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Let $M$ be the manifold of real positive definite $n \times n$ matrices, define a mapping $i:A \to \sqrt A$ (where $A\in M$ and $\sqrt A$ means the unique positive definite square root of $A$). Please show that $i$ is smooth.

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Consider $ A\mapsto A^2$ – user20266 Mar 6 '12 at 11:20
up vote 2 down vote accepted

New answer: The statement that your map is real analytic follows from this answer.

Old answer:

Hint. Let $f$ be your map, i.e. $f(A)^2=A$ and $f(A)$ is positive definite. Use also the notation $\sqrt A=f(A)$.

Then the statement results from the following two observations:

(a) By the theory of power series, there is, for each $\lambda > 0$, an open neighborhood of $\lambda I$ in $M$ on which $f$ is $C^\omega$ (i.e, real analytic). (Here $I$ denotes the identity matrix.)

(b) The action of $\text{GL}(n,\mathbb R)$ on the Grassmannian of $k$-dimensional subspaces of $\mathbb R^n$ is $C^\omega$.

Here are some more details:

(c) Let $A_0$ be positive definite, and let $V_1,\dots,V_p$ be its distinct eigenspaces. Then there is an open neighborhood $U$ of $A_0$ in $M$ such that, for $A$ in $U$, the $A$-eigenspace decomposition of $\mathbb R^n$ is a refinement of the decomposition $\mathbb R^n=\bigoplus AV_i$.

(d) In the above notation, define, for $i=1,\dots,p$, the endomorphism $g_i(A)$ of $\mathbb R^n$ by $$ g_i(A)v=Av\quad\text{if}\quad v\in AV_i, $$ $$ g_i(A)v=0\quad\text{if}\quad v\in AV_j\quad\text{for some}\quad j\neq i. $$ Then the $g_i$ are $C^\omega$ by (b), and $\sqrt A$ is the sum of the $\sqrt{g_i(A)}$.

(d) By shrinking $U$ if necessary, we can assume, in view of (a), that the $A\mapsto\sqrt{g_i(A)}$ are $C^\omega$.

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