# Given the linear transformation from $M_{n}(\mathbb{R}) \to M_{n}(\mathbb{R})$ defined by $S(A) = A + A^{T}$, how do you find the $\dim(\ker(S))$?

Given the linear transformation from $S: M_{n}(\mathbb{R}) \to M_{n}(\mathbb{R})$ defined by $S(A) = A + A^{T}$, where $A$ is a fixed $n \times n$ matrix, how do you find the $\dim(\ker(S))$?

I've found that $\ker(S) = \mbox{the set of all matrices whose transpose, negated, is itself}$, but how can I find the dimension? I'm having difficulty coming up with a basis.

-
Try to do it for $n = 2$ and $n = 3$. How many parameters you need to describe $2 \times 2$ anti-Hermitian matrices? $3 \times 3$? –  levap Oct 27 '12 at 3:42
For 2x2, I found a basis with dimension 1... –  dmonopoly Oct 27 '12 at 3:47

You already know that $\ker(S)=\{A \in M_n(\mathbb{R}):\ A^T=-A\}$. Now, let $A=(a_{ij})$ be a member of $\ker(S)$. Then $a_{ji}=-a_{ij}$ for every $i,j$. In particular we have $a_{ii}=0$ for every $i$. Hence, we only need $(n\times n-n)/2$ coefficients to describe an element of $\ker(S)$, i.e. $\dim\ker(S)=(n^2-n)/2$.
I would write down a few examples for smaller matrices and work out a general pattern. For example, the set of $2\times 2$ skew-symmetric matrices are of the form $$\begin{pmatrix}0 & a \\ -a & 0\end{pmatrix}$$ which is clearly of dimension $1$. (Why must the diagonals be $0$?). For $3 \times 3$ matrices, we have $$\begin{pmatrix}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{pmatrix}$$ If necessary, work out the $4 \times 4$ case yourself to see if you can spot the pattern.