A Little Linear Algebra Never Hurt Anyone

Posted by: hilbertthm90 | September 6, 2009

A Little Linear Algebra Never Hurt Anyone

I feel ashamed of how poor my linear algebra skills are. Today I spent far, far too long on a relatively simple old prelim problem, so I’ve decided to generalize it and make a post out of it.

Algebra Preliminary Exam 1995 Prob 7: Let V be a finite-dimesnional real vector space. Prove that any linear transformation $T: V\to V$ has an invariant subspace of dimension 1 or 2.

First, let’s throw away this real stuff and say that V is a vector space over a field k. Now we’ll prove the following: Suppose h is an irreducible polynomial that divides the minimal polynomial $m_T(x)$ . Then we’ll show that $T$ has an invariant subspace of dimension $deg(h)$ .

Note as a Corollary to the special case where $k=\mathbb{R}$ is the problem as it was originally stated, since any polynomial will factor into irreducible polynomials of degree 1 and 2 over $\mathbb{R}$ (the Intermediate Value Theorem).

First we show that $ker(h(T))$ is non-trivial. This is so that we can pick some non-zero $v\in ker(h(T))$ . Since $h(x) \big| m_T(x)$ we have that $m_T(x)=h(x)g(x)$ . Now $0=m_T(T)=h(T)g(T)$ . If $h(T)$ were invertible, then $g(T)=0$ . But $deg(g)<deg(m_T)$ which contradicts minimality of the minimal polynomial. Thus $h(T)$ is singular and has non-trivial kernel.

Now choose $v\in ker(h(T))$ . Suppose $deg(h)=n$ . Then we want to show that $S= span\{v, Tv, T^2v, \ldots, T^{n-1}v\}$ is an invariant subspace of dimension n. Well, the collection is linearly independent, otherwise we would have constants, some of which are non-zero, such that $\displaystyle \left(\sum_{k=0}^{n-1}c_kT^k\right)(v)=0$ . Thus $\displaystyle f(x)=\sum_{k=0}^{n-1}c_kx^k$ is a polynomial of degree n-1 such that $f(T)=0$ . By the division algorithm, there are $p(x), q(x)\in k[x]$ such that $fp+gq=1$ . But this means $0=[f(T)p(T)+g(T)q(T)]v=Id(v)$ a contradiction. Thus they are linearly independent.

So now we have a basis, and hence the dimension of $S$ is exactly n. It is invariant, since $T$ takes each basis element to another one except in the case of $T^{n-1}v$ . In this case, if $h(x)=a_0+\cdots a_nx^n$ we get that $T(T^{n-1}v)=T^nv=-(a_0/a_n)-(a_1/a_n)Tv-\cdots - (a_{n-1}/a_n)T^{n-1}v\in S$ . So we're done with the generalized problem.

Unfortunately, I essentially reproved the entire theory of Jordan Forms and Rational Canonical Forms in an attempt to get this before realizing that although both of those give nice decompositions into invariant subspaces, you can't force either of them to be small like the problem wanted.

Posted in algebra | Tags: cyclic decomposition, invariant subspace, linear algebra, linear transformation, vector space

Responses

I think you could also prove this using the following argument: take $V$ as a module over $k[X]$ . Then we can use the structure theorem over principal ideal domains to write
$V = k[X]/(p_1) \oplus \dots \oplus k[X]/(p_k)$ . One of these polynomials $p_k$ must be divisible by $h$ , say $p_k = gh$ . Then we can write $(g)/(p_k)$ as an $k[X]$ subspace of $k[X]/(p_k)$ (and consequently of $V$ ) which is annihilated by $h(X)$ and is of dimension at most $deg \ h$ .
By: Akhil Mathew on September 7, 2009
at 3:06 am
So that was exactly my first attempt at proving this. Note that the Structure Theorem actually tells us more: that $p_j\big| p_{j+1}$ and hence $p_1$ is the minimal polynomial and each $p_j$ has the same irreducible factors (and the characteristic polynomial is $p_1\cdots p_k$ ).

So that is how I jumped to the phrase “irreducible polynomial dividing the minimal polynomial,” since if $p_k$ is divisible by $h$ then so is $p_1$ .

I actually think these two solutions are fundamentally the same. The way I posted just unwinds definitions too far and ends up obscuring that what you posted is really what I constructed.
By: hilbertthm90 on September 7, 2009
at 1:52 pm
Actually, unless I’m screwing something up, $p_k$ is the minimal polynomial, not $p_1$, since the minimal polynomial needs to generate the annihilator of $V$ as a $k[X]$-module. That Grove’s algebra book defines the invariant factors as satisfying $p_j|p_{j-1}$ makes this all the more annoying to keep track of.
By: Toby on September 12, 2009
at 5:48 pm
Hmm…Whichever way you define it, it is just the smallest one that is dividing all the rest, right? I defined $p_j\big| p_{j+1}$ , so $p_1$ was minimal for me.
By: hilbertthm90 on September 12, 2009
at 6:52 pm
Isn’t it the largest one?
By: Akhil Mathew on September 12, 2009
at 7:42 pm
Oh right. Sorry. Bigger polynomials means smaller ideals, so $(p_1)\supset (p_2)\supset\cdots \supset (p_k)$ and hence in order to kill everything you need $p_k$ .

I guess to not mess this up, I can always just think of abelian groups. If it decomposes as $\mathbb{Z}/2\oplus \mathbb{Z}/4\oplus \mathbb{Z}/8$ then 2 won’t kill everything, but 8 will.
By: hilbertthm90 on September 12, 2009
at 8:31 pm