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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# 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 ${\mathbf{ℝ}}$?

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 $\mathrm{ℝ}$

See the 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, if is 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:

$\left({\mathrm{A}}^{2}\right)={\left(\mathrm{AA}\right)}^{\mathrm{T}}\phantom{\rule{0ex}{0ex}}={\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\phantom{\rule{0ex}{0ex}}=\left(-\mathrm{A}\right)\left(-\mathrm{A}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{A}}^{2}$

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

## Step 2: Determine the certainty of A2

b.) We have

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

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

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

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

## Step 3: 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 $1×1$ matrix which we can think of as a scalar. Now, taking transposes, since ${\mathrm{a}}^{\mathrm{T}}\mathrm{b}$ is $1×1$, 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 $\mathrm{a}=\overline{)\mathrm{v}}$ and localid="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

$-{\mathrm{v}}^{\mathrm{T}}\overline{)\mathrm{\lambda }}\overline{\mathrm{v}}=\mathrm{\lambda }{||\mathrm{v}||}^{2}\phantom{\rule{0ex}{0ex}}-\overline{)\mathrm{\lambda }}{||\mathrm{v}||}^{2}=\mathrm{\lambda }{||\mathrm{v}||}^{2}\phantom{\rule{0ex}{0ex}}-\mathrm{\lambda }=\mathrm{\lambda }$

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 $\mathrm{ℝ}$.