Here’s a nice little thing we can do with Jordan normal forms: show that every invertible linear endomorphism on a vector space of finite dimension over an algebraically closed field has a square root so that . 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 . This decomposes into Jordan blocks, none of which have eigenvalue zero since we’re assuming that is invertible. The transformation 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 . So we may as well restrict our attention to a single block , with .
We can write this block as , where is a nilpotent matrix. In fact, it’s the matrix with just above the diagonal and zeroes everywhere else. Since we’re working over an algebraically closed field, the scalar must have a square root. So if 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 . Yes, that was purely a product of our work on analysis over , 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 . But now if we put in for we notice something: after a while, all the powers of are zero, since is nilpotent! That is, we don’t have a power series, but just a nice polynomial in . And then if we multiply this polynomial by itself, we get a bigger polynomial. But once we take into account the fact that is nilpotent, the only terms that survive are .
To be a little more explicit, we’re trying to find a square root of , where . So we work out the Taylor series above and write down the transformation
Squaring this transformation gives .