How can I find the values a, b, c, d, e, f so that the following matrix becomes diagonalizable?
A = $\left( \begin{array}{ccc} 0 & a & b & c\\ 0 & 0 & d & e\\ 0 & 0 & 1 & f\\ 0 & 0 & 0 & 2 \end{array} \right) $
I know I need to have a basis of $\mathbb{R}^{4}$ formed by eigenvectors of A. I find the characteristic polynomial using the determinant. Thus my $\lambda^2(\lambda - 1)(\lambda -2) $.
Therefore $\lambda \in \{0, 1, 2\}$. I found the respective matrix for each one, and I understand I need to get a dimension of 2 for 0 because its multiplicity is 2, while it has to be 1 for $\lambda$ 1 and 2.
$A - I0$ = $\left( \begin{array}{ccc} 0 & a & b & c\\ 0 & 0 & d & e\\ 0 & 0 & 1 & f\\ 0 & 0 & 0 & 2 \end{array} \right) $
$A - I1$ = $\left( \begin{array}{ccc} -1 & a & b & c\\ 0 & -1 & d & e\\ 0 & 0 & 0 & f\\ 0 & 0 & 0 & 1 \end{array} \right) $
$A - I2$ = $\left( \begin{array}{ccc} -2 & a & b & c\\ 0 & -2 & d & e\\ 0 & 0 & -1 & f\\ 0 & 0 & 0 & 0 \end{array} \right) $
I can't figure it out, can someone hint me?
Your matrix $A-0I$ can be row-reduced to $$\pmatrix{0&a&b&c\cr0&0&1&f\cr0&0&0&2\cr0&0&0&0\cr}\ .$$ Now $A$ is diagonalisable if and only if this matrix has $2$ non-leading columns, and that occurs for one specific value of $a$. The values of $b,c,d,e,f$ are irrelevant.
See if you can finish the problem from here.