Subspaces of Symmetric, Skew-Symmetric Matrices

Linear algebra problems and solutions

Problem 143

Let $V$ be the vector space over $\R$ consisting of all $n\times n$ real matrices for some fixed integer $n$. Prove or disprove that the following subsets of $V$ are subspaces of $V$.

(a) The set $S$ consisting of all $n\times n$ symmetric matrices.

(b) The set $T$ consisting of all $n \times n$ skew-symmetric matrices.

(c) The set $U$ consisting of all $n\times n$ nonsingular matrices.

 
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Hint.

Recall that

  • a matrix $A$ is symmetric if $A^{\trans}=A$.
  • a matrix $A$ is skew-symmetric if $A^{\trans}=-A$.

Proof.

To show that a subset $W$ of a vector space $V$ is a subspace, we need to check that

  1. the zero vector in $V$ is in $W$
  2. for any two vectors $u,v \in W$, we have $u+v \in W$
  3. for any scalar $c$ and any vector $u \in W$, we have $cu \in W$.

(a) The set $S$ consisting of all $n\times n$ symmetric matrices.

We will prove that $S$ is a subspace of $V$. The zero vector $O$ in $V$ is the $n \times n$ zero matrix and it is symmetric. Thus the zero vector $O\in S$ and the condition 1 is met.


To check the second condition, take any $A, B \in S$, that is, $A, B$ are symmetric matrices.
To show that $A+B \in S$, we need to check that the matrix $A+B$ is symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}=A+B
\end{align*}
since $A, B$ are symmetric. Thus $A+B$ is also symmetric, and $A+B \in S$. Condition 2 is also satisfied.


Finally, to check condition 3, let $A \in S$ and let $r\in R$. We show that $rA \in S$, namely, we show that $rA$ is symmetric.
We have
\begin{align*}
(rA)^{\trans}=rA^{\trans}=rA
\end{align*}
since $A$ is symmetric. Thus $rA$ is symmetric and hence $rA \in S$.
Thus condition 3 is met.
By the subspace criteria, the subset $S$ is a subspace of the vector space $V$.

(b) The set $T$ consisting of all $n \times n$ skew-symmetric matrices.

We will prove that $T$ is a subspace of $V$.
The zero vector $O$ in $V$ is the $n \times n$ matrix, and it is skew-symmetric because
\[O^{\trans}=O=-O.\] Thus condition 1 is met.


For condition 2, take arbitrary elements $A, B \in T$. The matrices $A, B$ are skew-symmetric, namely, we have
\[A^{\trans}=-A \text{ and } B^{\trans}=-B \tag{*}.\] We show that $A+B \in T$, or equivalently we show that the matrix $A+B$ is skew-symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans} \stackrel{*}{=} -A+(-B)=-(A+B).
\end{align*}
Therefore the matrix $A+B$ is skew-symmetric and condition 2 is met.


To prove the last condition, consider any $A \in T$ and $r \in \R$.
We show that $rA$ is skew-symmetric, and hence $rA \in T$.
Using the fact that $A$ is skew-symmetric ($A^{\trans}=-A$), we have
\[(rA)^{\trans}=rA^{\trans}=r(-A)=-rA.\] Hence $rA$ is skew-symmetric and condition 3 is satisfied.

By the subspace criteria, the subset $T$ is a subspace of the vector space $V$.

(c) The set $U$ consisting of all $n\times n$ nonsingular matrices.

We claim that $U$ is not a subspace of $V$.
As the zero vector of $V$ is the $n \times n$ matrix and the zero matrix is singular, the zero vector is not in $U$. Hence condition 1 is not met, and thus $U$ is not a subspace.


Another reason that $U$ is not a subspace is that the addition is not closed. For example,
if $A$ is a nonsingular matrix (say, $A$ is $n\times n$ identity matrix), then $-A$ is also nonsingular matrix but their addition $A+(-A)=O$ is nonsingular, hence it is not in $U$.


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