So if there exists a P such that is diagonal, then A is diagonalizable. where are eigenvectors of A. So, A is diagonalizable if it has 3 distinct eigenvalues.
No. It is diagonalizable iff it has three linearly independent eigenvectors.
A is orthogonally diagonalizable if there exists an orthonormal set of 3 eigenvectirs if A. Also, it is orthogonally diagonalizable if there exists P such that is diagonal.
It should be " is orthog. diag. if there exists orthogonal P such that..."
So for (a), I need to show if there is a matrix P that diagonalizes A. First finding the eigenvalues of A:
This should be , and then we can use the theorem that says that a matrix is diagonalizable iff its
minima polynomial is a product of different linear factors, so it must be that has a two different roots in the definition field of the vector space we're
talking about, and since we're talking , apparently, of the complex field, then we must have that none of these two different roots equals , which
already is a root of the first factor...
Finally, remember (or prove) that a matrix orthogonally diagonalizable iff it is symmetric, and in our case that happens when...
And I'm stuck here, because I don't know the values of r,p,q. Can anyone help?