Is the dimension of this subspace 1?

Question

"Let $V=M_{2\times2}(\mathbb{R})$ denote the vectors space of all $2\times2$ matrices with real number entries. Determine which of the following subsets are subspaces of $V$. If it is a subspace, find its dimension

F) All $2\times2$ matrices $A$ such that $A^T=−A$, where $A^T$ is the transpose of $A$.

I found that this was indeed a subspace, however, I'm having problem determining its dimension. I found the number of free variables to be one, so, the dimension should be one right? Also, what is the basis of this subspace so I can make sure that I'm on the right track .

Answer 1

A matrix $A$ such that $A^T = -A$ is called skew-symmetric or antisymmetric. Such a matrix must have $0$ on its diagonal elements, and all the elements above the diagonal can be chosen freely as these determine the elements below (or vice-versa).

So indeed, for a $2 \times 2$ real skew-symmetric matrix, there is only one free choice and all such matrices are of the form $$\begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$$ for $a \in \mathbb R$. Thus the space does in fact have dimension $1$.

In general, the space of $n \times n$ skew-symmetric matrices over $\mathbb R$ has dimension $\frac{n(n-1)}{2}$ (try to convince yourself of this, if you'd like).

Answer 2

Consider the following subset of $V=M_{2\times2}(\mathbb{R})$: $$\{A\in M_{2\times2}(\mathbb{R})|A^T=-A\}.$$ Suppose that $$A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$$, where $a,b,c,d\in\mathbb{R}$. Then $$A^T=-A\Longleftrightarrow\begin{pmatrix} a & c\\ b & d \end{pmatrix}=-\begin{pmatrix} a & b\\ c & d \end{pmatrix}\Longleftrightarrow a=0,d=0,b=-c.$$ Therefore, $$\{A\in M_{2\times2}(\mathbb{R})|A^T=-A\}\Longleftrightarrow\left\{\left.\begin{pmatrix} 0 & -c\\ c & 0 \end{pmatrix}\right|c\in\mathbb{R}\right\}.$$ Note that $$\left\{\left.\begin{pmatrix} 0 & -c\\ c & 0 \end{pmatrix}\right|c\in\mathbb{R}\right\}=\left\{\left.c\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\right|c\in\mathbb{R}\right\}.$$

Thus, the vector $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$ spans $\{A\in M_{2\times2}(\mathbb{R})|A^T=-A\}$. Moreover, the vector $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$ linearly independent. Therefore, $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$ is a basis of $\{A\in M_{2\times2}(\mathbb{R})|A^T=-A\}$. Therefore, $\{A\in M_{2\times2}(\mathbb{R})|A^T=-A\}$ has dimension equals to $1$.