Nilpotent matrix over a division algebra

Question

Suppose I have an $n\times n$ nilpotent matrix $A$. If the entries are from any field, then I can show that all eigenvalues are zero and the trace is zero. Indeed, if we consider the algebraic closure of the field then the Jordan normal form $J$ of $A$ must be nilpotent, so all its eigenvalues are zero, which is of course in the original field itself. Also $\operatorname{tr}(J)$ is zero, and using the fact that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, it follows that $A$ (which is similar to $J$) must also have zero trace.

But now I want to consider the case where the entries of $A$ are from a finite dimensional associative division algebra $D$ over a field $K$. If $K$ is algebraically closed then we are back in the case above since the only finite dimensional division algebra over an algebraically closed field is the field itself. But I'm having some difficulty with the case where $K$ is not necessarily algebraically closed - are the above still true?

For simplicity let's assume that $K$ has characteristic $0$ but is not necessarily algebraically closed. The proof above (for a field) does not seem applicable in this case - at least I can't convince myself of it. I can't use the idea of algebraic closure, so I do not know if there exists any eigenvalues in $D$. Also, since commutativity does not in general hold in $D$, I do not know if the trace is invariant under a change of basis.

The difficulty seems to be that I don't know what results continue to hold for a division algebra. Any ideas of a good way to think about this?

Answer 1

In the quaternions, $\begin{pmatrix} i & j \\ -j & i \end{pmatrix}^2 = 0$ and has nonzero trace.

Answer 2

Be sure to see Jyrki Lahtonen's link in the comments.

In short, lack of commutativity makes life hard for definitions of determinants and roots of polynomials.

Still, if you just want to look at eigenvalues this way $$Ax= x\lambda$$ then it is obvious that nilpotent matrices have eigenvalues all $0$. (Even over a noncommutative ring with no nonzero nilpotent elements, this would work. Of course, that is a rather outlandish place to be thinking of eigenvalues...)

Answer 3

I think it is wise to banish the notions of eigenvalue and trace from your mind and vocabulary when doing linear algebra over a division ring (as I said in a comment, this applies also to determinants, characteristic polynomials, homothecies, and anything multilinear); while some aspects of these notions can be salvaged in some or other weak form, those behave so differently than in the commutative setting, that they are best considered something different from the original notion.

This does not leave much of your question standing. However I think that whatever you really wanted to know about nilpotent matrices (or endomorphisms of finite dimensional (say) right modules over $D$) is implied by the fact that these have a Jordan normal form, just as in the commutative case (where it requires no algebraically closed ground field either).

In fact the usual construction of a Jordan basis for a nilpotent endomorphism should work perfectly fine for a finite dimensional right module over a division ring $D$, which I shall call a right $D$-vector space (since the notion much closer to that of vector spaces than to that of modules). As I don't really know what the usual construction of a Jordan basis actually is, I'll give one here.

Let $V$ be a finite dimensional right $D$-vector space, and $\phi\in\operatorname{Hom}_D(V,V)$ a linear endomorphism (written on the left) that is nilpotent. We prove the existence of a basis on which the matrix of $\phi$ is a Jordan normal form, induction on $d=\dim_D(V)$. If $d=0$ we take the empty basis and are done.

Otherwise let $m$ be minimal such that $\phi^m=0$; then $m>0$ and there exists a vector $x\in V$ with $\phi^{m-1}(x)\neq0$. Our Jordan basis will start with $x,\phi(x),\ldots,\phi^{m-1}(x)$, which are clearly linearly independent and span a $\phi$-stable subspace $W$, and which will give the first Jordan block of size $m$. We need to find a $\phi$-stable complementary subspace to $W$.

Let $\nu\in V^*=\operatorname{Hom}_D(V,D)$ be chosen with $\nu(\phi^{m-1}(x))\neq0$, which is certainly possible (consider any basis of $V$ containing $\phi^{m-1}(x)$, and take the coordinate function for that basis vector). The usual transpose of $\phi$, given by $\phi^\top:V^*\to V^*:\alpha\mapsto\alpha\circ\phi$ is an endomorphism of the left $D$-vector space $V^*$. One has $(\phi^\top)^m=0$ while $(\phi^\top)^{m-1}(\nu)(x)=\nu(\phi^{m-1}(x))\neq0$, so the $m$ vectors $\nu,\phi^\top(\nu),\ldots,(\phi^\top)^{m-1}(\nu)$ are linearly independent in $V^*$ for the same reasons as above, and their span is $\phi^\top$-stable. The intersection of the kernels of $\nu,\phi^\top(\nu),\ldots,(\phi^\top)^{m-1}(\nu)$ then defines a subspace $C$ of codimension $m$ in $V$, which forms a direct sum with $W$ (for any nonzero $w\in W$ one finds that at least one of the mentioned images of $\nu$ does not vanish on it). By construction $C$ is $\phi$-stable. We can then apply the induction hypothesis to $C$ to furnish the remainder of our Jordan basis.