If $N$ is a $k \times k$ elementary nilpotent matrix, i.e. $N^k = 0$ but $N^{k-1} \ne 0$, then show that $N^\top$ is similar to $N$. Now use the Jordan form to prove that every complex $n \times n$ matrix is similar to its transpose.

I have figured out the second part, and am struggling with connecting nilpotency to being similar to the transpose.

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    $\begingroup$ It's customary on this site to include your progress in the question itself. $\endgroup$ – Olivier Bégassat Dec 13 '13 at 8:02
  • $\begingroup$ Which parts of this exercise do you feel confident of answering? What specifically have you tried and gotten "stuck" at? $\endgroup$ – hardmath Dec 13 '13 at 13:54
  • $\begingroup$ I actually just figured out the second part. I am struggling to prove the first statement: "If N is a k×k elementary nilpotent matrix, i.e. Nk=0 but Nk−1≠0, then show that N⊤ is similar to N." I can't seem to figure how the nilpotency factors in to the problem $\endgroup$ – GeorgeShi Dec 13 '13 at 17:58

To show that a complex $k \times k$ elementary nilpotent matrix is similar to its transpose, it suffices to show their Jordan Canonical Forms are equal.

But a complex nilpotent matrix $N$ has only zero complex eigenvalues, hence its Jordan Canonical Form is zero along the diagonal. Were the maximum size of its Jordan Blocks to be $k'\times k'$ for some $k' < k$, then $N^{k'} = 0$, contradicting the elementary nilpotency of $N$. Hence $N$ has exactly one Jordon block (of size $k \times k$) in its Jordan Canonical Form, with zeros along the diagonal.

By the same argument, $N^t$ being elementary nilpotent, its Jordan Canonical Form also has exactly one Jordon Block with zeros along the diagonal. Thus $N$ and $N^t$ are similar, for their Jordon Canonical Forms of $N$ and $N^t$ are the same.


In fact every square matrix is similar to its transpose. This applies in particular to nilpotent matrices.

Added. I'll admit that I guessed that the general proof is not what they wanted you to give for this result, but that rather it was meant that you use the nilpotent case as a step in proving the general case (that any square matrix is similar to its transpose). Yet I think that the direct general proof (which uses the Smith normal form algorithm over $F[X]$) is really the right way to go, as not only it shows that a fine decomposition (according to eigenvalues) is not necessary for this result, it also gives similarity over any field containing the entries of both matrices, rather than over a field in which their characteristic polynomials split.

Anyway, I think the idea was to proceed in the following steps:

  1. If two matrices are similar, than so are their transposes (a simple step, but needed because conjugation and transposition do not commute);

  2. therefore if a matrix $A$ is similar to a matrix $B$ that is similar to $B^\top$, then $A$ is also similar to $A^\top$;

  3. if the characteristic polynomial of $A$ splits (which is always the case over the complex numbers), then the vector space acted on decomposes as the direct sum of its generalised eigenspaces, and $A$ is similar to a block diagonal matrix where every block has a single eigenvalue;

  4. in a block diagonal matrix $A'$, if every block is replaced by another matrix similar to it, then the resulting block diagonal matrix is similar to$~A'$;

  5. the previous steps allow reducing to the case where $A$ only has a single eigenvalue$~\lambda$, in which case $A=\lambda I+N$ with $N$ nilpotent;

  6. one can reduce to the case of nilpotent matrices, since a conjugation that sense $N$ to $N^\top$ will also send $A=\lambda I+N$ to $A^\top=\lambda I+N^\top$;

  7. since nilpotent matrices are similar to their Jordan normal form, which is a block diagonal form composed of nilpotent Jordan block, the argument of point 4. allows further reduction to the case of a single nilpotent Jordan block;

  8. finally a nilpotent Jordan block is conjugate to its transpose by the permutation matrix for the order-reversing permutation (i.e., the identity reflected horizontally or vertically).

That is quite a long reduction, although fairly straightforward to guess, as it follows the best known reduction for arbitrary (complex) matrices, namely to their Jordan normal form. In fact the reduction is longer than necessary, only because I needed to bring nilpotent matrices into the picture due to the way the question is formulated. It would be possible to replace step 3. directly by the Jordan normal form, so that step 5. could reduce to a single Jordan block instead of to $A=\lambda I+N$ with $N$ nilpotent; step$~$8. works perfectly well for any Jordan block (the nilpotent property is not used). Even so I did not manage to completely match the question, which asks you to used the Jordan normal form after doning the nilpotent case, while it seems to be needed in order to do the nilpotent case.

  • $\begingroup$ I am supposed to prove them separately. I think I have found an answer, please let me know if this works: A matrix is nilpotent if and only if its eigenvalues are all 0. Also, matrices have the same eigenvalues as their transposes. Therefore, the transpose of N also has all 0 eigenvalues and is therefore nilpotent as well. So, if we say N^t = P^(-1)NP, then we can raise both sides to the k-th power, giving (N^t)^k = P^(-1)N^kP, and both sides are 0, confirming the similarity of N and N^t $\endgroup$ – GeorgeShi Dec 13 '13 at 18:54
  • $\begingroup$ @GeorgeShi No, what you write is not a proof. You appear to want to prove that all nilpotent matrices of the same size are similar, but this is definitely false. For one thing the zero matrix is nilpotent and it is not similar to any other matrix. $\endgroup$ – Marc van Leeuwen Dec 13 '13 at 19:39
  • $\begingroup$ do you know how i can approach this then? i am a bit stuck. hints or proofs would be much appreciated $\endgroup$ – GeorgeShi Dec 13 '13 at 20:40

You can obtain $N^\top$ by reversing the order of all rows and all columns of $N$. So, ...


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