91Ó°ÊÓ

A matrix is said to be a skew matrix, when it is equal to the negative of its transpose of the given matrix.

Simply, A= -A^T(or) A^T= -A

If the eigenvalues are non-positive, then the matrix is said to be a negative semidefinite matrix.

Consider ${\mathrm{A}}^{\mathrm{T}}\mathrm{A}$has all non-negative eigenvalues. If $\mathrm{A}$is skew-symmetric, then

$$\begin{gathered} {\mathrm{A}}^{\mathrm{T}}\mathrm{A}=\left(-\mathrm{A}\right)\mathrm{A} \\ =-{\mathrm{A}}^2 \end{gathered}$$

This means that $-{\mathrm{A}}^2$is symmetric and has all non-negative eigenvalues.

Also, ${\mathrm{A}}^2$is symmetric as $-{\mathrm{A}}^2$is symmetric.

Let $\mathrm{λ}$be an eigenvalue of ${\mathrm{A}}^2$. Then, for some vector $\stackrel{˙}{\mathrm{v}}$, we have the following:

role="math" localid="1664188506373" $$\begin{gathered} {\mathrm{A}}^2\stackrel{˙}{\mathrm{v}}=\mathrm{λ}\stackrel{˙}{v} \\ \left(-{\mathrm{A}}^2\right)\stackrel{˙}{v}=\left(-\mathrm{λ}\right)\stackrel{˙}{v} \end{gathered}$$

So, $\mathrm{λ}$must be non-negative since it is an eigenvalue of $-{\mathrm{A}}^2$. Thus,$-\mathrm{λ}$ is non-positive.

This implies that ${\mathrm{A}}^2$is symmetric and has all non-positive eigenvalues. Hence, ${\mathrm{A}}^2$is negative semi-definite.

The given statement is TRUE.