How to find dimensions of skew-symmetric diagonal matrices.

Question

Take for example that $V$ is a subspace of $\Bbb R^{3x3}$.

I know that dimension $V$ for a skew-symmetric matrix would be $3$.

How would the dimensions change for $V$ if it were to be a skew-symmetric DIAGONAL matrix?

Thank You

Answer 1

Hint: All diagonal entries of any skew-symmetric matrix must be $0$.

Answer 2

Well, unless I'm misunderstanding what you mean, this is fairly trivial. A matrix $A$ is skew-symmetric if $A^\dagger=-A,$ which means in particular that all diagonal entries will be $0.$ However, diagonal matrices have all non-diagonal entries equal to $0,$ so what does that leave us with?

Answer 3

A skew-symmetric diagonal matrix $K$ must be $0$; for, writing

$K = [k_{ij}], \tag 1$

we have, since $K$ is diagonal, that

$k_{ij} = 0, \; i \ne j; \tag 2$

since $K$ is skew, we also have

$K = -K^T, \tag 3$

which in terms of (1) yields

$k_{ij} = -k_{ji}; \tag 4$

if $i = j$, this reads

$k_{ii} = -k_{ii}; \tag 5$

thus

$k_{ii} = 0, \; 1 \le i \le n, \tag 6$

which tells us, in concert with (2), that

$k_{ij} = 0, \; \forall \; i, j, \; 1 \le i, j \le n; \tag 7$

therefore,

$K = [k_{ij}] = [0] = 0. \tag 8$

It follows that $V = \{0\}$ in this case, so apparently

$\dim V = 0. \tag 9$