I'm trying to figure out what the dimension of the set of self-adjoint operators on V would be, or in more concrete terms:
Let $dim V =n$. Let $S(V)$ denote the set of self-adjoint linear operators on V. What is its dimension?
The only thing I know that somewhat resembles this might be that $dim L(V) = (dim V)^2$ but I'm not sure how I might be able to apply that to this problem.
Any tips or assistance would be greatly appreciated. Thanks!
EDIT: Also, V is a finite dimensional inner product (or Hermitian) space.
Best Answer
Use matrix representations to figure out the answer. Let $\mathcal B$ be an orthonormal basis of $V$, and let $u\in\mathrm{End}_{\Bbb C}(V)$ be an endomorphism of $V$. Then $u$ is selfadjoint iff its matrix in the basis $\mathcal B$ is hermitian i.e. $$u\text{ is selfadjoint iff }^t\overline{\mathrm{Mat}(u,\mathcal B)}=\mathrm{Mat}(u,\mathcal B).$$ What is the (real) dimension of the subspace of hermitian matrices $H_n\subset M_n(\Bbb C)$? Consider a matrix $(u_{ij})\in M_n(\Bbb C)$: it is hermitian iff for all $i,j\in\lbrace 1,\dots,n\rbrace,~\overline{u_{ji}}=u_{ij}$. You can thus choose the upper coefficients ($i<j$) freely in $\Bbb C$ , the diagonal coefficients $(i=j$) have to be real but can be chosen arbitrarily, and the coefficients below the diagonal ($i>j$) are completely determined by those above the diagonal. Therefore the real dimension of $H_n$ is $2\frac{n(n-1)}{2}+n=n^2$.