# The Unapologetic Mathematician

## Square Roots

Here’s a neat thing we can do with the spectral theorems: we can take square roots of positive-semidefinite transformations. And this makes sense, since positive-semidefinite transformations are analogous to nonnegative real numbers, and every nonnegative real number has a unique nonnegative real square root. So we expect that every positive-semidefinite transformation will have a unique positive-semidefinite square root.

So we start by writing down the spectral decomposition

$\displaystyle P=U\Lambda U^*$

Where $\Lambda$ is diagonal. And since $P$ is positive-semidefinite, every diagonal entry of $\Lambda$ — every eigenvalue of $P$ — is a nonnegative real number. We can arrange them nonincreasing order, with the largest eigenvalue in the upper left, and so on down to the lowest eigenvalue (maybe ${0}$) in the lower right corner. Since the eigenvalues are uniquely determined, with this arrangement $\Lambda$ is uniquely determined. If there are repeated eigenvalues, $U$ might not be completely determined, since we have some freedom in picking the basis for degenerate eigenspaces.

Anyhow, since each entry in $\Lambda$ is a nonnegative real number, we can replace each one with its unique nonnegative square root. We call this new matrix $\Sigma$, and observe that $\Sigma^2=\Lambda$. Now we can define $S=U\Sigma U^*$, and calculate

$\displaystyle S^2=U\Sigma U^*U\Sigma U^*=U\Sigma^2U^*=U\Lambda U^*=P$

So $S$ is a square root of $P$. Since the eigenvalues of $S$ (the diagonal entries of $\Sigma$) are nonnegative real numbers, $S$ is positive-semidefinite.

On the other hand, what if we have some other positive-semidefinite square root $S'=U'\Sigma'U'^*$. Saying that it’s a square root of $P$ means that

$\displaystyle S'^2=U'\Sigma'U'^*U'\Sigma'U'^*=U'\Sigma'^2U'^*=U\Lambda U^*=P$

That is, we must have

$\displaystyle\Lambda=U^*U'\Sigma'^2U'^*U=\left(U^*U'\right)\left(\Sigma'^2\right)\left(U^*U'\right)^*$

The matrix $\Sigma'^2$ is diagonal, and its entries — the squares of the diagonal entries of $\Sigma$ — must be the eigenvalues of $\Lambda$. And so the entries of $\Sigma'$ are the same as those of $\Sigma$, though possibly in a different order. The rearrangement, then, is the content of conjugating by $U^*U'$. That is, we must have

$\displaystyle\Sigma=\left(U^*U'\right)\left(\Sigma'\right)\left(U^*U'\right)^*=U^*\left(U'\Sigma'U'^*\right)U$

and so

$\displaystyle U\Sigma U^*=U'\Sigma'U'^*$

And so we really have the exact same square root again. This establishes the uniqueness of the positive-semidefinite square root.

August 20, 2009 - Posted by | Algebra, Linear Algebra

## 2 Comments »

1. Good. Now let $P_1$ and $P_2$ be positive definite (hermitian) transformations, and define the positive definite transformation $P$ by the equation $P_1 P_2 = U P$ where $U$ is unitary. In short, $P$ is the positive definite part of $P_1 P_2$ in the (right) polar decomposition. Exercise: show $P$ is given by the following explicit formula: $P = (P_2 P_1^2 P_2)^{1/2}$.

One can think of this as defining a very natural (nonassociative) binary operation on the set of all positive definite transformations. It has properties I won’t describe here, but which are of considerable interest to those working in quasigroups and loops.

Comment by Michael Kinyon | August 20, 2009 | Reply

2. […] the radical denotes the unique positive-semidefinite square root any positive-semidefinite transformation […]

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