1
$\begingroup$

This question already has an answer here:

$\mathbf{B = P^{^-1} A P} \iff$ ($\mathbf{B}$ is similar to $\mathbf{A}$)

I'm a little confused about matrix similarity. Let's say we have the following matrix $A$:

\begin{bmatrix}2&5\\4&1\end{bmatrix}

Its eigenvalues are $6$ and $-3$. Respective eigenvectors $\left<(5,4)\right>, \left<(-1,1)\right>$.

How do I find out whether its similar to another matrix (B)?

Let's say matrix $B$:

\begin{bmatrix}4&5\\1&4\end{bmatrix}

Trying to actually find the regular $P$ so that the equation would be satisfied seems like way too much calculating.

My idea is to try and diagonalize both of them - and if I end up with same eigenvalues on the diagonal (in whatever order), then they are both similar? (using the transitive property, since similarity is an equivalence).

Thank you in advance.

$\endgroup$

marked as duplicate by Kamil Jarosz, Pragabhava, user296602, SchrodingersCat, hardmath Jan 25 '16 at 18:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

Your idea is correct and you can even use it to find the matrix $P$. Diagonalizing $A$, you will find $P_1$ such that $P_1^{-1} A P = D_1$ where $D_1$ is diagonal and the diagonal entries are the eigenvalues of $A$. Diagonalizing $B$, you will find $P_2$ such that $P_2^{-1} B P_2 = D_2$ with $D_2$ diagonal. If the matrices $A$ and $B$ have the same eigenvalues (counting with multiplicities), then you can arrange that $D_1 = D_2$ and then

$$ A = P_1 D_1 P_1^{-1} = P_1 D_2 P_1^{-1} = P_1 P_2^{-1} B P_2 P_1^{-1} = P^{-1} B P $$

where $P = P_2 P_1^{-1}$.

If $A$ and $B$ don't have the same eigenvalues, or $A$ is diagonalizable and $B$ is not or the other way around, then $A$ and $B$ won't be similar. If both $A$ and $B$ are not diagonalizable, you can bring them to a canonical form instead.

$\endgroup$
  • $\begingroup$ Thank you very much! You have explained it very clearly! Does that also mean that once a matrix is not diagonalizable - it is not similar to any matrix? $\endgroup$ – P. Lance Jan 25 '16 at 14:39
  • $\begingroup$ Nope - a non-diagonalizable matrix cannot be similar to a diagonalizable matrix but can be similar to many other non-diagonalizable matrices. $\endgroup$ – levap Jan 25 '16 at 14:48
  • $\begingroup$ Just checked it, you are right! Non-diagonalizable matrices are always similar to other non-diagonalizable matrices (multiplying non-D by any choice of regular inv(P) and P will result in another non-D matrix). Thanks a lot, really appreciate the help! :) $\endgroup$ – P. Lance Jan 25 '16 at 15:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.