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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.

Thanks in advance.

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    For 2x2, I found a basis with dimension 1...2012-10-27

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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. $\dim\ker(S)=(n^2-n)/2$.

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    But couldn't these coe$f$ficients be rearranged in a variety of ways? Let's consider the a, b, and c from the 3x3 example in @EuYu 's answer here. They could be... (a,b,c)=(+,+,+) or (+,+,-) or (+,-,+) or (-,+,+). That's 4. I think the formula might be dim ker(S) = 2^{ (n^2-n)/2 }/2 .2012-10-27
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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.