dimension of subspace hermitian matrix

Let $T\in H(m+1;\mathbb{R})$ be a symmetric matrix, s.t. $T^2=T$ and $tr(T)=1$.

Now let $U:=\{ S\in H(m+1;\mathbb{R}): TS+ST=S\}$ be a subset of the vector space of all hermitian matrices.

I want to show, that the dimension of U is $m$. For some special T (with a lot of zero's) I could show this. But I have no idea, how to show this for a general fixed T?

Do you have an idea?

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What is $H(m+1,\mathbb{R})$? – 1015 Feb 1 '13 at 16:30
Im sorry, this is the set of symmetric matrices (or hermitian if you want) – Braten Feb 1 '13 at 18:13

An easy way to do this is to diagonalize $T$ in an orthonormal basis and use matrix block multiplication.

The projection $T=T^2$ is diagonalizable in an orthonormal basis Since its trace is $1$, you can find $P$ invertible and symmetric (unitary, actually) such that: $$P^{-1}TP=T_0=\left( \matrix{1 & 0\\ 0& 0}\right),$$ where the lower-right $0$ is the null square matrix of size $m$.

Note that that the isomorphism $X\longmapsto P^{-1}XP$ transforms $U$ into: $$U_0=\{S\in H(m+1;\mathbb{R})\;:\; T_0S+ST_0=S\}.$$ So $\dim U=\dim U_0$.

Now write a general $S\in H(m+1;\mathbb{R})$ $$S=\left( \matrix{ a & b\\ b^t & d }\right)$$ where $b\in \mathbb{R}^m$, $b^t$ is its transpose, and $d$ is in $H(m,\mathbb{R})$.

Now compute by blocks $$T_0S+ST_0=\left(\matrix{2a & b \\ b^t & 0}\right).$$ Equating with $S$ yields $a=0$ and $d=0$.

So clearly $U_0$ is isomorphic to $\mathbb{R}^m$, since such an $S$ is determined by the vector $b$.

Hence $\dim U=\dim U_0=m$.

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