The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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


  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 […]

    Pingback by The Uniqueness of Polar Decomposition « The Unapologetic Mathematician | August 21, 2009 | Reply

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