Question

Ifis a symmetric matrix, what can you say about the definiteness of${\mathbf{A}}^{\mathbf{2}}$? When is${\mathbf{A}}^{\mathbf{2}}$ positive definite?

Short answer

$A^2$is always positive semidefinite$A^2$ is positive definite if and only if A is invertible

Step-by-step solution

Define symmetric matrix:

Equal matrices have the same dimension, only square matrices are symmetric. If is a matrix, then it is symmetric if the transpose of matrix $\mathrm{A}={\mathrm{A}}^{\mathrm{T}}$ is same. If "positive" is replaced by "nonnegative" and "invertible matrix" is replaced by "matrix," a matrix is positive semi-definite.

Symmetricn×n Matrices:

B is a symmetric $n\times n$matrix. B is said to be positive semidefinite. If ${\overrightarrow{\mathrm{x}}}^{\mathrm{T}}\mathrm{B}\overrightarrow{\mathrm{x}}\ge 0$ for the values of . Let A be symmetric. Then $A^2$is symmetric

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

If A is symmetric A = ${\mathrm{A}}^{\mathrm{T}}$

role="math" localid="1659620083828" $$\begin{gathered} ={\overrightarrow{\mathrm{x}}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\mathrm{A}\overrightarrow{\mathrm{x}} \\ ={\left(\mathrm{A}\overrightarrow{\mathrm{x}}\right)}^{\mathrm{T}}\left(\mathrm{A}\overrightarrow{\mathrm{x}}\right) \\ ={||\mathrm{Ax}||}^2 \end{gathered}$$

Since $={||\mathrm{Ax}||}^2\ge 0,{\mathrm{A}}^2$is always positive semidefinite For ${\overrightarrow{\mathrm{x}}}^{\mathrm{T}}{\mathrm{A}}^2\overrightarrow{\mathrm{x}}={||\mathrm{A}\overrightarrow{\mathrm{x}}||}^2{\mathrm{A}}^2$is positive semidefinite if ${||\mathrm{A}\overrightarrow{\mathrm{x}}||}^2\ge 0$for the values of $\overrightarrow{x}\ne \overrightarrow{0}$it and if ${||\mathrm{A}\overrightarrow{\mathrm{x}}||}^2\ne 0$for all $\overrightarrow{\mathrm{x}}\ne \overrightarrow{0}$

Then the value for $A^2$is positive definite if and only if A is invertible.