Linear Algebra

Find a Spanning Set for the Vector Space of Skew-Symmetric Matrices

Problem 714

Let $W$ be the set of $3\times 3$ skew-symmetric matrices. Show that $W$ is a subspace of the vector space $V$ of all $3\times 3$ matrices. Then, exhibit a spanning set for $W$.

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Proof.

To prove that $W$ is a subspace of $V$, the $3\times 3$ zero matrix $\mathbf{0}$ is the zero vector of $V$, and since $\mathbf{0}$ is clearly skew-symmetric, $\mathbf{0}\in W$. Next, let $A,B\in W$. Then $A^{T}=-A$ and $B^{T}=-B$. Therefore,
\[
(A+B)^{T}
=A^{T}+B^{T}
=-A+(-B)
=-(A+B).
\] Thus $A+B$ is skew-symmetric, and therefore $A+B\in W$. Next, take any $A\in W$ and $r$ in the scalar field. Then
\[
(rA)^{T}
=rA^{T}
=r(-A)
=-rA.
\] Thus $rA$ is skew symmetric, which implies $rA\in W$. Therefore, $W$ is a subspace of $V$.

To exhibit a spanning set for $W$, first note that
\[
W=
\left\{
A\left|
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
,\;a,b,c\in\R
\right.\right\}.
\] Let
\[
A_{1}=
\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
,\;
A_{2}=
\begin{bmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{bmatrix}
,\;
A_{3}=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{bmatrix}
.
\] Then any $A\in W$ can be written as
\[
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
=aA_{1}+bA_{2}+cA_{3},
\] which is a linear combination of $A_{1},A_{2},A_{3}$. Thus $\{A_{1},A_{2},A_{3}\}$ is a spanning set for $W$.

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