Counting the number of linear maps

Question

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?

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 08 '13 at 12:03

score 2 · Answer 1 · answered May 08 '13 at 12:11

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?...

answered May 08 '13 at 12:11

DonAntonio

214,715

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 08 '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 08 '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 08 '13 at 13:23

score 0 · Answer 2 · answered May 08 '13 at 14:13

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.

Counting the number of linear maps

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