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- Apr 14, 2013

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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 \begin{equation*}\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac\end{equation*}

I have show that $\mu_a$ is a linear operator on $\mathbb{K}$-vector space $\mathbb{K}^{m\times n}$ :

Let $\lambda \in \mathbb{K}$, $c, c_1, c_2\in \mathbb{K}^{m\times n}$.

$\mu_a$ is homogeneous : \begin{equation*}\mu_a\left (\lambda c\right )=a\left (\lambda c\right )=\lambda \left (ac\right )=\lambda \mu_a(c)\end{equation*}

$\mu_a$ is additive : \begin{equation*}\mu_a\left (c_1+c_2\right )=a\left (c_1+c_2\right )=\lambda c_1+\lambda c_2= \mu_a(c_1)+ \mu_a(c_2)\end{equation*}

Next I want to show that $\text{Sp}(\mu_a)=n\text{Sp}(a)$, $\det (\mu_a)=\det (a)^n$ and $P_{\mu_a}=P_a^n$, and that if $a$ is diagonalizable then $\mu_a$ is diagonalizable. ($P$ is the characteristic polynomial.)

The spectrum is the set of eigenvalues. What is meant by $\text{Sp}(\mu_a)=n\text{Sp}(a)$ ? The eigenvalues of the map $\mu_a$ are the same as the eigenvalues of the matrix of the map, or not?