# The Unapologetic Mathematician

## 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 $1_V+N+N^nP(N)=1_V+N$.

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

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