Let $K$ be a field s.t. $charac(K)\not= 2$. Let $A\in\mathcal{M}_n(K)$ be such that $rank(A)\geq n/2$. Can one find a matrix $B$ such that $B$ is similar to $A$ and $A+B$ is invertible ?

  • 1
    $\begingroup$ I want to assume that $A$ is in Jordan canonical form with all of the zero eigenvalues grouped at the lower/right part of the main diagonal, and then let $U$ be the "backwards-identity" matrix ($ij$th entry is 0 unless $i+j=n+1$, in which case it is 1), so that conjugation by $U$ is equivalent to rotating the matrix 180${}^\circ$, and set $B=U^{-1}AU$. Morally speaking, this wants to be a rank-$n$ matrix. Obstacles: one might accidentally have two eigenvalues add to 0 on the main diagonal; the resulting matrix has nonzero entries just above and below the main diagonal - still invertible? $\endgroup$ – Greg Martin Aug 4 '12 at 2:11
  • $\begingroup$ Greg, what is JNF when K is not algebraically closed? $\endgroup$ – Yemon Choi Aug 4 '12 at 3:55
  • $\begingroup$ Greg, surely a modification of this works, if instead of insisting on rotating the matrix 180 degrees, you allow yourself a more general permutation of the Jordan blocks. This should take care of the algebraically closed case, and I don't see how to do the general case. $\endgroup$ – Aaron Tikuisis Aug 4 '12 at 18:10

Initially my assumption was that the question was primarily about finite fields $K$, but after reading the comments it appears to me that there is some confusion about the infinite case as well. So I'll prove the claim for $|K|=\infty$. I would guess the assertion is also true if $K$ is finite, but I don't know how to show it.

If one assumes $K$ to be algebraically closed, then it's very easy to see that one can find an appropriate $B$: Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ (eigenvalues with multiplicity $m$ are supposed to appear $m$ times in that list), then for any permutation $\sigma\in S_n$ we can conjugate $A$ to a matrix of the form $$ \textrm{diag}(\lambda_{\sigma(1)}, \ldots, \lambda_{\sigma(n)}) + N $$ where $N$ is a strictly upper triangular matrix. (This should be clear but I'll give a short explantion anyhow: View $K^n$ as a $K[x]$ module where $x$ acts as $A$. Then finding a basis that conjugates $A$ into the form given above corresponds to finding a filtration $0= V_0 < V_1 <\ldots< V_n= K^n$ of the $K[x]$-module $K^n$ such that the quotient $V_i/V_{i-1}$ is isomorphic to the simple $K[x]$-module $K[x]/(x-\lambda_{\sigma(i)})$. But by the structure theorem on f. g. modules over PID's, $K^n$ is a direct sum of modules each having just one isomorphism type of simple modules as composition factors. So choosing such a filtration is no problem: in each step just pick a maximal submodule in the desired isotypic direct summand and leave the other summands unchanged.)

We may assume that $A$ is of the above shape with $\sigma=\textrm{id}$. So just choose a permutation $\sigma$ such that $\lambda_i+\lambda_{\sigma(i)} \neq 0$ for all $i$ (this can be done: just choose $\sigma$ to be the transposition that swaps each $\lambda_i$ which is zero with some $\lambda_i$ which is non-zero and fixes the rest; the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues; also $char(K)\neq 2$ is essential: it guarantees $\lambda_i\+\lambda_i\neq 0$ whenever $\lambda_i\neq 0$). Then $A+B$ can becomes $$ \textrm{diag}(\lambda_1+\lambda_{\sigma(1)}, \ldots, \lambda_n+\lambda_{\sigma(n)}) + \textrm{something strictly upper triangular} $$ and you are done.

As for the case of a non-algebraically closed (but still infinite) $K$: Note that the algebraically closed case implies that the rational function $$ det(A+TAT^{-1})=\frac{det(AT+TA)}{det(T)} \in K(T_{ij}) $$ is non-zero. But $GL(n,K)$ is Zariski dense in $GL(n,\bar K)$ (see here), and so the above rational funktion must be non-vanishing on $GL(n,K)$, which shows there is a $T\in GL(n,K)$ such that $det(A+TAT^{-1})\neq 0$.

| cite | improve this answer | |
  • $\begingroup$ Thanks Florian. Necessarily $char(K)\not=2$. Else, consider the matrix $A=diag(I_p,0_{n-p})$ where $p>\dfrac{n}{2}$. Then there are no matrices $X$ s.t. $AX+XA$ is invertible. I have a question about the Zariski-density. In our instance, the group is reductive but is it connected ? It remains to show the result when $K$ is a finite field. Does there exist an obvious obstruction ? $\endgroup$ – loup blanc Aug 7 '12 at 4:26
  • $\begingroup$ That $GL(n,\overline{K})$ is connected follows by identifying it with a closed subvariety of $\mathbb{A}^{n^2 + 1}$ rather than as an open subvariety of $\mathbb{A}^{n^2}$ since as a closed subvariety, it can be seen as the one defined by the vanishing of the ideal generated by $\textrm{det}T - 1$ where $T$ is the "extra" variable in the polynomial ring. That this is a prime ideal just means that it is an irreducible polynomial, which is an easy exercise (it has almost nothing to do with how the determinant looks). $\endgroup$ – Tobias Kildetoft Aug 7 '12 at 6:33
  • 1
    $\begingroup$ Hi Florian, I reread your proof and unfortunately it is false. Indeed your assertion: "the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues" is not true. In particular, your proof does not work if $A$ is a nilpotent matrix. The Greg's proof does not work any more; choose, for instance, A=JordanBlockMatrix([[0, 4], [0, 2], [0, 3]]) (according to the maple notation). In fact, the asked result is true for the previous matrix. To the moderator (S. Carnahan): I'd want to edit a question (to Tobias) in this file. I don't know how to do. Thanks. $\endgroup$ – loup blanc Aug 15 '12 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.