# Linear Algebra : prove geometric multiplicities are the same

In summary, to prove that the geometric multiplicities of the eigenvalues of A and B are the same, it is necessary to show that they have the same characteristic polynomial, which can be done by showing that the determinant of A-tI is equal to the determinant of B-tI.
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

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.

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)

jack_bauer said:
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.

## 1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of vectors, linear transformations, and matrices. It also involves solving systems of linear equations and studying properties of vector spaces.

## 2. What are geometric multiplicities?

Geometric multiplicities refer to the number of linearly independent eigenvectors associated with a specific eigenvalue of a matrix. They represent the number of directions in which a linear transformation stretches or shrinks a vector.

## 3. Why is it important to prove that geometric multiplicities are the same?

Proving that geometric multiplicities are the same for a matrix means that the matrix is diagonalizable. This is an important property in linear algebra as it simplifies calculations and allows for more efficient solutions to problems involving matrices.

## 4. How do you prove that geometric multiplicities are the same?

To prove that geometric multiplicities are the same, you need to show that the eigenvectors associated with a specific eigenvalue are linearly independent. This can be done by showing that the determinant of the matrix formed by the eigenvectors is non-zero.

## 5. Can geometric multiplicities be different for different eigenvalues?

Yes, geometric multiplicities can be different for different eigenvalues. This means that a matrix may not be diagonalizable, as there are not enough linearly independent eigenvectors to span the entire vector space. However, if the algebraic and geometric multiplicities are the same for all eigenvalues, then the matrix is diagonalizable.

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