Is negative square root defined for a positive semidefinte matrix?

Question

If I have a positive semidefinite matrix $A$, is an operation like $A^{-\frac{1}{2}}$ defined? I know it is defined for positive definite matrices but for the positive semidefinite case, do the zero eigenvalues cause trouble or do we just consider them to be $0$?

Answer 1

For any real symmetric matrix $A$ and any real function $f\colon \mathbb R\to\mathbb R$ we can define $f(A)$ as follows. Let $A=Q^TDQ$ be an orthogonal diagonalization of $A$ (which exists by the spectral theorem, see here), so that $D$ is a diagonal matrix with the eigenvalues of $A$ appearing on the diagonal. Then define $$ f(A):=Q^Tf(D)Q, $$ where $f(D)$ means we form a diagonal matrix by applying $f$ to each diagonal entry of $D$.

Now you are interested in a function from $\mathbb R^+\to \mathbb R$, but we can apply the same approach in this case under the additional hypothesis that all eigenvalues are strictly greater than zero - i.e., a positive definite matrix, rather than positive semidefinite as you wrote.

Since you are insisting on considering the positive semidefinite case, you will need to make some nonstandard definitions. The one that seems least contrived to me is to extend the matrix entries to the extended reals $\{-\infty\}\cup\mathbb R\cup \{\infty\}$, and consider the function $f\colon \mathbb [0,\infty)\to (0,\infty]$ defined by $$ f(x)=\begin{cases}x^{-1/2},&x>0\\ \infty,&x=0. \end{cases} $$