Real Eigenvalues and 3 Orthogonal Eigenvectors for Matrix (c,d)

[robierob12]

Homework Statement

For which real numbers c and d does the matrix have real eigenvalues and three orthogonal eigenvectors?

120
2dc
053

Homework Equations

im having trouble getting started on this one.

Ive tried using solving for the eigenvalues pretending that c and d are constants but that doesn't seem to help any. Can anyone nudge me in the correct direction.

The Attempt at a Solution

 

[Dick]

If the matrix has real eigenvalues and three orthogonal eigenvectors, then the corresponding linear transformation is self-adjoint. What does that mean in terms of its matrix representation?

 

[robierob12]

so... would I find the co-factor matrix of

123
2dc
053

take the transpose to find the adj.

which I did actually.

new values aprear in the d c spots... it is its own adjoint so those would be the fill in values?

I think I am off track

 

[Dick]

Not the adjugate, the adjoint. The conjugate transpose. You are making this much harder than it should be.

 

[robierob12]

I don't know why this one is so hard for me...
does this have to do with the real spectral theorem?

 

[Dick]

It's the converse of the spectral theorem. The spectral theorem says that if an operator has a certain property then its eigenspace has a certain property. This problem says if the eigenspace has a certain property then the operator has a certain property. If you write the linear transformation corresponding to the matrix as M, then self adjoint means (Mx).y=x.(My). Can you show that's true in this case? What might that have to do with a certain matrix being hermitian?

 

[robierob12]

I don't think that I've seen the operations the way that showing them...
I do see something now though.

Wont the matrix

120
2dc
053


have to be

120
2d5
053

It will have to symetric?

I don't see how to find d besides by trying different d until the found eigenvectors are orthogonal. Any real number choice for d should give real eigenvalues by the spectral theorem?

 

[Dick]

Correct. Any real choice for d will work.

 

[robierob12]

That easy?

120
205
053

120
215
053

120
225
053

120
235
053

will all have orthogonal eigenvectors?

I'll try a few to see if it works.

 

[Dick]

You don't trust the spectral theorem? That's healthy skepticism! Check away.

 

[robierob12]

One more question... is it possible to choose d so that one of eigenvalues is repeated makeing it so there are not actually three othogonal eigenvectors. Because I want to say right now that any choice of d will work.

 

[robierob12]

I appreciate the help on this one... thanks'a bunch

 

[Dick]

A repeated eigenvalue does not mean you don't have three orthogonal eigenvectors. The zero matrix has them. Take (1,0,0), (0,1,0) and (0,0,1). They are orthogonal and all have eigenvalue 0. You're welcome!

 

[robierob12]

nice...

even if it wasnt so, gram-schmidt could make them into orthonormal set I suppose.

 

[Dick]

Yes. The problem comes with matrices like [[1,1],[0,1]]. 1 is a double eigenvalue - but there is only one linearly independent eigenvector.