Smoothness of a Matrix Suppose $A$ and $B$ are square positive definite matrices over a complex field. Given a matrix $M^2$ such that   $M^2 = B +tA$ for $t\geq0$, why is it true that $M$ is a smooth function of $t\geq0$. In particular, why does it matter that t is nonnegative?
Here is what I tried so far: I have that $A = \dot M M + M \dot M$ and since A is positive definite and $M$ is positive definite  (since a square root of a positive definite matrix is positive definite), we have that $\dot M$ is positive definite. I am not sure how to proceed from here...
 A: Matrix square roots are not unique in general (e.g. the identity matrix $I$ has infinitely many square roots). However, if the context is about positive definite matrices, when we speak of a square root of a positive definite matrix $X$, we usually mean a positive definite square root $X^{1/2}$, and such a square root always exists and is unique.
When $t\ge0$, the positive definite square root of $B+tA$ is always well defined. It is still well defined when $t<0$ is sufficiently small, but just how small is sufficient depends on $A$ and $B$.
Now suppose $t\ge0$. Let $V$ be the real vector space of all Hermitian matrices and let $P\subset V$ be the open set of all positive definite matrices. Define $f:V\to V$ by $f(M)=M^2$. Clearly, $f$ is continuously differentiable. Its derivative is also nonzero on $P$ because the equation $MH+HM=0$ has only the trivial solution $H=0$ when $M\in P$. Therefore, by inverse function theorem, the square root function $f^{-1}$ exists and is (locally) continuously differentiable on $P$. It follows that $g(t)=f^{-1}(B+tA)=\sqrt{B+tA}$ is also continuously differentiable.
From the equation $M^2=B+tA$, we see that $\dot{g}(t)$ is the solution of the equation $M\dot{g}(t)+\dot{g}(t)M=A$. In vector form, this means $\operatorname{vec}(\dot{g}(t)) = (I\otimes M+M\otimes I)^{-1}\operatorname{vec}(A)$. Since $M$ is positive definite, entries of $(I\otimes M+M\otimes I)^{-1}$ are smooth rational functions of entries of $M$. Therefore $\dot{g}(t)$ is a smooth function of $M$ and hence $g$ is a (locally) smooth function of $t$.
Actually it can be proved that $M$, its eigenvalues and its eigenbasis can be followed analytically, but the proof requires a much more careful analysis because the paths of the eigenvalues can cross each other and eigenspaces can be unstable. For more details, see Tosio Kato, Perturbation Theory for Linear Operators, 2/e, Springer, Berlin, 1980. Kato's book and Nicolas Higham's Functions of Matrices: Theory and Computation and invaluable references on matrix functions. The MO question 116123 also contains helpful information.
