Let $a\in K$. Consider the matrix $$ M = \begin{bmatrix} 1-2a & -a & 0 & 0 \\ 4a & 1+2a & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \in M_{4\times4}(K) $$
a) Deduce the eigenvalues of $M$ from the characteristic polynomial of $M$.
b) Compute the minimal polynomial of $M$.
a) Using $\det(x\operatorname{Id}-M)$, I got a fairly long computation so I shall post only the final lines:
$$(x^2-2x-8a^2+1)(x^2-3x+2)=0.$$
If we develop, we get: $x^4-5x^3+9x^2-7x-8a^2x^2+24a^2x-16a^2+2$, and the real zeros are either $\pm1,2$ after we check, we get $1,2$.
Therefore, the eigenvalues are $x_1=1$ and $x_2=2$.
Regarding b): is minpoly of $M$ in this case the same as the charpoly of $M$? I do not know how to compute it with both $x$ and $a$ in the equations.
c) Prove $M$ is diagonalizable if and only if $a=0$.
First, check this property: $$ A = \begin{bmatrix} A_{11} & A_{12} \\O & A_{22}\end{bmatrix},\\ B = \begin{bmatrix} B_{11} & B_{12} \\O & B_{22}\end{bmatrix},\\ AB = \begin{bmatrix} A_{11}B_{11} & A_{11}B_{12}+A_{12}B_{22} \\O & A_{22}B_{22}\end{bmatrix},\\\\ det(A) = det(A_{11})\,det(A_{22}) $$ when $A_{11}$, $B_{11}$ are $m \times m$, $A_{12}$, $B_{12}$ are $m \times n$, $A_{22}$,$B_{22}$ are $n \times n$. (It can be proved by dirty calculations)
In a), your characteristic polynomial is actually wrong, and you don't need to develop.
You can get and solve $(x^2-2x+1)(x^2-3x+2)=0 : \\(x-1)^3(x-2)=0 \Leftrightarrow x=1 \: or\: x=2$
Now b) is clear : A's minimal polynomial should be a form of $(x-1)^i (x-2)$ for some $i\geq1$, and you can find out $i = \cases{1 \:\:\: {if} \:\:\:a=0 \\2 \:\:\:otherwise}$ using property above.
c) is trivial : $A$ is diagonalizable iff its minimal polynomial's roots are those of its characteristic polynomial with multiplicity 1.