# Eigenvalue Proof

## Homework Statement

Prove that similar square matricies have the same eigenvalues with the same algebraic multiplicities.

C^-1PC=Q

## The Attempt at a Solution

Am I supposed to show that (P-$$\lambda$$I)x=(C^-1PC-$$\lambda$$I)x?

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Dick
Homework Helper
The eigenvalues are the roots of the characteristic polynomial. Can you show Q=C^(-1)PC and P have the same characteristic polynomial? I.e. det(Q-xI)=det(P-xI)?

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Am I supposed to show that (P-$$\lambda$$I)x=(C^-1PC-$$\lambda$$I)x?
Well, that's equivalent to C-1PC=Q, isn't it?

solution:
What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since $$det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q} -\lambda\mathbf{I} )\mathbf{C})$$
Now use det(AB)=det(A)*det(B) and your done.
cheers

Dick
Homework Helper
Well, that's equivalent to C-1PC=Q, isn't it?

solution:
What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since $$det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q} -\lambda\mathbf{I} )\mathbf{C})$$
Now use det(AB)=det(A)*det(B) and your done.
cheers
Hi, 1777-1855. The forum rules prohibit posting complete solutions (i.e. doing the problem for the poster). That's awfully close. Try and give a hint first and see if they can figure out the rest themselves, ok?? It's more fun that way and they learn more.

The eigenvalues are the roots of the characteristic polynomial. Can you show Q=C^(-1)PC and P have the same characteristic polynomial? I.e. det(Q-xI)=det(P-xI)?
So if I prove those two determinants are equal, it necessarily follows that the two matricies have the same eigenvalues and algebraic multiplicities, because the characteristic polynomials are the same?

Dick