Dimensions of Subspace

Question

Suppose $V$ is a subspace of $\mathbb{R}^{3 \times 3}$ consisting of skew-symmetric diagonal matrices. What is $\mathrm{dim} \; V$?

The skew-symmetric matrix implies that the transpose of the matrix is its negative. I.e $A^\top = -A$.

Taking into account that $A$ may be a diagonal skewed matrix, how does that determine the dimension of $V$. Would $\dim V$ simply be 3...

Answer 1

To properly answer the question, you need to find a basis for skew symmetric matrices. All such matrix is of the form $$\pmatrix{0&a&b\\ -a&0&c\\-b&-c&0}$$ A basis would consist of specific such matrices for $(a=1,\,b=c=0)$, $\ (b=1,\,a=c=0)$ and $(c=1,\,a=b=0)$.

Answer 2

Since it is diagonal the $\dim V$ is at most $3$. Also skew-symmetry imposes that$$A^T=-A$$since $A$ is diagonal we have $A^T=A$ therefore $$A=-A$$ or $$A=0$$which means that $$\Large \dim V=0$$