Dimension of vector space of 2x2 skew symmetric matrices

Question

I had a question about the dimension of this subspace. This was related to a problem that had a case of $n\times n$ matrices, but I accidentally read it as the special case of $2\times 2$, but never the less the answer to this question should help me with the general case.

Question

Consider the set of all skew-symmetric $M_{2\times 2}$ matrices, call it $W$. A matrix $A$ is an element of this set provided that $-A = A^{t}$. Find a basis for $W$, what is the dimension of $W$?

Attempted Answer

If we consider $$A = \begin{pmatrix} a &b \\ c&d \end{pmatrix}$$

Then $$A^{t} = \begin{pmatrix} a & c \\ b & d\end{pmatrix}.$$

So, for a matrix to be in our subspace, we would have

$$\begin{pmatrix}-a & -b \\ -c & -d\end{pmatrix} = \begin{pmatrix}a & c \\ b & d\end{pmatrix}$$

This implies that $-a = a$ and $-d = d$, so $a$ and $d$ are $0$ and that $b = -c$.

Therefore the resulting set of matrices will look like $$\begin{pmatrix}0 & -c\\c & 0\end{pmatrix}.$$

This is where I run into trouble. I'm not sure if the dimension of this subspace is $1$ or $2$. If I factor the $c$ out, then I have the matrix $$ \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}, $$ and by taking any scalar multiple of this matrix, I can create any matrix in the subspace of $2\times 2$ skew-symmetric matrices. It's also linearly independent so it should be a basis.

My confusion arises in that I could also decompose the matrix further into the two matrices

$$\begin{pmatrix}0 & -c \\ 0 & 0\end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 0 & 0 \\c & 0\end{pmatrix}$$

These are both linearly independent vectors and span the subspace, so it's a basis, and I was wondering if that means the dimension of the subspace should be $2$. But I wonder if the fact that the choice of $c$ affects both matrices means that this decomposition isn't necessary and that since you can't choose two different weights for these matrices that the dimension of the subspace must in fact be one.

Sorry that was a long post, and maybe not the cleanest, still getting used to Latex. Thanks for any help.

Answer 1

A skew symmetric matrix has each $(ij)^{th}$ entry is equal to the $(-ji)^{th}$ entry.

The possible number of base elements is, $$(n-1)+(n-2)+(n-3)+ \dotsm +[n-(n-1)]$$ $$=(n-1)+(n-2)+(n-3)+ \dotsm +1$$ (the sum of the integers from $1$ to $n-1$) $$=\frac{n(n-1)}{2}$$ which is the dimension of the given $n\times n$ skew symmetric matrix

Answer 2

$\begin{bmatrix}0 & -c \\ 0 & 0\end{bmatrix}$ is not a...

Answer 3

$\{\begin{bmatrix}0 & -1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \}$ spans a bigger space, which is:

$$\{\begin{bmatrix}0 & a \\ b & 0\end{bmatrix} \space | \space a,b \in F\}$$

And one can easily show that these 2 matrices are independent and there is no smaller set which spans this space, and hence It's dimension is $2$.

The space you are after is a subspace of the above, satisfying $a=-b$. This is a new restriction and so It's safe to assume that the dimension of this subspace is smaller, and has to be $1$ (Because we know skew symmetric matrices exist).

Alternatively, you said so your self:

$$\{\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \}$$ spans your space; Every $2 \times 2$ skew symmetric matrix is some single scalar times this matrix.