A real matrix $A$ has the singular value decomposition $A = U \Sigma V^T$. If $A = \begin{bmatrix} 3 & 0 \\ 0 & -4 \end{bmatrix}$, the singular values $\sigma_1$ and $\sigma_2$ (ordered such that $\sigma_1 \ge \sigma_2$) of matrix $A$ are:

(A) $\sigma_1 = 4, \quad \sigma_2 = 3$

(B) $\sigma_1 = 3, \quad \sigma_2 = -4$

(C) $\sigma_1 = 16, \quad \sigma_2 = 9$

(D) $\sigma_1 = 4, \quad \sigma_2 = 0$

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### 1. Strategy & Theoretical Principles By definition, the singular values $\sigma_i$ of an $m \times n$ real matrix $A$ are the non-negative square roots of the eigenvalues of $A^T A$: $$\sigma_i = \sqrt{\lambda_i(A^T A)}$$ Singular values are strictly non-negative real numbers ($\sigma_i \ge 0$). ### 2. Analytical Formulation & Calculation Compute $A^T A$: $$A^T A = \begin{bmatrix} 3 & 0 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & -4 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 16 \end{bmatrix}$$ The eigenvalues of $A^T A$ are $\lambda_1 = 16$ and $\lambda_2 = 9$. Taking the positive square roots: $$\sigma_1 = \sqrt{16} = 4$$ $$\sigma_2 = \sqrt{9} = 3$$ ### 3. Distractor Forensics - **Option A:** Correct singular values $\{4, 3\}$. - **Option B:** Confuses eigenvalues $\{3, -4\}$ with singular values (which can never be negative). - **Option C:** Square of the singular values ($A^T A$ eigenvalues). ### 4. Exam Trap & Standard Reference - **Trap Alert:** Singular values are always non-negative: $\sigma_i = |\lambda_i|$ for diagonal matrices. - **Standard Reference:** Strang, G., *Linear Algebra and Its Applications*, Chapter 6.

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