Calculate the matrix from minimal polynomial and eigenspaces I need to find a matrix $A \in M(6,\mathbb C)$ that satisfies following:


*

*$e_1+e_2+e_3\in \ker(L_A-3\cdot \operatorname{id})^3 \setminus \ker(L_A-3*id)^2$

*$\operatorname{span}(e_1+e_4,e_5+e_6)= \operatorname{Eig}(L_A, 2)$

*$\mu_A=(t-3)^3(t-2)^2$


I think that $A$ is similar to the matrix in Jordan form
$$
        \begin{pmatrix}
        2 & 0 & 0&0&0&0 \\
        0&2&1&0&0&0 \\
        0&0&2&0&0&0\\
        0&0&0&3&1&0\\
        0&0&0&0&3&1\\ 
        0&0&0&0&0&3\\
        \end{pmatrix}
$$
But how do I find a basis to transform this matrix to $A$ without knowing what the matrix looks like? I only have 3 vectors, how do I compute the rest?
 A: So before we start, note that the problem doesn't say find the matrix, so the answer may not be unique.
We have our Jordan Canonical form so in order to get a matrix so that those qualities are true, we can craft a similarity transformation that'll get us to such a matrix as is specified by the problem.
The first thing to note is that for a matrix $A$, $A = PBP^{-1} $, with $P$ being a representation of the basis. This property clearly still applies for $B$ in Jordan canonical form, so we can apply this here as well.
So, what we need is an invertible matrix such that the first two columns are eigenvectors $(e_1+e_4),(e_5+e_6) $, (or vice versa since the conditions didn't specify), and the sixth column is $(e_1+e_2+e_3)$. So, basically, we have:
$$ P =
\begin{bmatrix}
1 & 0 & ... & 1 \\
0 & 0 & ... & 1 \\
0 & 0 & ... & 1 \\
1 & 0 & ... & 0 \\
0 & 1 & ... & 0 \\
0 & 1 & ... & 0 \\
\end{bmatrix} $$
And now it's basically a fill in the blanks problem so that you make an inveritble $P$. Because there are only three vectors, it has quite a broad solution set. After you have your $P$, once you calculate $PJP^{-1} $, you'll have your new matrix $A$.
