Solving the matrix equation $AP=PB$, where $A$ and $B$ are similar

Question

Currently I'm stuck with this problem:

The following matrices are similar to each other.

$$A=\begin{bmatrix}-3&0\\8&4\end{bmatrix},\qquad B=\begin{bmatrix}-4&4\\-2&5\end{bmatrix},\qquad C=\begin{bmatrix}k&1\\m&-2\end{bmatrix}.$$

How do you find the matrix $P$ such that $AP = PB$?

And how can you calculate the values of $k$ and $m$?

Thank you in advance.

Answer 1

Let $P=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Solve the equation $AP=PB$ and you will get $b=-\frac a2$ and $d=-4a-4c$. So$$P=\begin{pmatrix}a&-\frac a2\\c&-4a-4c\end{pmatrix}\text{ and }\det P=-\frac{a(8a+7c)}2.$$So, if you take, say, $a=1$ and $c=-1$, $P$ is invertible.

For the other question, take $k=3$ and $m=6$, so that $C$ has the same trace and the same determinant as $A$ and $B$.

Answer 2

Hint:

If you can diagonalize matrix $A$ and $B$, then we have

$$A=P_ADP_A^{-1}$$

$$D=P_A^{-1}AP_A$$

$$B=P_BDP_B^{-1}$$

$$D=P_B^{-1}BP_B$$

Hence, $$P_A^{-1}AP_A=P_B^{-1}BP_B$$

Hint for the second question, consider the trace and determinant.

Answer 3

Hint. Two similar matrices has the same same characteristic polynomial. Recall that the characteristic polynomial of a $2\times 2$ matrix $M$ is $$\det(M-\lambda I)=\lambda^2-\mbox{tr}(M)\lambda+\det(M).$$