Showing whether a matrix is diagonalizable over $C$?

Question

I have the following matrix \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} and I want to determine whether it is diagonalizable over the field of complex numbers.

If we calculate the characteristic polynomial, we get $p(x) = -x^3+11x^2+4x-1$. Unforunately, this has no rational roots, and I don't know of any way to determine the complex roots of this polynomial as is, since it is a cubic.

My other thought was that, this is a symmetric matrix. Are there any substantial results about whether or not a symmetric matrix is diagonalizable over the field of complex numbers?

Answer 1

A real symmetric matrix is diagonalizable over the reals, which trivially implies that it is diagonalizable over the complex numbers.

In general, for complex matrices, the corresponding result is that a Hermitian matrix is diagonalizable (all the eigenvalues will also be real, which is a nice bonus). "Hermitian" means it's equal to its own conjugate transpose.

Answer 2

Even if you can’t find the roots explicitly, knowing that the roots are distinct would be enough to conclude that the matrix is diagonalizable (if a! eigenvalue have algebraic multiplicity 1 then it also must have geometric multiplicity 1).

For a cubic it’s easy to tell that it has distinct roots: any irrational roots have to come with at least one conjugate, so a repeated irrational root can only occur with degree 4 or higher. So the fact that there are no rational roots already implied the roots are distinct.

More generally, we can always tell whether a polynomial $f(x)$ has repeated roots without solving for those roots: just by taking the polynomial GCD of $f$ and the derivative $f’$. The set of repeated roots is equal to the roots of the GCD (in particular, if there are no repeated roots then the polynomial GCD will be $1$).

Of course, the symmetry of this matrix is a much quicker way to see this is diagonalizable. I just wanted to point out that your approach could still work.

Answer 3

Since the given matrix is symmetric hence it is diagonalizable. Also because it has three distinct latent roots $\lambda= 11.3448, -0.515729, 0.170915$ coming from $|A-\lambda I|=0$. These latent roots are the eigenvalues of $A$ and the diagonal element of the diagonalized matrix.

Answer 4

Two facts you need to know about the symmetric matrix in general:

Any symmetric matrix

         1) has only real eigenvalues;
         2) is always diagonalizable;

The proof is:

1) Let λ ∈ C be an eigenvalue of the symmetric matrix A. Then Av = λv, $v ̸\neq 0$, and

$v^∗Av = λv^∗v$, $v^∗ = \bar{v}^T.$$

But since A is symmetric

$λv^∗v = v^∗Av = (v^∗Av)^∗ = \bar{λ}v^∗v.$

Therefore, λ must be equal to $\bar{λ}$ !

Also, you don't need to find complex root of this characteristic polynomial because it doesn't have one (all roots are real).

2) If the symmetric matrix A is not diagonalizable then it must have generalized eigenvalues of order 2 or higher. That is, for some repeated eigenvalue $λ_i$ there exists $v ̸= 0$ such that

$(A − λ_iI)2v = 0$, $\;\;(A − λ_iI)v \neq 0$

But note that $0 = v^∗(A − λ_iI)2v = v^∗(A − λ_iI)(A − λ_iI) ̸\neq0$,

which is a contradiction. Therefore, as there exist no generalized eigenvectors of order 2 or higher, A must be diagonalizable.

Therefore by this result, the matrix in the question is diagonalizable.