So if there exists a P such that $\displaystyle P^{-1}AP$ is diagonal, then A is

**diagonalizable**. $\displaystyle P=[P_1 P_2 P_3]$ where $\displaystyle P_1,P_2,P_3$ 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 $\displaystyle P^TAP$ 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:

$\displaystyle det(\lambda I -A)=0$

$\displaystyle \lambda^3-q \lambda-rp \lambda -rpq = 0$

This should be $\displaystyle x^3-qx^2-prx+pqr=(x-q)(x^2-pr)$ , 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 $\displaystyle x^2-pr$ 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 $\displaystyle q$ , 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... Tonio
And I'm stuck here, because I don't know the values of r,p,q. Can anyone help?