Diagonalization of Matrix I have a problem that says to find the values of k for which the matrix $A$ is NOT diagonalizable over $\mathbb{C}$. I know that I need to find the zeros for the characteristic polynomial and check for which values their geometric multiplicities equal 1. I'm really just stuck on the calculation.
The matrix $$A=\left(\begin{matrix}0&1&0\\0&0&1\\0&k-6&7-k\end{matrix}\right)$$
I'm getting the characteristic polynomial using the determinant of the characteristic equation as $$-x^3+(7-k)x^2+(k-6)x$$
So I'm just unsure of what the calculations should look like.
 A: There might be a better way, but note that $-x^3+(7-k)x^2+(k-6)x=0$ has solutions
$$\{x=0,x=1,x=6-k\}$$
Since eigenspaces of algebraic multiplicity one have geometric multiplicity one, the only possible cases where an eigenspace could potentially be deficient is when $$6-k=0\text{ or }1.$$
Then you can just check by hand whether the nullspaces of $A-(6-k)I$ are deficient for those two cases. Indeed, in both cases, $A$ turns out to be not diagonalizable. In Mathematica:
m[k_] := ( {
    {0, 1, 0},
    {0, 0, 1},
    {0, k - 6, 7 - k}
   } );
Eigenvectors[m[5]]
Eigenvectors[m[6]]

which yields
{{1, 1, 1}, {0, 0, 0}, {1, 0, 0}}
{{1, 1, 1}, {1, 0, 0}, {0, 0, 0}}

and the presence of {0,0,0} in both results indicates that a diagonalization does not exist.
A: by solving your charateristics equation one root is x=0(expect this)  there may be either two real and different roots or two equal roots.
case 1:  for two different roots as well as x=0 indicate three different roots
           hence the matrix is always digonalisable.
case 2:  if other two roots are equal.
          matrix may be digonalisable or may not be.
        let us do-
             for k=5;
           roots-  0,1,1
for 1 eigenvector is-
matrix (A-I) has rank 2. so number of linearly independent vector is (3-2=1)
so we have only 2 eigenvectors from 0 and
                               from 1.
so matrix is not digonalisable for k=5
