Trace of map & matrices Let $1\leq m,n\in \mathbb{N}$ and let $\mathbb{K}$ be a field.
For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$
I want to show that $trace(\mu_a)=n\cdot trace(a)$ .
I have done the following:
Let $\lambda$ be the eigenvalues of $\mu_a$ then we have that $\mu_a(c)=\lambda c$.
From $\mu_a(c)=\lambda c$ we get $ac=\lambda c$.
So if $\lambda$ is an eigenvalue of $\mu_a$, tthere is a non-zero $c\in\mathbb{K}^{m\times n}$ with $\mu_a(c)=\lambda c$.
The columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$.
The matrix $c$ has $n$ columns.
So for each eigenvalue $\lambda$ of $a$ there are $n$ eigenvectors, so the multiplicity of $\lambda$ is $n$.
The trace of a matrix is the sum of teh eigenvalues considering the multiplicity.
Since each eigenvalue of $\mu_a$ has a multiplicity of $n$, it follows that $\text{trace}(\mu_a)=\sum_i n\cdot \lambda_i=n\cdot \sum_i\lambda $.
Since $\lambda_i$ is the eigenvalue of $a$, it follows that $\text{trace}(a)=\sum_i\lambda_i$.
Therefore we get $\text{trace}(\mu_a)=n\cdot \text{trace}(a)$.
Is everything correct?
 A: Your analysis is correct in the case that $a$ is diagonalizable. If $a$ is not diagonalizable, it suffices to show that the dimension of the generalized eigenspace of $\mu_a$ associated with $\lambda$ is $n$ times the dimension of the generalized eigenspace of $a$ associated with $\lambda$.
We can show this in turn by noting that this generalized eigenspace of $a$ is $\ker(a - \lambda I)^m$, and
$$
(\mu_a - \lambda I)^k = \mu_{(a - \lambda I)^k}.
$$
A: With a matrix representation for $\mu_a$ you may compute the trace somewhat simpler as the sum of its diagonal elements, without having to worry about diagonalising the map. To get such a representation you may use the trick of introducing a $\delta$ (over indices) and rewrite in the following index form (like a disguised variant of the comment of @Omnomnomnom):  $$(\mu_a(c))_{ij}= \sum_{k} a_{ik}c_{kj} = \sum_k\sum_l a_{ik}\delta_{jl} c_{kl} = \sum_{k,l} M_{ij,kl} c_{kl}.$$
Here $M_{ij,kl}= a_{ik} \delta_{jl}$ is just a 'normal' $mn \times mn$ matrix and the trace is given by $${\rm tr\;} \mu_a = {\rm tr\;} M = \sum_{ij} M_{ij,ij} = \sum_{i=1}^m \sum_{j=1}^n a_{ii}\delta_{jj} = {\rm tr\; } a  \times n.$$
