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A $5\times 5$ matrix $A$ satisfies the equation $(A-2I)^3(A+2I)^2=0$. Assuming there are at least $2$ linearly independent eigenvectors for $2$, write all possible Jordan canonical forms.

My Question:

The eigenvalues of $A$ are $2$ and $-2$. If we know the number of linearly independent eigenvectors for $2$, can we determine the number of linearly independent eigenvectors for $-2$?

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marked as duplicate by Jyrki Lahtonen, Claude Leibovici, max_zorn, Jose Arnaldo Bebita-Dris, Arnaud D. Aug 16 '18 at 10:09

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  • $\begingroup$ the eigenvectors of $2$ and $-2$ are linearly independent. i don't know what else you can conclude. $\endgroup$ – abel Jan 7 '15 at 2:46
  • $\begingroup$ Can't you use the Cayley Hamilton theorem to find out the multiplicity of the eigenvalues? mathworld.wolfram.com/Cayley-HamiltonTheorem.html $\endgroup$ – Mary Jan 7 '15 at 2:46
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No: If we denote by $J_k(\lambda)$ the Jordan block of eigenvalue $\lambda$ and size $k \times k$, then both $$ J_1(2) \oplus J_1(2) \oplus J_1(2) \oplus J_1(-2) \oplus J_1(-2) \qquad\text{and} \qquad J_1(2) \oplus J_1(2) \oplus J_1(2) \oplus J_2(-2) $$ satisfy the equation $(A - 2I)^3 (A + 2I)^2 = 0$, and both have three linearly independent eigenvectors of eigenvalue $2$, but the first has two linearly independent eigenvectors of eigenvalue $-2$ and the second has only one.

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    $\begingroup$ why not $J_1(2)+J_2(2)+J_2(-2),J_1(2)+J_2(2)+J_1(-2)+J_1(-2)?$ $\endgroup$ – abel Jan 7 '15 at 2:53
  • $\begingroup$ Ok so the answer must be the union of Travis's and abel's answers. Isn't it? $\endgroup$ – Extremal Jan 7 '15 at 2:55
  • $\begingroup$ No reason to prefer one over the other. Of course, it's sufficient to give a single example to show that it's not enough to know the number of linearly independent eigenvectors of eigenvalue $2$. $\endgroup$ – Travis Jan 7 '15 at 5:07

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