I'm teaching myself linear algebra and came upon this exercise. It is probably not very hard, but I just have hard time solving it.
$$\text{Which values for constants a, b, c, d, e, f make the matrix diagonalizable?}\\ A= \begin{pmatrix} 1 & a & b & c \\ 0 & 2 & d & e \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$
I have tried wrestling this for many hours now and I need some help. What I have been able to do so far, that I know of is correct:
The characteristic polynomial is $$p(\lambda)=(\lambda-2)^3(\lambda-1)$$ and therefore the eigenvalues and eigenvectors for those are $$\lambda_1=2,\space \lambda_2=1$$
Also the two eigenvectors are $$v_1=[1,0,0,0]^T \text{ and } v_2=[a,1,0,0]^T$$
That is what I have for certain.
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Then there are some other things I've been thinking, but which I'm not sure of:
I assume that in $$D=P^{-1}AP$$ the D would be $$ D= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$
as the eigenvalues should be on the diagonal and the eigenvalue 2 is repeated.
One great point of uncertainty is that I seem to be missing $2$ vectors from the $P$ (so that it would be $4\times 4$ matrix) and I don't seem to find a way to constructing them. I know that the 2 missing vectors should be linearly independent, but only vectors I could think of were vectors $[0,0,1,0]^T$ and $[0,0,0,1]^T$ and that lead to following when $D=P^{-1}AP$ is used:
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -a & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & a & b & c \\ 0 & 2 & d & e \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \begin{pmatrix} 1 & a & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 & 2a & b & c \\ -a & 2-2a^2 & d-ab & e-ac \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$
And if that is correct, all constants have to be $0$, otherwise it is not diagonalizable. But I'm very skeptical that this would be the correct answer. I think I need somehow link the additional two eigenvectors to the eigenvalue $2$, but I don't know how to do that.
I'm very grateful for all help.