The Unapologetic Mathematician

Mathematics for the interested outsider

Square Roots

Here’s a nice little thing we can do with Jordan normal forms: show that every invertible linear endomorphism S:V\rightarrow V on a vector space V of finite dimension d over an algebraically closed field \mathbb{F} has a square root T:V\rightarrow V so that T^2=S. Over at the Secret Blogging Seminar, they discussed this without using the Jordan normal form, and they dealt with the nilpotent blocks later, which has a similar feel to what we’re doing.

First, we find the Jordan normal form for S. This decomposes V into Jordan blocks, none of which have eigenvalue zero since we’re assuming that S is invertible. The transformation S acts without mixing up the blocks, so if we can find a square root for each block then we can put those square roots together into a square root for the whole of S. So we may as well restrict our attention to a single block J_n(\lambda), with \lambda\neq0.

We can write this block as \lambda(1_V+N), where N is a nilpotent matrix. In fact, it’s the matrix with \frac{1}{\lambda} just above the diagonal and zeroes everywhere else. Since we’re working over an algebraically closed field, the scalar \lambda must have a square root. So if 1_V+N has a square root, we’ll be done.

Now, it might seem like a really weird digression, but let’s look at the Taylor series for the function \sqrt{1+x}. Yes, that was purely a product of our work on analysis over \mathbb{R}, but let’s just consider it formally. It’s an infinite series

\displaystyle\sqrt{1+x}=1+a_1x+a_2x^2+\ldots

which has the formal algebraic property that if we multiply it by itself (and wave our hands frantically at all the convergence issues) we’ll get the polynomial 1+x. But now if we put N in for x we notice something: after a while, all the powers of N are zero, since N is nilpotent! That is, we don’t have a power series, but just a nice polynomial in N. And then if we multiply this polynomial by itself, we get a bigger polynomial. But once we take into account the fact that N is nilpotent, the only terms that survive are 1_V+N.

To be a little more explicit, we’re trying to find a square root of 1_V+N, where N^n=0. So we work out the Taylor series above and write down the transformation

\displaystyle1_V+a_1N+a_2N^2+\ldots+a_{n-1}N^{n-1}

Squaring this transformation gives 1_V+N+N^nP(N)=1_V+N.

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March 10, 2009 - Posted by | Algebra, Linear Algebra

3 Comments »

  1. “and wave our hands frantically at all the convergence issues” works well for people who get the right answers, and badly for those who don’t. Can someone say “Renormalization”? I like your fast and fun exposition on the formal structures underlying the matter.

    Comment by Jonathan Vos Post | March 11, 2009 | Reply

  2. Well, part of the trick is that a power series is two different things masquerading as one. Actually, I think I’ll turn this into a bit of a post on its own.

    Comment by John Armstrong | March 11, 2009 | Reply

  3. Problem: the Taylor series for sqrt(1+x) has rational coefficients which may not be defined over all algebraically closed fields. In particular, you’ll have trouble with the algebraic closure of F_2. In characteristic 2, squaring commutes with adding the identity matrix, so the non-singularity assumption is of no benefit.

    Comment by ebw | April 30, 2009 | Reply


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