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Q29E

Expert-verifiedFound in: Page 1

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**If A**** is a symmetric nxn**** matrix such that ${{\mathbf{A}}}^{{\mathbf{n}}}{\mathbf{=}}{\mathbf{0}}$****, then A**** must be the zero matrix.**

The given statement is TRUE.

If $\lambda $ is an eigenvalue of A, then ${\mathrm{\lambda}}^{\mathrm{n}}$ is an eigenvalue of ${\mathrm{A}}^{\mathrm{n}}$.

The only eigenvalue of ${\mathrm{A}}^{\mathrm{n}}$ is 0. Hence,${\lambda}^{n}=0$ which implies that $\lambda =0$.

This shows that the only eigenvalue of A is 0.

Since A is symmetric, every singular value $\sigma $ of A satisfies $\mathrm{\sigma}=\left|\mathrm{\lambda}\right|$ for some eigenvalue $\lambda $ of A. This implies that every singular value of A is 0.

If $\mathrm{A}=\mathrm{U}\sum _{}{\mathrm{V}}^{\mathrm{T}}$ is the singular value decomposition of A, then the statement implies that $\sum _{}$ is the zero matrix. Hence,

$\mathrm{A}=\sum {\mathrm{V}}^{\mathrm{T}}\phantom{\rule{0ex}{0ex}}=\mathrm{A}\left(0\right){\mathrm{V}}^{\mathrm{T}}\phantom{\rule{0ex}{0ex}}=0$

The given statement is TRUE.

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