Sorry, I wasn't careful there...

This theorem answers part

**(c)**. If p=r, A is symmetric and hence orthogonally diagonalizable. This is because for a symmetric matrix, the eigenvectors from different eigenspaces are orthogonal.

Now to clarify part

**(a)**. I'm reading your first post but I'm still not sure if I understand it correctly...

A is diagonalizable whenever q is not a root of

from

?

Again, I didn't write that. The following is part of what I wrote in my first post: "... then we must have that none of these two different roots

equals , which already is a root of the first factor..." .This is a necessary condition, not a sufficient one, meaning: if the matrix is diagonalizable **then **it

must be that but **not necessarily** the other way around ...**sigh**

Remember: the characteristic and the minimal pol's of a matrix have the very same irreducible factors, and the matrix is diag. iff its minimal pol. is

a product of different linear factors, and now you __must__ check the different cases:

1) if then the characteristic pol. is , so:

1a) if also then is nilpotent and thus it is diag. iff it is the zero matrix (i.e, iff also ) ;

1b) if then it must be that the minimal pol. is the zero matrix. But you

can easily check this is possible iff

2) if so:

2a) if , then we get that , so that is diagonalizable iff

the zero matrix. Check that in one case this happens but not in the other one (unless I erred. Check this carefully)

2b) if then all the roots are different and thus is diag.

Pay attention to the fact that in the whole point (2) above we must have that is an element of the definition field, so if we work over the rationals this means must be a (rational) square!
& for part

**(b)**, A is diagonalizable, but not orthogonally diagonalizable whenever q is not a root of

and A is not symmetric. Is this correct?