Rank of a linear transformation T Let, $n$ be a positive integer & let $M_n(\mathbb R)$ be the space of all $n\times n$ real matrices. If $T:M_n(\mathbb R)\to M_n(\mathbb R)$ is a linear transformation such that $T(A)=0$ , whenever $A\in M_n(\mathbb R)$ is symmetric or skew-symmetric, then what is the rank of $T$ ?
Here we have to find $N(T)$.
Clearly $Dim(M_n(\mathbb R))=n^2$.
So, $N(T)=n^2 -R(T)$.
I want to know the dimension of $R(T)$.
 A: Let $\DeclareMathOperator{Span}{Span}\DeclareMathOperator{Sym}{Sym}\Sym_n(\Bbb R)$ and $\DeclareMathOperator{Skew}{Skew}\Skew_n(\Bbb R)$ be the subspaces of $\DeclareMathOperator{Mat}{Mat}\Mat_n(\Bbb R)$ consisting of $n\times n$ symmetric and skew-symmetric matrices respectively.
Note that
\begin{align*}
\dim\Sym_n(\Bbb R)&=\frac{n(n+1)}{2}&\dim\Skew_n(\Bbb R)&=\frac{n(n-1)}{2}
\end{align*}
(can you prove this?). It follows that
$$
\dim\Mat_n(\Bbb R)=n^2=\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=\dim\Sym_n(\Bbb R)+\dim\Skew_n(\Bbb R)
$$
Thus $$\Mat_n(\Bbb R)=\Sym_n(\Bbb R)\oplus\Skew_n(\Bbb R)$$
This means any linear operator $T:\Mat_n(\Bbb R)\to\Mat_n(\Bbb R)$ that vanishes on both $\Sym_n(\Bbb R)$ and $\Skew_n(\Bbb R)$ must vanish on all of $\Mat_n(\Bbb R)$ and thus has rank zero.
Note. Another way to see that $$\Mat_n(\Bbb R)=\Sym_n(\Bbb R)\oplus\Skew_n(\Bbb R)$$ is to observe that for every $A\in\Mat_n(\Bbb R)$ we have
$$
A=\frac{1}{2}(A+A^\top)+\frac{1}{2}(A-A^\top)
$$
with
\begin{align*}
\frac{1}{2}(A+A^\top)\in\Sym_n(\Bbb R) &&
\frac{1}{2}(A-A^\top)\in\Skew_n(\Bbb R)
\end{align*}
and that 
$$
\Sym_n(\Bbb R)\cap\Skew_n(\Bbb R)=\{\Bbb O\}
$$
where $\Bbb O$ is the zero matrix.
