# Homework Help: Similar Matrices & Geometric Multiplicity

1. Feb 27, 2010

### Whatever123

1. The problem statement, all variables and given/known data
Prove that if two matrices are similar then they have the same eigenvalues with the same algebraic and geometric multiplicity.

2. Relevant 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...).

3. 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-$$\lambda$$I = C$$\breve{}$$BC - $$\lambda$$I
A-$$\lambda$$I = C$$\breve{}$$BC - $$\lambda$$C$$\breve{}$$C since C$$\breve{}$$C=I
A-$$\lambda$$I = C$$\breve{}$$[B-$$\lambda$$]C
det(A-$$\lambda$$I) = det(C$$\breve{}$$[B-$$\lambda$$I]C)
det(A-$$\lambda$$I) = det(C$$\breve{}$$det(B-$$\lambda$$I)det(C)
det(A-$$\lambda$$I) = det(C$$\breve{}$$)det(C)det(B-$$\lambda$$I)
Therefore, det(A-$$\lambda$$I) = det(B-$$\lambda$$I)
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.

2. Feb 27, 2010

### 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?

3. Feb 27, 2010

### Whatever123

So now that I've proven that A, B have same eigenvalues $$\lambda$$1-$$\lambda$$n 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 = $$\lambda$$x where $$\lambda$$ is the corresponding eigenvalue.

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

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?

4. Feb 27, 2010

### 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?

5. Feb 27, 2010

### 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!