# 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 $W$.

## 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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