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


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


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

March 10, 2009 Posted by | Algebra, Linear Algebra | 3 Comments