Question

Recall that a real square matrix A is called skew symmetric if${\mathbf{A}}^{\mathbf{T}\mathbf{=}\mathbf{-}\mathbf{A}}$.

a. If A is skew symmetric, is${\mathbf{A}}^{\mathbf{2}}$skew symmetric as well? Or is${\mathbf{A}}^{\mathbf{2}}$symmetric?

b. If is skew symmetric, what can you say about the definiteness of ${\mathbf{A}}^{\mathbf{2}}$? What about the eigenvalues of ${\mathbf{A}}^{\mathbf{2}}$?

c. What can you say about the complex eigenvalues of a skew-symmetric matrix? Which skew-symmetric matrices are diagonalizable over $\mathbf{ℝ}$?

Short answer

Therefore the solution is

a.${\mathrm{A}}^2$is symmetric.

b.${\mathrm{A}}^2$is a negative definite.

c. The zero matrix is the only skew-symmetric matrix that is diagonalizable over$\mathrm{ℝ}$

Step-by-step solution

 Step 1: Determine the transpose of a symmetric matrix is the same.

a.) A symmetric matrix is equal to its transpose

$\mathrm{A}={\mathrm{A}}^{\mathrm{T}}$

and a skew symmetric matrix is a matrix whose transpose is equal to its negative

${\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}$

Now, ifis a skew symmetric matrix, then as we just mentioned:${\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}$, now to check if${\mathrm{A}}^2$would be skew symmetric as well, we have:

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

therefore, ${\mathrm{A}}^2$is not skew symmetric, it is just symmetric.

Determine the certainty of A2

b.) We have

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

from part (a) we mentioned A that is said to be a skew matrix if ${\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}$, so

$\begin{array}{l}-{\overrightarrow{x}}^T\left(-A^T\right)A\overrightarrow{x}\\ =\left(A\overrightarrow{x}\right).\left(A\overrightarrow{x}\right)\\ ={||A{\overrightarrow{x}}^T||}^2\\ \le 0\end{array}$

therefore, for all$\mathrm{x}.\mathrm{q}\left(\overrightarrow{\mathrm{x}}\right)\le 0$ , which implies that ${\mathrm{A}}^2$is a negative definite (its eigenvalues are less than or equal to 0).

Determine the skew symmetric matric

c.) Let be an eigenvector corresponding to the eigenvalue,

$\mathrm{Av}=\mathrm{\lambda v}$

multiply both sides by${\mathrm{v}}^{-\mathrm{T}}$

${\mathrm{v}}^{-\mathrm{T}}\mathrm{Av}={\mathrm{\lambda v}}^{-\mathrm{T}}\mathrm{v}$

note: the bar" -" above v is for complex conjugation.

${\mathrm{v}}^{-\mathrm{T}}\mathrm{Av}=\mathrm{\lambda }{||\mathrm{v}||}^2$

for the left-hand side suppose we have column vectors of the same size, a and b, then${\mathrm{a}}^{\mathrm{T}}\mathrm{b}$is a$1\times 1$matrix which we can think of as a scalar. Now, taking transposes, since${\mathrm{a}}^{\mathrm{T}}\mathrm{b}$is $1\times 1$, it is its own transpose, so

${\mathrm{a}}^{\mathrm{T}}\mathrm{b}={\left({\mathrm{a}}^{\mathrm{T}}\mathrm{b}\right)}^{\mathrm{T}}={\mathrm{b}}^{\mathrm{T}}{\left({\mathrm{a}}^{\mathrm{T}}\right)}^{\mathrm{T}}={\mathrm{b}}^{\mathrm{T}}\mathrm{a}$

here let$\mathrm{a}=\overline{\mathrm{v}}$andlocalid="1659623457872" $\mathrm{b}=\mathrm{Av}$, then for our left-hand side

$\begin{array}{l}{\left(\mathrm{Av}\right)}^{\mathrm{T}}\overline{\mathrm{v}}=\mathrm{\lambda }{||\mathrm{v}||}^2\\ {\mathrm{v}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\overline{\mathrm{v}}=\mathrm{\lambda }{||\mathrm{v}||}^2\end{array}$

since A is skew-symmetric, we have${\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}$. Substituting,

$-{\mathrm{v}}^{\mathrm{T}}\mathrm{A}\overline{\mathrm{v}}=\mathrm{\lambda }{||\mathrm{v}||}^2$

taking the conjugate of $\mathrm{Av}=\mathrm{\lambda v}$yields $\mathrm{A}\overline{\mathrm{v}}=\mathrm{\lambda }\overline{\mathrm{v}}$, notice we can substitute this into the left-hand side above

$$\begin{gathered} -{\mathrm{v}}^{\mathrm{T}}\overline{\mathrm{\lambda }}\overline{\mathrm{v}}=\mathrm{\lambda }{||\mathrm{v}||}^2 \\ -\overline{\mathrm{\lambda }}{||\mathrm{v}||}^2=\mathrm{\lambda }{||\mathrm{v}||}^2 \\ -\mathrm{\lambda }=\mathrm{\lambda } \end{gathered}$$

Let $\mathrm{\lambda }=\mathrm{a}+\mathrm{ib}$,

$-\mathrm{a}+\mathrm{ib}=\mathrm{a}+\mathrm{ib}$which implies $\mathrm{a}=0$and $\mathrm{\lambda }=\mathrm{bi}$. We also conclude that the zero matrix is the only skew-symmetric matrix that is diagonalizable over $\mathrm{ℝ}$.