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 $W$.

 
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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:

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 $A^{\trans}=-A$.

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

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