Find $m$ such that matrix equation holds.

Question

I have the matrix $A = \begin{pmatrix} 3 & -1\\ -5 & 2 \end{pmatrix}$. I need to find real values of $m$ for which there exists a matrix $B\in M_2(\mathbb{R})$ which is not the $0$ matrix such that $AB = mB$ holds.

I tried to set up a matrix $\begin{pmatrix} x & y\\ z & t \end{pmatrix}$, and then do the multiplication $AB$, and solve the system $AB = mB$ for m. But I cannot figure out what m is.

Answer 1

Maybe I misunderstand your question, but are you not asking for which $m$ does there exists a matrix $B$ such that $(A-m Id)B=0$, where $Id$ is the $2\times 2$ identity matrix? Well, this is the same as to say that $A-m Id$ has zero determinant, otherwise it would be invertible and then B had to be the zero matrix. So, solving the determinant of $A- m Id$ and setting it to zero leads to the equation $(m-2)(m-3)-5=0$, which you can easily solve. For these two concrete values of $m$, I guess it should be easy then to find some matrix $B$ with the desired property.

Answer 2

Hints:

1).Compute the two eigenvalues ($\alpha=\frac{5+\sqrt{21}}{2}$ and $\beta=\frac{5-\sqrt{21}}{2}$) and find corresponding eigenvectors of $A$, call them $u$ and $v$ (so $Au=\alpha u$ and $Av=\beta v$).

2). Then you can have $B=\begin{bmatrix}\uparrow & \uparrow \\ u & 0\\ \downarrow & \downarrow\end{bmatrix}$ or $B=\begin{bmatrix}\uparrow & \uparrow \\ u & u\\ \downarrow & \downarrow\end{bmatrix}$ etc..

This will give $$AB=\alpha B.$$

Likewise you can do it with $\beta$.