Gram-Shmidt getting PtAP is diagonal

[nautolian]

Homework Statement

in light of the gram-schmidt orthogonalization process, if A: Rⁿ -> Rⁿ and we have a basis of Rⁿ of eigenvectors of A, can't we just orthonormalize them and get a matrix P such that P^-1=P^T and thus P^TAP is diagonal?

Homework Equations

The Attempt at a Solution

I believe the answer to this is yes, but I'm not sure how to prove this. Any ideas would be appreciated, thanks!

 

[Dick]

Homework Statement

in light of the gram-schmidt orthogonalization process, if A: Rⁿ -> Rⁿ and we have a basis of Rⁿ of eigenvectors of A, can't we just orthonormalize them and get a matrix P such that P^-1=P^T and thus P^TAP is diagonal?

Homework Equations

The Attempt at a Solution

I believe the answer to this is yes, but I'm not sure how to prove this. Any ideas would be appreciated, thanks!

No, not true. If your eigenvectors have different eigenvalues and aren't already orthogonal, you can apply gram-schmidt but the results won't be eigenvectors.

 

[nautolian]

Why aren't those eigenvectors? Because you've changed the vectors too much?

 

[Dick]

Why aren't those eigenvectors? Because you've changed the vectors too much?

Sure, you changed them too much. Why would you think they would be eigenvectors? A linear combination of two eigenvectors is not necessarily an eigenvector.

 

[nautolian]

I thought it might be an eigenvector because i was under the impression for some reason that the orthonormalization would no necessarily mean using the Gram-Shmidt process, but rather just normalizing it (dividing by sqrt(sum of squares of eigenvector values)). Is this not the case?

 

[Dick]

I thought it might be an eigenvector because i was under the impression for some reason that the orthonormalization would no necessarily mean using the Gram-Shmidt process, but rather just normalizing it (dividing by sqrt(sum of squares of eigenvector values)). Is this not the case?

That's pretty confused. Look, take the matrix [[1,1],[0,2]]. The eigenvalues are 1 and 2. Find the eigenvectors. It's easy. Now you can normalize them, but you can't force them to be orthogonal.