Find P Such That P^-1AP=B: Similar Matricies

[dangish]

1a) First find if A and B are similar (ie: A~B).
b) If so find P such that P(^-1)AP=B. (P^-1 is the inverse of P)

Ok so I'm not going to give the matricies because I don't know how to write them out properly on this and It doesn't really matter anyways.

First I found if A and B were similar, which to the best of my knowledge has to do with the determinant. ie: If the determinant of A and B are equal then A~B, is this correct?

Since I found they were similar, I went on to part b and this is where I am stuck. I have looked through all my notes and the book notes and none of them seem to ever solve for p, they just get to a certian point in the problem and write out, "therefore P(^-1)AP=B" and it makes no sense to me.

Some advice on a method to go about finiding P would be much appreciated, it's exam time! Thanks in advance.

 

[HallsofIvy]

No, it is not correct. For example, the matricies
$$\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}$$
and
$$\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$$
have the same determinant (4) but are not similar.

$$\begin{bmatrix}8 & 0 \\ 0 & 3\end{bmatrix}$$
and
$$\begin{bmatrix}6 & 0 \\ 0 & 4\end{bmatrix}$$
have the same determinant but are not similar.

What is true is is the other way- if two matrices are similar, then they have the same determinant.

Two matricies that have the same eigenvalue and same corresponding eigenvectors are similar.

 

[dangish]

So when given two matricies, the only way to tell if they are similar is to check their eigenvalues and eigenvectors?

 

[dangish]

I found they have the same characteristic polynomials, do I need to continue to find if they have the same eigenvectors?

 

[HallsofIvy]

Yes. Again,
$$\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}$$
and
$$\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$$
have the same characteristic polynomial, $(\lambda- 2)^2$,
but are not similar.

(They NOT have the same minimal polynomial, $(\lambda- 2)^2$ for the first and $\lambda- 2$ for the second.)