Okay, so I need to find the smallest positive real number C such that for any two-dimensional vector v, the norm of the matrix multiplied by v is less than or equal to C times the norm of v. Hmm, that sounds like finding the operator norm of the matrix. The operator norm is the smallest C that satisfies this inequality for all vectors v. 

First, let me recall that the operator norm of a matrix A is the maximum value of ||A v|| / ||v|| over all non-zero vectors v. So, to find the smallest C, I need to compute the maximum of ||A v|| / ||v||, which is the operator norm of A. Then, C will be that maximum value. 

Given the matrix A:
\[ A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \]

I need to compute the operator norm of A. For this, I remember that the operator norm when using the standard Euclidean norm (which is the case here since the problem mentions the magnitude of a vector as ||a||, which is the Euclidean norm) is equal to the largest singular value of the matrix. So, perhaps I can compute the singular values of A and take the largest one as C.

Alternatively, another approach is to consider the maximum of ||A v|| / ||v|| for all unit vectors v. Since the problem is asking for the smallest C such that this inequality holds for all v, C would be the maximum of ||A v|| when ||v|| = 1. Therefore, finding the maximum of ||A v|| over all unit vectors v will give me the operator norm C.

Let me try both approaches. Maybe starting with the singular value decomposition (SVD) since that gives the singular values directly. The SVD of A will be A = U Σ V^T, where Σ is a diagonal matrix with singular values σ1 ≥ σ2 ≥ ... ≥ 0. The largest singular value σ1 is the operator norm.

Alternatively, if I don't remember SVD, maybe I can compute the eigenvalues of A^T A, since the singular values are the square roots of the eigenvalues of A^T A. The largest singular value would then be the square root of the largest eigenvalue of A^T A.

Let me try that. First, compute A^T:

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

Then, compute A^T A:

A^T A = \[ \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 2*2 + 0*0 & 2*3 + 0*(-2) \\ 3*2 + (-2)*0 & 3*3 + (-2)*(-2) \end{pmatrix} \]

Calculating each element:

First row, first column: (2)(2) + (0)(0) = 4 + 0 = 4

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

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

Second row, second column: (3)(3) + (-2)(-2) = 9 + 4 = 13

So, A^T A is:

\[ \begin{pmatrix} 4 & 6 \\ 6 & 13 \end{pmatrix} \]

Now, I need to find the eigenvalues of this matrix. The eigenvalues λ satisfy the characteristic equation det(A^T A - λ I) = 0.

So, compute the determinant:

|4 - λ   6       |

|6       13 - λ  |

= (4 - λ)(13 - λ) - (6)(6) = 0

Compute (4 - λ)(13 - λ) = 4*13 - 4λ -13λ + λ^2 = 52 -17λ + λ^2

Subtract 36 (since 6*6=36):

λ^2 -17λ +52 -36 = λ^2 -17λ +16 = 0

So, the quadratic equation is λ^2 -17λ +16 = 0.

Solving for λ:

Using quadratic formula: λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2

Therefore, λ1 = (17 +15)/2 = 32/2 = 16

λ2 = (17 -15)/2 = 2/2 = 1

So, the eigenvalues of A^T A are 16 and 1. Therefore, the singular values of A are sqrt(16) = 4 and sqrt(1) = 1. Hence, the largest singular value is 4. Therefore, the operator norm of A is 4, which means the smallest C is 4.

Wait, but let me verify this because sometimes there might be a mistake in calculations. Let me check the determinant again.

First, A^T A:

First element: 2*2 + 0*0 = 4

Second element: 2*3 + 0*(-2) = 6

Third element: 3*2 + (-2)*0 = 6

Fourth element: 3*3 + (-2)*(-2) = 9 +4=13

Yes, that's correct. Then the determinant calculation:

(4 - λ)(13 - λ) - 36 = (4*13 -4λ -13λ +λ²) -36 = (52 -17λ + λ²) -36 = λ² -17λ +16. Correct.

Then solving λ² -17λ +16=0, discriminant 289 -64=225. Square root is 15. So, (17 ±15)/2. 17+15=32/2=16, 17-15=2/2=1. Correct. So eigenvalues 16 and 1. Therefore, singular values 4 and 1. So operator norm is 4. So C=4.

