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I have the next problem:

Find all the invariant subspaces over $\mathbb{R^4}$ of the following endomorphism given by the matrix A. The matrix A doesn't matter. What it does is the next:

I found the characteristic polynomial: \begin{equation} P_{A}(x)=(x-2)^2(x^2+x+1). \end{equation}

Then, I saw that there was two invariant subspaces (if I am not mistaken): \begin{equation} \ker(f-2id) \qquad \ker(f^2+f+id). \end{equation}

Now, I have to find the dimension and a basis for each subspace, is that right?

Thank you!

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  • $\begingroup$ $\ker(A-2Id)^2$ might be an invariant subspace too, if $\ker(A-2Id)^2\neq \ker(A-2Id)$. $\endgroup$ – Daniel May 20 '14 at 18:49
  • $\begingroup$ Yep, but $\ker(A-2id)$ has dimension 2, so the chain is over, because every subspace from this will be equal to $\ker(A-2id)$. The point is: I hace to find a basis and everything of $\ker(f^2+f+id)$ like in the other subspace? $\endgroup$ – Relure May 20 '14 at 18:54
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Consider the matrices $ M_1 = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & A_{22} \end{array} \right)$ and $ M_2 = \left( \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & A_{22} \end{array} \right)$ where $P_{A_{22}}(x) = x^2 + x + 1$, in both the cases i.e. $P_{M_1} = P_{M_1} = P_A$. But the invariant spaces for matrix $M_1$ are $\langle e_1 \rangle, \langle e_2 \rangle, \langle e_1,e_2 \rangle$ and $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & A_{22} \end{array} \right)$ invariant space. But in the case of $M_2$, $\langle e_2 \rangle$ is not an invariant space where $ e_1 = \left( \begin{array}{c} 1 & 0 & \dots \end{array} \right) ^ T$ . Hence there exists a maximum number of invariant spaces and minimum number of invariant spaces possible for matrices with this particular characteristic polynomial. As for the basis for each subspace, if only the Jordan form is considered, then for the above mentioned Jordan form (with $A_{22}$ expanded of course!) the bases would be as described before.

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