Diagonalizing a matrix given the eigenvalues

[Sunwoo Bae]

Homework Statement

Diagonalize the matrix, given that its eigenvalues are λ= 2,5

Relevant Equations

For matrix A,
D=P-^1AP
A= PDP^-1

The following matrix is given.
Since the diagonal matrix can be written as C= PDP^-1, I need to determine P, D, and P^-1.
The answer sheet reads that the diagonal matrix D is as follows:

I understand that a diagonal matrix contains the eigenvalues in its diagonal orientation and that there must be three eigenvalues in the case of this problem as the number of eigenvalues equals the dimension of the matrix (3).
However, how do you determine rather 2 or 5 gets a multiplicity of 2?

Thank you!

 

[Orodruin]

What is the characteristic polynomial of C?

 

[Steve4Physics]

In case it is not clear why you are told the eigenvalues (2, 5), it is because the person setting the question is very kind and doesn’t want you to waste your time solving cubic equations!

As @Orodruin says, start by finding the characteristic polynomial (which is a cubic).

From the eigenvalues, you know the characteristic polynomial must be either (λ-2)²(λ-5)=0 or (λ-5)²(λ-2)=0. And it’s easy to choose.

 

[Gaussian97]

Since you know what the eigenvalues are, and you know that the matrix can be diagonalized. Then you simply need to compute the eigenvectors, then either 2 or 5 will have 2 eigenvectors while the other will have only one. Then that gives you which one has multiplicity 2.

 

[Mark44]

Since you know what the eigenvalues are, and you know that the matrix can be diagonalized.

That's not necessarily true. If there are repeated eigenvalues, as in this problem, whether a repeated eigenvalue has more than one eigenvector depends on whether the eigenvalue multiplicity is algebraic multiplicity (the number of times the eigenvalue appears in the characteristic equation) or geometric multiplicity (the dimension of the eigenspace associated with the eigenvalue).

If the geometric multiplicity in this problem is only 1, then the matrix can't be diagonalized, at least by the usual means. (I vaguely remember that there is some technique for matrices with deficient eigenvectors, but don't recall the details off the top of my head.)

 

[Gaussian97]

That's not necessarily true. If there are repeated eigenvalues, as in this problem, whether a repeated eigenvalue has more than one eigenvector depends on whether the eigenvalue multiplicity is algebraic multiplicity (the number of times the eigenvalue appears in the characteristic equation) or geometric multiplicity (the dimension of the eigenspace associated with the eigenvalue).

If the geometric multiplicity in this problem is only 1, then the matrix can't be diagonalized, at least by the usual means. (I vaguely remember that there is some technique for matrices with deficient eigenvectors, but don't recall the details off the top of my head.)

That's why I said, "Since you know what the eigenvalues are, and you know that the matrix can be diagonalized". If the matrix can be diagonalized, then its geometrical multiplicity must be equal to the algebraic multiplicity.

 

[Mark44]

That's why I said, "Since you know what the eigenvalues are, and you know that the matrix can be diagonalized".

Just wanted to clarify that for someone who didn't have access to the answer sheet.

 

[Gaussian97]

Just wanted to clarify that for someone who didn't have access to the answer sheet.

Yes, sure enough, the different multiplicities are a useful thing to keep in mind. But just to clarify, there's no need to access the answer sheet.
Even if they haven't yet proved that all symmetric matrices are diagonalizable and that the statement asks you to diagonalize the matrix and therefore it must be diagonalizable. There's no need of computing the algebraic multiplicity for this problem. Either the geometric multiplicities sum to 2, in which case the matrix is not diagonalizable or they sum to 3, in which case you know which eigenvalue has multiplicity 2.

 

[vela]

If the geometric multiplicity in this problem is only 1, then the matrix can't be diagonalized, at least by the usual means. (I vaguely remember that there is some technique for matrices with deficient eigenvectors, but don't recall the details off the top of my head.)

You can use generalized eigenvectors to put the matrix into Jordan canonical form in this case.

But just to clarify, there's no need to access the answer sheet.

That's of course assuming the person who wrote the problem didn't make a mistake. It's a reasonable assumption to start with, but if you were to find each eigenvalue produced only one eigenvector, then @Mark44's note can save you from beating your head against the wall trying to figure out how diagonalize an undiagonalizable matrix.