SOLUTION: Find the singular values σ1 ≥ σ2 of A = [-3 -1; 3 -1]. If σ1 = sqrt(18), and σ2 = sqrt(2). Find unit vectors v⃗ 1 and v⃗ 2 such that ∥Av⃗ 1∥=σ1 and ∥Av⃗ 2∥=

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Question 1161297: Find the singular values σ1 ≥ σ2 of A = [-3 -1; 3 -1].
If σ1 = sqrt(18), and σ2 = sqrt(2). Find unit vectors v⃗ 1 and v⃗ 2 such that ∥Av⃗ 1∥=σ1 and ∥Av⃗ 2∥=σ2.
Answer by CPhill(2285) (Show Source):
You can put this solution on YOUR website!
To find the singular values and the corresponding right singular unit vectors $\vec{v}_1$ and $\vec{v}_2$, we analyze the symmetric matrix $A^T A$.
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### Step 1: Calculate $A^T A$
Given:

$$A = \begin{bmatrix} -3 & -1 \\ 3 & -1 \end{bmatrix}$$
Compute $A^T$:

$$A^T = \begin{bmatrix} -3 & 3 \\ -1 & -1 \end{bmatrix}$$
Now, calculate $A^T A$:

$$A^T A = \begin{bmatrix} -3 & 3 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -3 & -1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} (-3)(-3) + (3)(3) & (-3)(-1) + (3)(-1) \\ (-1)(-3) + (-1)(3) & (-1)(-1) + (-1)(-1) \end{bmatrix} = \begin{bmatrix} 18 & 0 \\ 0 & 2 \end{bmatrix}$$
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### Step 2: Find the Singular Values $\sigma_1 \ge \sigma_2$
The singular values of $A$ are the square roots of the eigenvalues of $A^T A$.
Since $A^T A = \begin{bmatrix} 18 & 0 \\ 0 & 2 \end{bmatrix}$ is already diagonal:
* $\lambda_1 = 18 \implies \sigma_1 = \sqrt{18} = 3\sqrt{2}$
* $\lambda_2 = 2 \implies \sigma_2 = \sqrt{2}$
This verifies the given values: **$\sigma_1 = \sqrt{18}$** and **$\sigma_2 = \sqrt{2}$**.
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### Step 3: Find the Unit Vectors $\vec{v}_1$ and $\vec{v}_2$
The vectors $\vec{v}_1$ and $\vec{v}_2$ are the normalized eigenvectors of $A^T A$ corresponding to the eigenvalues $\lambda_1 = 18$ and $\lambda_2 = 2$, respectively.
1. **For $\lambda_1 = 18$ ($\sigma_1 = \sqrt{18}$):**
$$(A^T A - 18I)\vec{v}_1 = \mathbf{0} \implies \begin{bmatrix} 0 & 0 \\ 0 & -16 \end{bmatrix} \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

This yields $-16y_1 = 0 \implies y_1 = 0$, while $x_1$ can be any non-zero real number. Choosing $x_1 = 1$ and normalizing:
$$\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad (\text{or } \begin{bmatrix} -1 \\ 0 \end{bmatrix})$$

2. **For $\lambda_2 = 2$ ($\sigma_2 = \sqrt{2}$):**
$$(A^T A - 2I)\vec{v}_2 = \mathbf{0} \implies \begin{bmatrix} 16 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

This yields $16x_2 = 0 \implies x_2 = 0$, while $y_2$ can be any non-zero real number. Choosing $y_2 = 1$ and normalizing:
$$\vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad (\text{or } \begin{bmatrix} 0 \\ -1 \end{bmatrix})$$

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### Verification:
* **For $\vec{v}_1$:**
$$A\vec{v}_1 = \begin{bmatrix} -3 & -1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 3 \end{bmatrix}$$

$$\Vert{}A\vec{v}_1\Vert{} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \mathbf{\sqrt{18}} = \sigma_1$$

* **For $\vec{v}_2$:**
$$A\vec{v}_2 = \begin{bmatrix} -3 & -1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$

$$\Vert{}A\vec{v}_2\Vert{} = \sqrt{(-1)^2 + (-1)^2} = \mathbf{\sqrt{2}} = \sigma_2$$

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### Final Answer:
* **Singular Values:** $\sigma_1 = \sqrt{18}$ and $\sigma_2 = \sqrt{2}$
* **Unit Vectors:** $\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ *(Note: negative counterparts $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ are also valid answers)*