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Given a $12 \times 12$ matrix with only one eigenvalue $\lambda = 7$, if we're given that
$nullity (A-7I) = 4, nullity (A-7I)^2 = 7, nullity (A-7I)^3 = 10, nullity (A-7I)^4 = 12$ how would I find a Jordan form of $A$? I don't understand the solution at all, they say the jordan form will consist of

$4$ blocks, with sizes $4,4,3,1$. How would you find this?

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For an eigenvalue $\lambda$, the nullity of $A-\lambda I$ is the number of Jordan blocks.

The nullity of $(A-\lambda I)^2$ is the number of Jordan blocks plus the number of Jordan blocks of size at least $2$.

The nullity of $(A-\lambda I)^3$ is the number of Jordan blocks plus the number of Jordan blocks of size at least $2$ plus the number of Jordan blocks of size at least $3$. Etc.

In your example you have $4$ Jordan blocks. Then $7$ is $4$ plus the number of Jordan blocks of size $\ge2$. So there are three of those and so exactly one of size $1$.

Then $10$ is $4+3$ plus the number of Jordan blocks of size $\ge3$. So all three Jordan blocks of size $\ge2$ actually are of size $\ge3$. There are no size $2$ Jordan blocks. Etc.

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  • $\begingroup$ When you say "number of Jordan blocks plus ...", are you referring to these jordan blocks as size 1? Edit: Nvm, i think I misread $\endgroup$ – Natash1 May 21 '17 at 8:03
  • $\begingroup$ In your last line: Should it read $10$ is $4 + 3$ plus the number of Jordan blocks of size $\geq 3$? $\endgroup$ – Natash1 May 21 '17 at 8:10

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