Geometric multiplicity of the largest eigenvalue

Question

Let $$ A= \begin{bmatrix} a & 2f & 0 \\ 2f & b & 3f \\ 0 & 3f & c \end{bmatrix}, $$ where $a$, $b$, $c$, $f$ are real numbers and $f\neq 0$. Find the geometric multiplicity of the largest eigenvalue of $A$.

I don't think I have to use the characteristic equation. Or do I?

Answer 1

We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)

Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + \sqrt{13}$, $λ_2 = 1 - \sqrt{13}$ , $λ_3=1$.

More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= \begin{bmatrix} 0 & 2 & 0 \\ 2 & 0 & 3 \\ 0 & 3 & 0 \end{bmatrix}$)

you have $λ_1 = n+f\sqrt{13}$, $λ_2 = n -f\sqrt{13}$ , $λ_3=n$.

Evidently at least for some pattern of values multiplicity is just $1$.

Answer 2

let $\lambda $ be an eigen value then $$A - \lambda I = \begin{bmatrix} a-\lambda & 2f & 0 \\ 2f & b-\lambda & 3f \\ 0 & 3f & c-\lambda \\ \end{bmatrix} $$ then $ Rank(A-\lambda I )$ = 2 because

Minor wrt $a_{31}$ = $\begin{vmatrix} 2f & 0 \\ b-\lambda & 3f \\ \end{vmatrix} = 6f^{2} \ne 0$ as $ f \ne 0$

So geometric multiplicity of $ \lambda = 3 - Rank(A-\lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ \lambda$.

Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.