# Eigenvalues eigenvectors

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.

• Hint: Consider the matrix $X$ whose $i^{\text{th}}$ column is $a_j$ and the remaining entries are all $0$. – Michael Burr Feb 19 '19 at 13:29
• @JoséCarlosSantos I don't think he did. You can put any eigenvector in any column. – Theo Bendit Feb 19 '19 at 13:59

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.

• 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 $L(X) = AX = \lambda X$. – Theo Bendit Feb 19 '19 at 13:57
• @TheoBendit Thanks for pointing out. I was reading to careless.. – user526015 Feb 19 '19 at 14:02

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)?

• I also have to consider that multiplicity of eigenvalues is possible? – GIFT Feb 19 '19 at 20:58
• 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. – Michael Burr Feb 19 '19 at 22:11
• I expect n eigenvalues, as I understand i=1....n and $a_j$ are different – GIFT Feb 20 '19 at 9:50
• 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. – Michael Burr Feb 20 '19 at 10:37
• Thanks a lot for your help – GIFT Feb 21 '19 at 10:50