Question

Ifis skew symmetric, then${\mathbf{A}}^{\mathbf{2}}$must be negative semi-definite.

Short answer

The given statement is TRUE.

Step-by-step solution

Definition of a Skew matrix

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.

To Find TRUE or FALSE

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

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

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

Final Answer

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.