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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?

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The division algebra itself can be embedded to the algebra of $m\times m$ matrices over its center (you get smaller $m$, if you allow matrix entries from a maximal subfield of $D$). Then you can map $n\times n$ matrices with entries from $D$ to $mn\times mn$ matrices. True, you have to be careful in defining things like eigenvalues... –  Jyrki Lahtonen Sep 4 '12 at 7:50
    
Oh, and I should have pointed you to this question right away. –  Jyrki Lahtonen Sep 4 '12 at 9:35
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@JyrkiLahtonen: one odd thing about such embeddings is that they do not have to preserve the trace. For instance the 1×1 quaternion matrix [i] has trace i, but its 4×4 matrix over R has trace 0. Even the reduced trace of j over C is 0. Roughly this is because ±j cannot be distinguished over C or R, so they come in conjugate pairs and cancel. –  Jack Schmidt Sep 4 '12 at 12:44
    
Correct. If we use a maximal subfield, then the trace of the bigger matrix becomes the reduced trace of the trace of the original matrix, so some information will be lost. –  Jyrki Lahtonen Sep 4 '12 at 14:11
    
@JyrkiLahtonen: Thanks for the link, interesting! I must confess that I do not yet understand a lot of it, as I'm only beginning to study abstract algebra, but I'll come back to it in time. I ask this question as it turns out to be somewhat of a stumbling block in a proof that I thought of for another problem. Time for a different approach I guess. –  under Sep 5 '12 at 3:57

3 Answers 3

up vote 6 down vote accepted

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

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+1 Pretty! Just to tie this together with the remarks under the OP I want to point out that the reduced trace of the trace of this matrix is zero. –  Jyrki Lahtonen Sep 4 '12 at 14:22
    
Neat counterexample, thanks. –  under Sep 5 '12 at 4:00

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...)

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+1 A detail we need to keep in mind is that multiplying a (column) vector $x\in V=D^n$ by a matrix $A\in M_n(D)$ from the left is not a $D$-linear mapping, unless $D$ acts on $V$ from the right (as you correctly showed). –  Jyrki Lahtonen Sep 4 '12 at 14:09
    
@JyrkiLahtonen: So let's assume that $D$ acts on $V$ from the right, so that multiplication from the left by $A$ is $D$-linear (which is the whole point of writing down a matrix). This is what the equation $Ax=x\lambda$ seems to assume. But this equation still itches me: if $x$ is such an eigenvector for $\lambda$, then a (non-central) scalar multiple $x\mu$ of $x$ will in general not be an eigenvector for $\lambda$ but for $\mu^{-1}\lambda\mu$. Somehow eigenvectors/spaces make no sense: we cannot expect the linear action of $M$ to correspond to the non-linear multiplication by $\lambda$. –  Marc van Leeuwen Sep 4 '12 at 14:44
    
@Marc: An interesting observation. Are we, as you pointed out, lead to think that eigenspaces don't really make sense in the non-commutative world? Or was your idea to move the eigenvalue back to the left (and I was jumping the gun)? –  Jyrki Lahtonen Sep 4 '12 at 14:52
    
@Jyrki: One cannot have scalars one the same side as matrices. You might think: use a scalar multiple of the identity matrix, but that notion is basis-dependent. As I see it eigenvalues, togther with determinants, traces, characteristic polynomials, homothecies and bilinear forms do not belong in the non-commutative world. –  Marc van Leeuwen Sep 4 '12 at 17:06
    
@MarcvanLeeuwen I think it's not so much "They do not belong" as much as it is "the things we are used to being the same are no longer the same." (Certainly a common theme of noncommutative algebra.) Things like subspaces that map into themselves still certainly have meaning, for example. Of course, those who are very attached to the commutative realm can be forgiven for despairing :) –  rschwieb Sep 4 '12 at 17:55

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.

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