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
Where is diagonal. And since is positive-semidefinite, every diagonal entry of — every eigenvalue of — 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 ) in the lower right corner. Since the eigenvalues are uniquely determined, with this arrangement is uniquely determined. If there are repeated eigenvalues, might not be completely determined, since we have some freedom in picking the basis for degenerate eigenspaces.
Anyhow, since each entry in is a nonnegative real number, we can replace each one with its unique nonnegative square root. We call this new matrix , and observe that . Now we can define , and calculate
So is a square root of . Since the eigenvalues of (the diagonal entries of ) are nonnegative real numbers, is positive-semidefinite.
On the other hand, what if we have some other positive-semidefinite square root . Saying that it’s a square root of means that
That is, we must have
The matrix is diagonal, and its entries — the squares of the diagonal entries of — must be the eigenvalues of . And so the entries of are the same as those of , though possibly in a different order. The rearrangement, then, is the content of conjugating by . That is, we must have
And so we really have the exact same square root again. This establishes the uniqueness of the positive-semidefinite square root.