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I need to find a matrix $A \in M(6,\mathbb C)$ that satisfies following:

  1. $e_1+e_2+e_3\in \ker(L_A-3\cdot \operatorname{id})^3 \setminus \ker(L_A-3*id)^2$
  2. $\operatorname{span}(e_1+e_4,e_5+e_6)= \operatorname{Eig}(L_A, 2)$
  3. $\mu_A=(t-3)^3(t-2)^2$

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?

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    $\begingroup$ Compute the generalized eigenvectors corresponding to the eigenvalues 2 and 3. $\endgroup$ – Mose Wintner May 30 '16 at 20:29
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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 $A$.

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