On Nilpotent Elements of $M_n (F)$

Question

For a field $F$, I have proved that $A \in M_n(F)$ is nilpotent iff $A^n=0$. Now I am curious about Division Rings. If we consider $F$ as a division ring then what happens? Does the result remain true?

Answer 1

Yes this holds for square matrices over division rings, essentially for the same reason as for those over fields. The only point to prove is that if $A^k=0$ for any $k$, then $A^n=0$ where $n$ is the size of the matrix.

Over division rings$~D$, one still has the notion of dimension of a vector space over$~D$, which I shall define here to be a right module over $D$. Column vectors over$~D$ form a vector space on which square matrices of the matching size operate from the left as $D$-linear operators. Now consider the sequence of subspaces that are the images of $A^0=I,A,A^2,A^3,\ldots$, in other words the subspaces spanned respectively by the columns of those matrices. The image of $A^{i+1}$ equals the image under $A^i$ of the subspace $\operatorname {Im}A$, which is naturally a subspace of the (full) image of $A^i$; in other words we have a weakly decreasing sequence of subspaces, in the strong sense of inclusion of each space in the previous one.

The hypothesis is that these subspaces eventually become the zero space. To prove that this implies that already $A^n=0$, I will show that it is impossible that two consecutive subspaces in the sequence have the same nonzero dimension; this claim obviously suffices, since it implies the dimension strictly decreases at each step until reaching $0$, which has to happen after at most $n$ steps. But if two consecutive subspaces in the sequence have the same dimension, then since we also have one inclusion, they must be equal. This means that left multiplication by $A$ restricted to this subspace is an isomorphism of vector spaces, and as a consequence repeated application of $A$ will continue mapping this subspace onto itself. Which contradicts that the fact the subspaces eventually become zero. Thus the claim is proved, and the proof complete.

The essential point of the argument is that dimension is defined, which is not true over arbitrary rings. But you can check that the whole theory leading to to conclusion that all bases of a finite dimensional vector space have the same cardinality is perfectly valid for "vector spaces" over division rings.

Answer 2

http://en.wikipedia.org/wiki/Division_ring

On a division ring every module is free and the usual concept of dimension and basis are as for fields(dimension is the lenght of a maximal set of linearly indipendent vectors, two basis have always same lenght, a proper subspace(in finite dimension) has littler dimension).

Assuming these things that you can easily prove by yourself, you can conclude in this way: consider your operator $A:F^n \to F^n, n>0$. It cannot be surjective, because in that case $A^m:F^n \to F^n$ is surjective for all m, but by ipothesis at some point is 0, so you should have n=0, wich is not the case. Then, applying the preceding remark on linear algebra over division ring, Im(A) is a free submodule with $dim_F(Im(A)) \leq n-1$, with of course $A:Im(A) \to Im(A)$. So by induction, you get that $A^{dim(Im(A))}Im(A)=0$, so $A^{n-1}Im(A)=0$, so $A^n=0$.