# 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.

(ii) For any vectors $\mathbf{u}, \mathbf{v}\in W$, the sum $\mathbf{u}+\mathbf{v}$ is in $W$.

(iii) For any vector $\mathbf{u}\in W$ and any scalar $c\in \R$, the scalar product $c\mathbf{u}\in W$.

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 $A^{\trans}=A$.

**Problem**.

Let $V$ be the vector space over $\R$ of all real $2\times 2$ matrices.

Let $W$ be the subset of $V$ consisting of all symmetric matrices.

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

**(b)**Find a basis of $W$.

**(c)**Determine the dimension of $W$.

The solution is given in the post ↴

The set of $2\times 2$ Symmetric Matrices is a Subspace

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