Linear Algebra

7 Problems on Skew-Symmetric Matrices

Problem 564

Let $A$ and $B$ be $n\times n$ skew-symmetric matrices. Namely $A^{\trans}=-A$ and $B^{\trans}=-B$.

(a) Prove that $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

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Proof.

(a) Prove that $A+B$ is skew-symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}=(-A)+(-B)=-(A+B).
\end{align*}
Hence $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

We compute
\begin{align*}
(cA)^{\trans}=cA^{\trans}=c(-A)=-cA.
\end{align*}
Thus, $cA$ is skew-symmetric.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

Using the properties of transpose, we have
\begin{align*}
(P^{\trans}AP)^{\trans}&=P^{\trans}A^{\trans}(P^{\trans})^{\trans}=P^{\trans}A^{\trans}P\\
&=P^{\trans}(-A)P=-(P^{\trans}AP).
\end{align*}
This implies that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

Note that since $A$ is real, we have $\bar{A}=A$.
Then we have
\begin{align*}
(\overline{iA})^{\trans}=(\bar{i}\bar{A})^{\trans}=(-iA)^{\trans}=(-i)A^{\trans}=(-i)(-A)=iA.
\end{align*}
It follows that $iA$ is Hermitian.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

We calculate
\begin{align*}
(AB)^{\trans}&=B^{\trans}A^{\trans}=(-B)(-A)\\
&=BA=-AB,
\end{align*}
where the last step follows from the assumption $AB=-BA$.
This proves that $AB$ is skew-symmetric.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

Observe that $\mathbf{v}^{\trans}A\mathbf{v}$ is a $1\times 1$ matrix, or just a number.
So we have
\begin{align*}
\mathbf{v}^{\trans}A\mathbf{v}&=(\mathbf{v}^{\trans}A\mathbf{v})^{\trans}=\mathbf{v}^{\trans}A^{\trans}(\mathbf{v}^{\trans})^{\trans}\\
&=\mathbf{v}^{\trans}A^{\trans}\mathbf{v}=\mathbf{v}^{\trans}(-A)\mathbf{v}=-(\mathbf{v}^{\trans}A\mathbf{v}).
\end{align*}
This yields that $2\mathbf{v}^{\trans}A\mathbf{v}=0$, and hence $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

Let us compute the length of the vector $A\mathbf{v}$.
We have
\begin{align*}
\|A\mathbf{v}\|&=(A\mathbf{v})^{\trans}(A\mathbf{v})=\mathbf{v}^{\trans}A^{\trans}A\mathbf{v}\\
&=\mathbf{v}^{\trans}(-A)A\mathbf{v}=-\mathbf{v}^{\trans}A^2\mathbf{v}\\
&=-\mathbf{v}\mathbf{0} &&\text{by assumption}\\
&=0.
\end{align*}
Since the length $\|A\mathbf{v}\|=0$, we conclude that $A\mathbf{v}=\mathbf{0}$.

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