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

## Problem 586

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 $W$.

Contents

## Proof.

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

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

We verify the following subspace criteria:

**Subspace Criteria**.

- The zero vector of $V$ is in $W$.
- For any $A, B\in W$, the sum $A+B\in W$.
- For any $A\in W$ and $r\in \R$, the scalar product $rA\in W$.

The zero vector in $V$ is the $2\times 2$ zero matrix $O$.

It is clear that $O^{\trans}=O$, and hence $O$ is symmetric.

Thus $O\in W$ and condition 1 is met.

Let $A, B$ be arbitrary elements in $W$.

That is, $A$ and $B$ are symmetric matrices.

We show that the sum $A+B$ is also symmetric.

We have

\begin{align*}

(A+B)^{\trans}=A^{\trans}+B^{\trans}=A+B.

\end{align*}

The second equality follows as $A, B$ are symmetric.

Hence $A+B$ is symmetric and $A+B\in W$.

Condition 2 is met.

To check condition 3, let $A\in W$ and $r\in \R$.

We have

\begin{align*}

(rA)^{\trans}=rA^{\trans}=rA,

\end{align*}

where the second equality follows since $A$ is symmetric.

This implies that $rA$ is symmetric, and hence $rA\in W$.

So condition 3 is met, and we conclude that $W$ is a subspace of $V$ by subspace criteria.

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

Let

\[A=\begin{bmatrix}

a_{11} & a_{12}\\

a_{21}& a_{22}

\end{bmatrix}\]
be an arbitrary element in the subspace $W$.

Then since $A^{\trans}=A$, we have

\[\begin{bmatrix}

a_{11} & a_{21}\\

a_{12}& a_{22}

\end{bmatrix}=\begin{bmatrix}

a_{11} & a_{12}\\

a_{21}& a_{22}

\end{bmatrix}.\]
This implies that $a_{12}=a_{21}$, and hence

\begin{align*}

A&=\begin{bmatrix}

a_{11} & a_{12}\\

a_{12}& a_{22}

\end{bmatrix}\\[6pt]
&=a_{11}\begin{bmatrix}

1 & 0\\

0& 0

\end{bmatrix}+a_{12}\begin{bmatrix}

0 & 1\\

1& 0

\end{bmatrix}+a_{22}\begin{bmatrix}

0 & 0\\

0& 1

\end{bmatrix}.

\end{align*}

Let $B=\{v_1, v_2, v_3\}$, where $v_1, v_2, v_3$ are $2\times 2$ matrices appearing in the above linear combination of $A$.

Note that these matrices are symmetric.

Hence we showed that any element in $W$ is a linear combination of matrices in $B$.

Thus $B$ is a spanning set for the subspace $W$.

We show that $B$ is linearly independent.

Suppose that we have

\[c_1v_1+c_2v_2+c_3v_3=\begin{bmatrix}

0 & 0\\

0& 0

\end{bmatrix}.\]
Then it follows that

\[\begin{bmatrix}

c_1 & c_2\\

c_2& c_3

\end{bmatrix}=\begin{bmatrix}

0 & 0\\

0& 0

\end{bmatrix}.\]
Thus $c_1=c_2=c_3=$ and the set $B$ is linearly independent.

As $B$ is a linearly independent spanning set, we conclude that $B$ is a basis for the subspace $W$.

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

Recall that the dimension of a subspace is the number of vectors in a basis of the subspace.

In part (b), we found that $B=\{v_1, v_2, v_3\}$ is a basis for the subspace $W$.

As $B$ consists of three vectors, the dimension of $W$ is $3$.

## Related Question (Skew-Symmetric Matrices)

A matrix $A$ is called **skew-symmetric** if $A^{\trans}=-A$.

**Problem**.

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.

Prove that $W$ is a subspace of $V$ and also find a basis and dimension of $W$.

The solution is given in the post ↴

Subspace of Skew-Symmetric Matrices and Its Dimension

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