Similar Matrices & Geometric Multiplicity

[Whatever123]

Homework Statement

Prove that if two matrices are similar then they have the same eigenvalues with the same algebraic and geometric multiplicity.

Homework Equations

Matrices A,B are similar if A = C\breve{}BC for some invertible C (and C inverse is denoted C\breve{} because I tried for a long time to figure out how to get an inverse sign in latex but couldn't figure it out...).

The Attempt at a Solution

To show that two similar matrices have the same eigenvalues with the same geometric multiplicities, I need to show that their characteristic polynomials are the same.
Let A,B be similar matrices. Then,
A = C(inverse)BC
A-\lambdaI = C\breve{}BC - \lambdaI
A-\lambdaI = C\breve{}BC - \lambdaC\breve{}C since C\breve{}C=I
A-\lambdaI = C\breve{}[B-\lambda]C
det(A-\lambdaI) = det(C\breve{}[B-\lambdaI]C)
det(A-\lambdaI) = det(C\breve{}det(B-\lambdaI)det(C)
det(A-\lambdaI) = det(C\breve{})det(C)det(B-\lambdaI)
Therefore, det(A-\lambdaI) = det(B-\lambdaI)
So, A, B have the same characteristic polynomials. This implies that they have the same eigenvalues with the same algebraic multiplicity. However, I do not think that this implies that they have the same geometric multiplicity because it doesn't same anything about the dimension of the eigenspace. Does it? Any suggestions on how I should try to show that the geometric multiplicity is the same?

So does anyone have any advice? I spent a while to get the first part, I just could use a nudge in the right direction for the next part... Thanks in advance.

 

[Dick]

No, algebraic multiplicity doesn't tell you anything about geometric multiplicity. The geometric multiplicity is the number of linearly independent eigenvectors corresponding to a given eigenvalue lambda. Hint: can you show if A=C^(-1)BC and x is an eigenvector of A, then Cx is an eigenvector of B with the same eigenvalue?

 

[Whatever123]

No, algebraic multiplicity doesn't tell you anything about geometric multiplicity. The geometric multiplicity is the number of linearly independent eigenvectors corresponding to a given eigenvalue lambda. Hint: can you show if A=C^(-1)BC and x is an eigenvector of A, then Cx is an eigenvector of B with the same eigenvalue?

So now that I've proven that A, B have same eigenvalues \lambda1-\lambdan and that they appear the same amount of times, consider the eigenvectors for those eigenvalues. As you said, let x be an eigenvector of A. Need to show that Cx is an eigenvector of B with the same eigenvalue...

If x is an eigenvector of A, then:
Ax = \lambdax where \lambda is the corresponding eigenvalue.

Since A=C^(-1)BC, Ax=C^(-1)BCx=\lambdax
Then, CAx=BCx=C\lambdax

So, BCx = Cx\lambda

Shows that Cx is an eigenvector of B with the same eigenvalue \lambda.

So since if x is an eigenvector of A with eigenvalue \lambda then Cx is an eigenvector of A with the same eigenvalue \lambda then the geometric multiplicities are the same? The same eigenvalue will correspond to a similar eigenvector and eigenspace so the geometric multiplicity will be the same?

 

[Dick]

Basically. The geometric multiplicity is the dimension of the eigenspace of lambda. You've shown C maps the eigenspace of A to the eigenspace of B. C^(-1) does the reverse. C is a nonsingular matrix. That shows the dimensions are the same, right?

 

[Whatever123]

Yes, I understand now. Thank you for your help. I figured out the first part and just needed that little hint to help me with the rest!