquadratic** in terms of $ x^4 $. Let’s make the substitution:

$$
y = x^4
$$

Then the polynomial becomes:

$$
y^2 + 3y - 4
$$

This is a quadratic in $ y $, which we can factor:

$$
y^2 + 3y - 4 = (y + 4)(y - 1)
$$

Substituting back $ y = x^4 $, we get:

$$
x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1)
$$

---

### Step 2: Factor Each Component

Now we factor each of the two components:

- $ x^4 - 1 $ is a **difference of squares**:

$$
x^4 - 1 = (x^2 - 1)(x^2 + 1)
$$

- $ x^2 - 1 $ is also a **difference of squares**:

$$
x^2 - 1 = (x - 1)(x + 1)
$$

So:

$$
x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)
$$

Now, we factor $ x^4 + 4 $. This is a special case that can be factored using the **Sophie Germain identity**:

$$
x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)
$$

---

### Step 3: Final Factorization

Putting it all together, the full factorization of the original polynomial is:

$$
x^8 + 3x^4 - 4 = (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2)
$$

Each of these factors is **monic**, has **integer coefficients**, and is **irreducible over the integers**. Therefore, this is the complete factorization.

---

### Step 4: Evaluate Each Factor at $ x = 1 $

We now compute the value of each factor at $ x = 1 $:

- $ (x - 1) \Rightarrow 1 - 1 = 0 $
- $ (x + 1) \Rightarrow 1 + 1 = 2 $
- $ (x^2 + 1) \Rightarrow 1^2 + 1 = 2 $
- $ (x^2 + 2x + 2) \Rightarrow 1^2 + 2(1) + 2 = 1 + 2 + 2 = 5 $
- $ (x^2 - 2x + 2) \Rightarrow 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1 $

---

### Step 5: Compute the Sum

Now, summing the values:

$$
0 + 2 + 2 + 5 + 1 = 10
$$

---

### Final Answer

$$
\boxed{10}
$$

Solution:
First, we can factor $x^8 + 3x^4 - 4$ as $(x^4 - 1)(x^4 + 4).$  Then
\[x^4 - 1 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x - 1)(x + 1),\]and by Sophie Germain,
\[x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2).\]Thus, the full factorization is
\[x^8 + 3x^4 - 4 = (x^2 + 1)(x - 1)(x + 1)(x^2 + 2x + 2)(x^2 - 2x + 2).\]Evaluating each factor at $x = 1,$ we get $2 + 0 + 2 + 5 + 1 = \boxed{10}.$