Alternatively, maybe I can approach this problem using the definition. Let me take a general vector v = [x; y], and compute ||A v||, then divide by ||v||, and find the maximum value over all non-zero v. Then take the square of the norm to make it easier.

Compute A v:

A v = [2x + 3y; 0x -2y] = [2x +3y; -2y]

Then, ||A v||² = (2x +3y)^2 + (-2y)^2 = (4x² + 12xy +9y²) +4y² =4x² +12xy +13y²

||v||² = x² + y²

Therefore, the ratio squared is (4x² +12xy +13y²)/(x² + y²). We need to find the maximum of this ratio over all x, y not both zero. Then the square root of that maximum will be the operator norm.

Let me denote this ratio as R = (4x² +12xy +13y²)/(x² + y²). We need to maximize R.

To maximize R, we can set up the problem as maximizing 4x² +12xy +13y² subject to x² + y² =1. Then, using Lagrange multipliers, or by diagonalizing the quadratic form.

Alternatively, we can write R as a quadratic form:

R = [x y] [4 6; 6 13] [x; y] / (x² + y²)

Which is exactly the Rayleigh quotient. The maximum value of the Rayleigh quotient is the largest eigenvalue of the matrix [4 6; 6 13]. Wait, but earlier we had A^T A = [4 6;6 13], so the eigenvalues of A^T A are 16 and 1, so the maximum Rayleigh quotient is 16, hence the square root is 4. Therefore, the maximum ratio R is 16, so sqrt(16)=4. Therefore, C=4.

Therefore, both approaches confirm that the operator norm is 4, so the smallest C is 4.

Alternatively, let's verify by taking specific vectors. For example, take a vector v where the ratio ||A v|| / ||v|| is maximized.

Suppose we take v = [1; 0]. Then A v = [2; 0], so ||A v|| = 2, ||v|| =1, so ratio is 2.

If we take v = [0;1], then A v = [3; -2], so ||A v|| = sqrt(9 +4) = sqrt(13) ≈3.6055, which is less than 4.

If we take v = [1;1], then A v = [2*1 +3*1; -2*1] = [5; -2], ||A v|| = sqrt(25 +4)=sqrt(29)≈5.385, ||v||=sqrt(2). So the ratio is sqrt(29)/sqrt(2)=sqrt(14.5)≈3.807, still less than 4.

Wait, but maybe there's a vector where the ratio is 4. Let's see. Suppose we take the singular vector corresponding to the singular value 4. Since the singular values are 4 and 1, the singular vector for 4 will be the one where ||A v|| =4 ||v||, so the ratio is 4.

Alternatively, let's find such a vector. Since the singular vectors are the eigenvectors of A^T A corresponding to the singular values squared. So, the eigenvectors of A^T A for eigenvalue 16 will correspond to the vector v such that A^T A v =16 v.

So, solving (A^T A -16 I) v =0.

A^T A -16 I = [4-16, 6; 6, 13-16] = [-12, 6; 6, -3]

So, the equations are:

-12 x +6 y=0

6 x -3 y=0

Both equations simplify to -12x +6y=0 => 2x - y=0 => y=2x

Therefore, the eigenvectors are scalar multiples of [1; 2]. Let's check:

Take v = [1;2]. Then ||v|| = sqrt(1 +4)=sqrt(5). Compute A v = [2*1 +3*2; 0*1 + (-2)*2] = [2+6; -4] = [8; -4]. Then ||A v|| = sqrt(64 +16)=sqrt(80)=4*sqrt(5). Therefore, ||A v|| / ||v|| =4*sqrt(5)/sqrt(5)=4. So indeed, the ratio is 4. Hence, this vector v is the one that achieves the maximum ratio. Therefore, the operator norm is indeed 4, so C=4.

Therefore, the answer is 4.

**Final Answer**
The smallest positive real number \( C \) is \(\boxed{4}\).
Therefore, the smallest positive real number $C$ for which the inequality holds is $\boxed{4}$.
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