Question

If Ais a symmetric nxnmatrix such that${\mathbf{A}}^{\mathbf{n}}\mathbf{=}\mathbf{0}$, then Amust be the zero matrix.

Short answer

The given statement is TRUE.

Step-by-step solution

Check whether the given statement is TRUE or FALSE

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,

$$\begin{gathered} \mathrm{A}=\sum {\mathrm{V}}^{\mathrm{T}} \\ =\mathrm{A}\left(0\right){\mathrm{V}}^{\mathrm{T}} \\ =0 \end{gathered}$$

Final Answer

The given statement is TRUE.