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I'm working on the problem:

Suppose $V$ is a $123$ dimensional vector space:

$i)$ How many linear maps $T:V \rightarrow V $ are diagonalizable and have $T^2 =0$?

$ii)$ How many linear maps $T:V \rightarrow V $ are not diagonalizable and have $T^2 =0$?

First of all, I reasoned that for part $i)$ we know that there are no diagonalizable nilpotents, hence the answer to part $i)$ is $0$. As for part $ii)$, I know that $T$ must have a Jordan normal form (I don't know to how type up matrices on this) that has Jordan blocks with 1's on the superdiagonal and everything else zero.How can I count how many such arrangements there are?

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1  
Try first the case were $\dim V$ is small, say up to 6 or 7 and figure out a way to generalize. –  Andrea Mori May 8 '13 at 12:03
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Is $0$ diagonalizable? –  wj32 May 8 '13 at 12:04

2 Answers 2

Hints:

1) There is one linear nilpotent map that is diagonalizable...

2) You need all the maps for which their characteristic polynomial is $\,x^{123}\,$ and his minimal polynomial is $\,x^2\,$ ...What's the maximal size a Jordan block can have for such a map/matrix? How many such blocks can you have in a $\,123\times 123\,$ matrix?...

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1) I forgot that the zero matrix is the exception. 2) Is the maximal size a Jordan block can have $123X123$? If so, you can only get one such block. –  Mel May 8 '13 at 12:27
    
No, that's not the maximal size of Jordan Block, but rather $\,2\,$...the power to which each eigenvalue appears in the minimal polynomial (i.e., the eigenvalue's algebraic multiplicity in the min. pol.) is the maximal size a J.B. has in the block corresponding to that eigenvalue. –  DonAntonio May 8 '13 at 12:36
    
Ah right. The only eigenvalues of a nilpotent linear map are $0$, so does this imply that the number of different arrangements of the Jordan blocks is 1 (when all the blocks are 2X2, the arrangements are identical) + 123 choose n (summed from n = 1 up to 123. This represents the arrangments of the n 1X1 Jordan blocks in this matrix). –  Mel May 8 '13 at 13:23

i) If $T$ is diagonalizable, it has a basis of eigenvectors. For each eigenvector $v$, $T^2v=\lambda^2v=0$ implies $\lambda=0$, hence $T=0$.

ii) Let $U_1$ be any sub vector space of $V$ with $62\le\dim U_1\le 121$. Let $U_2$ be a complement, i.e. such that $V=U_1\oplus U_2$ and $\phi\colon U_2\to U_1$ any injective linear map. Then $T(u_1,u_2)=(\phi(u_2),0)$ is a linear map $T\colon V\to V$ with $T^2=0$ and (since $T\ne 0$) is not diagonalizable. If the base field is infinite, the number of such $T$ is obviously infinite (and is in fact the same cardinality of the base field). If the base field is finite, you need to invest some combinatirics to find the exact number.

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