Let $V$ be an n-dimensional real inner product space. Let $M\in L(L(V))$ be defined by $M(S)=S+S^*$. Then determine the trace of $M$
I'm confused by this problem. I think $M$ is an operator $M: L(V)\to L(V)$, so it acts on the operators on $V$ and the result is also an operator on $V$
Then my first idea for dealing with this problem is to consider the basis for $L(V)$ since intuitively I think the basis of $L(V)$ is the matrix with "1" in only one entry and zero in other entries and there are $n\times n$ many of this kind of matrices. Then under an orthonormal basis, the adjoint of a matrix is just the conjugate transpose of that matrix, plus now the inner products space is real. Thus under an orthonormal basis, the matrix of its adjoint is just the transpose of itself.
Then I just choose an orthonormal basis for $V$ and then based on that basis for $V$ I can get the basis for $L(V)$. I denote $e_{ij}$ as the matrix that has "1" in $i$ row $j$ column and "0" everywhere else
Then based on the definition of $M$, it sends an operator to itself plus its adjoint, and hence I think if $M$ acts on $e_{11}$, it will just be $$M(e_{11})=e_{11}+e_{11}=2e_{11}$$ If $M$ acts on the basis $e_{12}$ it will give $$M(e_{12})=e_{12}+e_{21}$$
Then for $M$, it acts on the basis vectors for $L(V)$ which are $(e_{11},e_{12},\cdots,e_{1n},e_{21},\cdots,e_{2n},\cdots\cdots,e_{n1},\cdots,e_{nn})$.
Thus this will make the matrix of $M$ have some "2" on the diagonal, since when $M$ acts on $e_{ii}$, it will directly give $2e_{11}$; when $M$ acts on $e_{ij}$ it will only give "1" on the diagonal. Thus I think the trace of $M$ is $2n+1\cdot (n-1)$ Is this right? If not, any help on this problem? Thanks!
There are several ways for achieving that. One way is to find the spectral decomposition of the operator.
A basis for $\mathbb{R}^{n\times n}$ has $n^2$ components. Rather than evaluating on the standard basis $\{e_{ij}\}_{i,j=1}^n$ where $e_{ij}=e_ie_j^T$, it may be more interesting to find all the $n^2$ eigenvalues of the operator and just sum them all.
We have that $M(e_{ii}) = 2e_{ii}$, $i=1,\ldots,n$. So, $e_{ii}$ is an eigenmatrix and its associated eigenvalue is 2. We have $n$ of them.
We have that $M(e_{ij}-e_{ji}) = e_{ij}-e_{ji}+(e_{ij}-e_{ji})^T=e_{ij}-e_{ji}+e_{ji}-e_{ij}=0$. So, $e_{ij}-e_{ji}$ is an eigenmatrix and its associated eigenvalue is 0. We have $n(n-1)/2$ of them.
Finally, $M(e_{ij}+e_{ji}) = e_{ij}+e_{ji}+(e_{ij}+e_{ji})^T=2(e_{ij}+e_{ji})$. So, $e_{ij}-e_{ji}$ is an eigenmatrix and its associated eigenvalue is 2. We also have $n(n-1)/2$ of them.
All the multiplicities sum to $n^2$, so we are done and the trace is given by
$$\mathrm{trace}(M)=2n+2(n^2-n)/2=n^2+n.$$