# 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)

## 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.

(i) The zero vector $\mathbf{0}\in V$ is in $W$.
(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

### 1 Response

1. 10/17/2017

[…] The solution is given in the post ↴ Subspace of Skew-Symmetric Matrices and Its Dimension […]

##### Vector Space of Polynomials and a Basis of Its Subspace

Let $P_2$ be the vector space of all polynomials of degree two or less. Consider the subset in $P_2$ \[Q=\{...

Close