# An algebra of nilpotent linear transformations is triangularizable

How to prove "An algebra of nilpotent linear transformations is triangularizable" using linear algebra only?

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What do you mean by triangularizable? That each element can be represented by a triangular element? – Manos Apr 19 '12 at 1:23
@Manos: Yes, it is. – Sunni Apr 19 '12 at 1:31
Do you know the result that every linear operator $T$ on a vector space $V$ (real or complex) is nilpotent iff there is a basis $\mathcal{B}$ for $V$ such that $T$ in that basis is upper triangular? – user38268 Apr 19 '12 at 1:53
@BenjaminLim: Dear Benjamin, I believe that Sunni wants to simultaneously triangularize the matrices in the algebra, as in my answer. Regards, – Matt E Apr 19 '12 at 4:04
@MattE Ah ok thanks :D – user38268 Apr 19 '12 at 6:46

The following argument is a simplified version of an earlier argument posted here:

Let $A$ be the algebra in question, and let $V$ be the finite-dimensional vector space on which $A$ acts. Suppose that $V \neq 0$ (as we may, since otherwise $A = 0$ and there is nothing to prove). I claim that $A V$ is a proper subspace of $V$. Indeed, suppose not, i.e. suppose that $V = AV$, and choose a minimal generating set $v_1,\ldots, v_n$ for $V$ over $A$. Then in particular we may find $a_{i} \in A$ such that $v_1 = \sum_i a_{i} v_i$, and hence so that $$(1-a_{1})v_1 = \sum_{i> 1} a_{i}v_i.$$ Since $a_{1}$ is nilpotent, say $a_{1}^N = 0$, the operator $1 - a_{1}$ is invertible, with inverse $1 + a_{1} + \cdots + a_{1}^{N-1}$. Acting this on both sides of the displayed identity, we find that $v_1 = \sum_{i>1} b_{i} v_i$ for appropriately chosen $b_{i} \in A$, and so in fact our original generating set was not minimal ($v_1$ was superfluous), a contradiction.

Thus $AV$ is a proper subspace of $V$. Similary $A(AV)$ is a proper subspace of $AV$, and continuing, we find a strictly decreasing sequence of $A$-invariant subspaces $$V \supset AV \supset A^2V \supset \ldots \supset A^NV \supset \cdots$$ which must eventually reach $0$ (just for dimension reasons). Choosing a basis for $V$ compatible with this descending sequence, we obtain a triangularization of the action of $A$ on all of $V$. QED

This argument uses nothing more than the notion of dimension and basic algebra, and so in principal is just linear algebra, but of course it is more sophisticated than the term "linear algebra" might suggest. You could probably make it less sophisticated on a line-by-line basis at the expense of adding (many) more lines. My feeling is that the result you are asking about is sufficiently non-trivial as a statement of algebra that any proof is going to be either somewhat sophisticated (using a view-point of abstract algebra, even if it only uses facts from commutative algebra) or else quite painful to write down. But maybe I'm wrong!

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This reminds me a lot of the non-Cayley-Hamilton proof of Nakayama. – Dylan Moreland Apr 19 '12 at 4:43
@Dylan: Dear Dylan, That's what I had in mind when I wrote it. Cheers, – Matt E Apr 19 '12 at 4:43
@Sunni Just to expand a little on MattE's answer, to see that $a_1$ nilpotent implies that $1- a_1$ is invertible you can look at this link here: math.stackexchange.com/a/68201/5783 – user38268 Apr 19 '12 at 6:48
@MattE I know how if we have a nilpotent operator $T$ on some (real or complex) vector space, then looking at the ascending chain $0 \subset \ker T \subset \ker T^2, etc$ which must eventually terminate, choosing an appropriate basis from $\ker T$, $\ker T^2$, etc gives us a basis for which $T$ in that basis is upper triangular. However now we are doing this for an entire collection of nilpotent operators, I am not so clear how all this can be done at once. – user38268 Apr 19 '12 at 6:52
@Benjamin: Dear Benjanmin, Just to fix ideas, imagine that $V$ were $5$ dimensional, that $AV$ were three dimensional, that $A^2V$ were two-dimensional, and that $A^3V = 0$. Choose a basis $v_1,\ldots,v_5$ for $V$ so that $v_1$ and $v_2$ form a basis for $A^2 V$, and $v_1,\ldots , v_3$ is a basis for $AV$. Then each element of $A$ will have a matrix of the form $$\begin{pmatrix} 0 & 0 & *& *& *\\ 0&0&*&*&*\\ 0 & 0& 0& * & * \\ 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$ This should help illustrate the general assertion. Regards, – Matt E Apr 19 '12 at 12:18

For the record, the non-«linear algebra» proof of this is:

Consider the local artinian algebra $B=k1_V\oplus A\subseteq\mathrm{End}(V)$. Its radical is $A$ with $1$-dimensional semisimple quotient $B/A\cong k$, so the unique simple module is of dimension $1$ over the base field. The vector space $V$ is tautologically a $B$-module which has finite length, so it has a composition series and each subquotient, being simple, is of dimension $1$. Any ordered basis of $V$ such that each of its prefixes is a basis of one of the layers of this composition series triangularizes $A$.

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