# Similar matricies

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 doesnt 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.

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HallsofIvy
Homework Helper
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.

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

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

HallsofIvy
Homework Helper
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.)