Question

It's a fact that similar matrices have the same eigenvalues. Prove that matrices with the same eigenvalues aren't necessarily similar.

Answer 1

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Answer 2

By the way, \(A\) is similar to \(B\) if there is some matrix \(P\) that relates them like this:

\[A = P^{-1}BP\]

Answer 3

i don't get this, is this a question or a lesson?

Answer 4

A fun challenge question for anyone who's interested >:D

Answer 5

count me in with my algebra 1 education

Answer 6

if the matrices \(A,\;B\) have the same eigenvalues, then, the subsequent condition holds:
\[\Large \det \left( {A - \lambda I} \right) = \det \left( {B - \lambda I} \right)\]

Answer 7

Yup, that's true. Can you use that to show:

\[A \ne P^{-1} B P\]

Answer 8

taco @ShadowLegendX

Answer 9

for example we can consider two similar matrices A, B such that A is nilpotent and B is not nilpotent

Answer 10

How will that show that two matrices with the same eigenvalues are NOT similar though? I think you might have more success if you start the other way round with matrices that have the same eigenvalues, then show they lead to matrices that are not similar.

Answer 11

if:
\[\Large \begin{gathered}
{A^k} = 0 \hfill \\
B = {N^{ - 1}}AN \hfill \\
\end{gathered} \]
then:
\[\Large {B^k} = {N^{ - 1}}{A^k}N = 0\]
which is false by hypothesis

Answer 12

This shows that nilpotent matrices are not similar to a matrix that's not nilpotent. It's a nice proof I like this, but it doesn't have anything to do with what I'm asking.

Answer 13

I have to think better to this question, I will answer tomorrow, now I have to go
thanks for your suggestion @Kainui

Answer 14

Ahhh I think I see your reasoning now, but I think it needs a little more before it works.

I think you're saying A and B have the same eigenvalues and are similar. Then you show that they can't be similar by the fact that A is nilpotent and B is not.

The only thing left to show is that you can construct A and B with the same eigenvalues, but with one being nilpotent and the other not. I am not sure if this is possible or not.

Answer 15

there is a trivial example...you are working too hard

Answer 16

If B is the 2x2 identity matrix, what does A have to be if A and B were similar?

Answer 17

I agree that there's a simple way to do this, since we just have to find a single counter example. But I kinda like this more general answer that we're looking for it's kinda fun.

Here's the proof I had in mind:

\[\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = A\]

A has the same eigenvalues as the identity matrix but can't be similar since this is a contradiction:

\[A=P^{-1}IP = I \]