Matrices Questions - Symmetric, Minimal Polynomial, Bilinear Form, Jordan

[Ted123]

Is any symmetric matrix diagonalisable with an orthogonal change of basis?

Does the minimal polynomial of any real matrix split into distinct linear factors?

Is a real inner product an example of a bilinear form?

Could 2 complex matrices which are similar have different Jordan normal forms?

 

[micromass]

So, what did you try already??

 

[Ted123]

So, what did you try already??

I just want to know 'yes' or 'no' out of interest

 

[micromass]

For the first one, let A be symmetric. Let u be an eigenvector with eigenvalue \lambda and let w be an eigenvector with eigenvalue \mu. Can you calculate

<Au,w>

in several ways?

 

[Ted123]

I think YES, NO, YES, NO respectively. Is that correct?

 

[micromass]

I think YES, NO, YES, NO respectively. Is that correct?

Why do you think that?

 

[Ted123]

I think the 4th question is worded slightly ambiguously; it's probably safe to assume that it means "different up to reordering of the Jordan blocks" (in which case you're correct). But if it doesn't, then \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix} and \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} would be an example of similar matrices with different JNFs.

I know the first 3 are right - yes, no, yes but for the 4th one, do you think that the answer to the last one is no?

 

[micromass]

I think the 4th question is worded slightly ambiguously; it's probably safe to assume that it means "different up to reordering of the Jordan blocks" (in which case you're correct). But if it doesn't, then \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix} and \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} would be an example of similar matrices with different JNFs.

I know the first 3 are right - yes, no, yes but for the 4th one, do you think that the answer to the last one is no?

If you mean "different up to reordering of the Jordan blocks" then you are correct. Similar matrices have the same Jordan canonical form then.

This can be seen by putting A=S^{-1}BS. If C is the Jordan basis, then

CAC^{-1}

is the Jordan canonical form of A. Now, can you find easily the Jordan canonical form of B (just plug the expressions into each other).