How to show unitary decomposition is continuous It is a well-known fact that, for $A \in GL(n,\mathbb{C})$ with polar decomposition $A=U_AP_A$ for $U_A$ unitary and $P_A$ positive definite and Hermitian, the map $GL(n,\mathbb{C}) \rightarrow U(n)$ by $A \mapsto U_A$ is a homotopy equivalence. This is a fact I have seen many times. However, I have recently realized that I have no idea how to prove this map is continuous.
It's pretty clear that since $U_A=A P_A^{-1}=A (\sqrt{A^*A})^{-1}$, where * denotes the conjugate transpose, all one needs to prove is continuity for the matrix square root map which sends a positive-definite matrix to its principal square root since continuity for multiplication and inverses are well-known and continuity for * is obvious.
Can anyone provide a good proof for continuity of the matrix square-root map on the space of positive definite matrices? Of course if you can prove $A \mapsto U_A$ is continuous without that fact, that would also be a fine answer.
 A: The matrix square root on positive definite Hermitian matrices can be defined using the holomorphic functional calculus with the principal branch of the square root.
Let $\Gamma$ be a simple positively oriented contour surrounding the spectrum of positive definite Hermitian matrix $A$ but not intersecting $(-\infty,0]$.  Then for $B$ sufficiently close to $A$, the spectrum of $B$ is inside $\Gamma$ and $\sqrt{B} =  \dfrac{1}{2\pi i} \int_\Gamma \sqrt{z}\, (B - z I)^{-1}\ dz$, which is easily seen to be continuous in $B$.
A: Here is a tedious answer using the implicit function theorem applied to $\phi(X,Y) = X-Y^2 = 0$, with $X,Y \in \mathbb{C}^{n \times n}$.
It is straightforward to see that $\frac{\partial \phi(X,Y)}{\partial X}\Gamma = \Gamma$, and $\frac{\partial \phi(X,Y)}{\partial Y}\Delta = -(\Delta Y + Y \Delta)$. Since both are smooth, it follows that $\phi$ is smooth as well.
We need to show that $L=\frac{\partial \phi(X,Y)}{\partial Y}$ is invertible, at least in a neighborhood of a Hermitian $Y>0$.
If $Y>0$ and is Hermitian, it has a full set of eigenvectors $v_n$ corresponding to each (real) eigenvalue $\lambda_n$. It is easy to check that $L(v_n v_m^*) = -(\lambda_n+\lambda_m)v_n v_m^*$, and that $\{v_n v_m^*\}_{m,n}$ is a basis. Since $\lambda_n+\lambda_m>0$, for all $m,n$, $L$ is invertible. Furthermore, note that if $B$ is Hermitian, then the solution $\Delta$ to $L(\Delta) = B$ will also be Hermitian.
Consequently, if we have $\phi(\hat{X},\hat{Y}) = 0$, with Hermitian $\hat{X},\hat{Y} >0$, then there exists a unique $\eta$ defined in a neighborhood of $\hat{X}$, such that $\eta(\hat{X}) = \hat{Y}$, and $\phi(X,\eta(X)) = 0$, for $X$ in this neighborhood. Furthermore, $\eta$ is differentiable on this neighborhood.
However, we are not finished. It remains to show that if $X>0$ is Hermitian (and sufficiently close to $\hat{X}$ so that $\eta(X)$ remains positive definite), then $\eta(X)$ is Hermitian. Let $\Xi(t) = \hat{X}+t(X-\hat{X})$, and $H(t) = \eta(\Xi(t))$. Then $\dot{H}(t) = \frac{\partial \eta(\Xi(t))}{\partial X} (X-\hat{X}) = \frac{\partial \phi(\Xi(t),H(t))^{-1}}{\partial Y} (X-\hat{X})$, with $H(0) = \hat{Y}$, of course. If we let $f(H,t) = \frac{\partial \phi(\Xi(t),H)^{-1}}{\partial Y} (X-\hat{X})$, then this can be written as  $\dot{H} = f(H,t)$, $H(0) = \hat{Y}$. Note that if $H$ is Hermitian, then $f(H,t)$ is also Hermitian.
Now let $\Pi$ be the projection onto the subspace of Hermitian matrices, and consider the equation $\dot{J} = \Pi f(J,t)$, $J(0) = \hat{Y}$. Since $J(t) = J(0) + \int_0^t \Pi f(J(\tau),\tau) d \tau = \Pi(J(0) + \int_0^t f(J(\tau),\tau) d \tau)$, we see that $J(t)$ is Hermitian. It follows from this that $\dot{J} = f(J,t)$, and by uniqueness of solution we have that $H(t) = J(t)$, hence $H(t)$ is Hermitian, and so $\eta(X) = H(1)$ is Hermitian.
A: A short proof can be found in Problems 17 and 18 from 7.2 of Horn, Roger A., and Charles R. Johnson. Matrix Analysis. Cambridge University Press, 1990.
Proof. We first prove the following inequality: let $A
 , B\in\mathbb{C}^{n}$
  be Hermitian positive semidefinite matrices, with $A$
  not singular. Then
$$\|A-B\|_{2}\le\|A^{2}-B^{2}\|_{2}/\left[\lambda_{\min}(A)+\lambda_{\min}(B)\right]. \tag{1}\label{1}$$
 Indeed, define $E=A-B$
  and let $\lambda=|\lambda_{\max}(E)|.$ 
Since $E$ is also Hermitian, $\lambda=\|E\|_{2}$. 
Moreover, $A^{2}-B^{2}=A(A-B)+(A-B)A-(A-B)^{2}=AE+EA-E^{2}$,
 and consequently, for $x\in\mathbb{C}^{n}$
  with unit norm and satisfying $Ex=\lambda x$, one has that 
\begin{eqnarray}
\|A^{2}-B^{2}\|_{2} &\ge& |x^{*}\left(AE+EA-E^{2}\right)x| \\ 
 &=& |x^{*}AEx+x^{*}E(E+B)x-x^{*}E^{2}x| \\
 &=& |\lambda||x^{*}Ax+x^{*}Bx| \\
 &\ge& |\lambda|\left(\lambda_{\min}(A)+\lambda_{\min}(B)\right). \\
\end{eqnarray} 
By \eqref{1}, we have that \begin{eqnarray}
\|A^{1/2}-B^{1/2}\|_{2} &\le& \frac{\|A-B\|_{2}}{\lambda_{\min}(A^{1/2})+\lambda_{\min}(B^{1/2})} \\
&\le& \frac{\|A-B\|_{2}}{\lambda_{\min}(A^{1/2})} \\
&=& \|A-B\|_{2}\;\lambda_{\max}(A^{-1/2}) \\
&\le& \|A-B\|_{2}\|A^{-1/2}\|_{2}.
\end{eqnarray}
 Suppose that $\varepsilon>0$
  is given. Choosing $\delta=\varepsilon/\|A^{-1/2}\|_{2}$
 , if $\|A-B\|_{2}<\delta$
  then $\|A^{1/2}-B^{1/2}\|_{2}<\delta\|A^{-1/2}\|_{2}=\varepsilon$
 . 
