TITLE: Geometric multiplicity of the largest eigenvalue
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.