Geometric multiplicity of the largest eigenvalue 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?
 A: 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.
A: 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  $1$.
