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Good evening; can you help me with the problem?

Let an $ n\times n$ matrix A have eigenvalues $\alpha_{1},...\alpha_n$ and eigenvectors ${a}_{1},.. {a}_{n}$ Find eigenvalues and eigenvectors for operator $$L:{Mat}_{n\times m} \to {Mat}_{n\times m}, \; L(X) = AX;$$

I tried to present $$ x = v *u^T ,$$ where $v$ is $n\times n$ and $u^T$ is $n\times m$ , but what to do next, I don't understand.

I looked for examples of similar problems, but I found nothing.

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    $\begingroup$ Hint: Consider the matrix $X$ whose $i^{\text{th}}$ column is $a_j$ and the remaining entries are all $0$. $\endgroup$ – Michael Burr Feb 19 at 13:29
  • $\begingroup$ @JoséCarlosSantos I don't think he did. You can put any eigenvector in any column. $\endgroup$ – Theo Bendit Feb 19 at 13:59
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Let $X$ be an eigenvector for $L$, i.e. a non-zero matrix such that $$AX=\lambda X.$$ Write $X=(x_1,\dots, x_n)$, whith columns $x_1,\dots, x_n$. The above equation can then be read column-wise as $$Ax_i=\lambda x_i.$$ So you see that every column of $X$ has to be an eigenvector of $A$ with corresponding eigenvalue $\lambda$ or the zero vector.

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  • $\begingroup$ I believe the question is asking for eigenvectors/values for the operator on the space of matrices. In other words, finding scalars $\lambda$ and non-zero matrices $X$ such that $L(X) = AX = \lambda X$. $\endgroup$ – Theo Bendit Feb 19 at 13:57
  • $\begingroup$ @TheoBendit Thanks for pointing out. I was reading to careless.. $\endgroup$ – James Feb 19 at 14:02
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Let $A$ and $X=\begin{bmatrix}x_1&x_2&\dots&x_m\end{bmatrix}$ be as you defined, where $x_i$ is the $i^{\text{th}}$ column of $X$. Then, recall that $$ AX=\begin{bmatrix}Ax_1&Ax_2&\dots&Ax_m\end{bmatrix}. $$

An eigenvector for the operator you defined is a nonzero matrix $X$ so that $AX=\lambda X$ for some $\lambda$.

Let $X_j^i$ be the matrix consisting of all zeros except that vector $a_j$ is in column $i$. In other words, $$ X_j^i=\begin{bmatrix}0&0&\dots&0&a_j&0&\dots&0\end{bmatrix}. $$ Then, \begin{align} AX_j^i&=\begin{bmatrix}A0&A0&\dots&A0&Aa_j&A0&\dots&A0\end{bmatrix}\\ &=\begin{bmatrix}0&0&\dots&0&\lambda_ja_j&0&\dots&0\end{bmatrix}=\lambda_jX_j^i. \end{align}

Therefore, $X_j^i$ is an eigenvector for this map, are these all of the eigenvectors/eigenvalues (how many eigenvalues do you expect and how many have you found)?

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  • $\begingroup$ I also have to consider that multiplicity of eigenvalues is possible? $\endgroup$ – GIFT Feb 19 at 20:58
  • $\begingroup$ I'm not sure of your question. Multiple eigenvalues in $A$ is not a problem because you have a full set of eigenvectors. You will certainly have repeated eigenvalues for the given operator, but they have distinct eigenvectors. $\endgroup$ – Michael Burr Feb 19 at 22:11
  • $\begingroup$ I expect n eigenvalues, as I understand i=1....n and $a_j$ are different $\endgroup$ – GIFT Feb 20 at 9:50
  • $\begingroup$ Since $m\times n$ matrices have $mn$ entries, this linear map can be thought of as a map on $\mathbb{R}^{mn}$. This leads to many more eigenvalues, although there are repeats. $\endgroup$ – Michael Burr Feb 20 at 10:37
  • $\begingroup$ Thanks a lot for your help $\endgroup$ – GIFT Feb 21 at 10:50

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