Subspace of Skew-Symmetric Matrices and Its Dimension

Problem 166

Let $V$ be the vector space of all $2\times 2$ matrices. Let $W$ be a subset of $V$ consisting of all $2\times 2$ skew-symmetric matrices. (Recall that a matrix $A$ is skew-symmetric if $A^{\trans}=-A$.)

(a) Prove that the subset $W$ is a subspace of $V$.

(b) Find the dimension of $W$.

(The Ohio State University Linear Algebra Exam Problem)
 
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Proof.

(a) Prove that the subset $W$ is a subspace of $V$

To prove that $W$ is a subspace of $V$, we check the following subspace criteria.

The zero vector in $V$ is the $2\times 2$ zero vector
\[O=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\] Since have
\[O^{\trans}=O=-O,\] the zero vector $O$ is a skew-symmetric matrix. Thus $O$ is in $W$, and hence condition (i) is met.

For condition (ii), consider $A, B\in W$. This means that $A, B$ are skew symmetric matrices and thus we have
\[A^{\trans}=-A, B^{\trans}=-B \tag{*}.\]

To show that $A+B \in W$, we compute as follows.
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}\stackrel{(*)}{=}-A+(-B)=-(A+B).
\end{align*}
Thus we have $(A+B)^{\trans}=-(A+B)$, and it follows that the matrix $A+B$ is skew-symmetric.
Hence $A+B \in W$ and condition (ii) is also met.

To check condition (iii), consider $A\in W$ and $c\in \R$.
We want to show that $cA\in W$, that is, we want to show that $cA$ is a skew-symmetric matrix.

We see this by computing as follows.
\begin{align*}
(cA)^{\trans}&=cA^{\trans}\\
&=c(-A) \text{ since } A \text{ is skew-symmetric}\\
&=-(cA).
\end{align*}

Thus we obtained $(cA)^{\trans}=-(cA)$, and thus $cA$ is a skew-symmetric matrix as required.
Hence $cA\in W$ and condition (iii) is satisfied.

We confirmed all three subspace criteria (i)-(iii), and thus conclude that $W$ is a subspace of $V$.

(b) Find the dimension of $W$

Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix.

If $A$ is a skew-symmetric matrix, namely we have $A^{\trans}=-A$, we have
\[\begin{bmatrix}
a & c\\
b& d
\end{bmatrix}=-\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}.\] Comparing entries of the matrices, we obtain
\begin{align*}
a&=-a\\
b&=-c\\
d&=-d.
\end{align*}
It follows that $a=0, d=0$, and $c=-b$.

Thus any skew-symmetric matrix is of the form
\[A=\begin{bmatrix}
0 & b\\
-b& 0
\end{bmatrix}=b\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}.\]

Therefore, the subspace $W$ is spanned by
\[B=\left\{ \begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}\right\}\] and since the set $B$ consists of only one vector, it is linearly independent, and thus $B$ is a basis.
From this, we conclude that the dimension of $W$ is $1$.

Related Question (Symmetric Matrices)

Recall that a matrix $A$ is symmetric if $A^{\trans}=A$.

The solution is given in the post ↴
The set of $2\times 2$ Symmetric Matrices is a Subspace

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