Linear Algebra : prove geometric multiplicities are the same

[braindead101]

Let A and B be similar matrices. Prove that the geometric multiplicities of the eigenvalues of A and B are the same.

Some help I have gotten so far but still don't know how to proceed from there:
To prove that the geometric multiplicities of the eigenvalues of A and B are the same, we can show that, if B = P^-1 AP , then every eigenvector of B is of the form P^-1 v for some eigenvector v of A.

And i also know that for A and B to be similar matrices, these 5 properties must hold.
1. det A = det B
2. A and B have the same rank
3. A and B have the same characteristic polynomial
4. A and B have the same eigenvalues
5. A is invertible iff B is invertible

any help would be greatly appreciated

 

[HallsofIvy]

The "geometric multiplicity" of an eigenvalue is the dimension of the "eigenspace" (subspace of all eigenvectors) for that eigenvalue.

Now, are you saying that you can use the fact that "if B = P^-1 AP , then every eigenvector of B is of the form P^-1 v for some eigenvector v of A" or do you have to prove that (it's very easy)? If you can use that, then there is an obvious one-to-one correspondence. In particular, you can show that If ${v_1, v_2, ..., v_n}$ is a basis for the eigenspace of A, corresponding to eigenvalue $\lambda$, then $P^{-1}v_1, P^{-1}v_2, ..., P^{-1}v_n}$ is a basis for the eigenspace of B, corresponding to that same eigenvalue.

 

[jack_bauer]

To show that I think it might be sufficient to show that these two matrices (since they're similar) have the same characteristic polynomial. So...

There's an invertible matrix P such that
A= ((P^-1) B P)


det(A-tI)
det([P^-1 BP]-tI)
det([P^-1 BP]-[P^-1tI P])
det([P^-1(B-tI) P])
det(P^-1 P) det(B-tI)
= det(B-tI)

 

[Dick]

To show that I think it might be sufficient to show that these two matrices (since they're similar) have the same characteristic polynomial. So...

There's an invertible matrix P such that
A= ((P^-1) B P)


det(A-tI)
det([P^-1 BP]-tI)
det([P^-1 BP]-[P^-1tI P])
det([P^-1(B-tI) P])
det(P^-1 P) det(B-tI)
= det(B-tI)

Sharing the same characteristic polynomial only tells you the eigenvalues and algebraic multiplicities are the same. It tells you nothing about geometric multiplicity or eigenvectors.