# 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.

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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 $B$.

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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.

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