Matrix Square Root Taylor Series Does applying the taylor series for the square root to matrices give a meaningful definition for the square root of a matrix? (As in the definition of the matrix exponential?)
For symmetric matrices, the spectral theorem gives us a trivial way to find a square root, but does the taylor series agree for these matrices, and does it generalize the square root to a larger class of matrices?
 A: Let $A$ be a positive definite symmetric matrix (for positive semidefinite matrices there is a technical problem - see below). Then $A = UDU^*$ with a unitary matrix $U$ and a diagonal matrix $D$ which has the (positive) eigenvalues of $A$ on its diagonal. Now, let $f(x) = \sum_n a_n(x-x_0)^n$ be an expansion of the square root whose convergence interval contains all those eigenvalues $d_i$. Then
\begin{align*}
f(A)&= f(UDU^*) = \sum_n a_n(UDU^*-x_0)^n = \sum_n a_n(U(D-x_0)U^*)^n\\
&= U\left(\sum_n a_n(D-x_0)^n\right)U^* = U\left(\sum_n a_n\left(\operatorname{diag}_i(d_i-x_0)^n\right)\right)U^*\\
&= U\operatorname{diag}_i\left(\sum_n a_n(d_i-x_0)^n\right)U^* = U\operatorname{diag}_i\sqrt{d_i}U^* = A^{1/2}.
\end{align*}
If the series also converges in zero (which I don't know) the calculation goes through as above also for positive semidefinite matrices.
With the above approach you can also define a square root for diagonalizable matrices having their eigenvalues in the convergence disk. You can extend this definition even for all matrices having their eigenvalues in $D = \mathbb C\setminus (-\infty,0]$. For this, draw a closed path $\gamma$ in $D$ around all the eigenvalues of $A$ and define
$$
A^{1/2} := -\frac 1 {2\pi i}\int_\gamma \sqrt{\lambda}(A - \lambda I)^{-1}\,d\lambda.
$$
This is the Riesz-Dunford functional calculus. 
