# 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