Positive-definite bundle endomorphisms have smooth positive square roots

In each fibre [F1] gives a unique positive-definite self-adjoint square root $S_x=T_x^{1\mathrm{/}2}$. These fibre maps automatically define a bundle endomorphism set-theoretically; it remains to prove local smoothness.

Fix $x_0$ and a smooth orthonormal frame near it, so self-adjoint maps are symmetric matrices. For $\mathrm{\Phi }(R)=R^2$, the derivative at the positive matrix $S_{x_0}$ is $D{\mathrm{\Phi }}_{S_{x_0}}(H)=S_{x_0}H+H{S}_{x_0}$. In an orthonormal eigenbasis of $S_{x_0}$ its $(i,j)$ entry is $(s_i+s_j)H_{ij}$; every $s_i>0$, so this derivative is an isomorphism on symmetric matrices.

Apply [F2] to $R^2-T_x=0$. It produces a unique smooth symmetric solution $R(x)$ near $x_0$ with $R(x_0)=S_{x_0}$. After shrinking, positivity persists; fibrewise uniqueness in [F1] then gives $R(x)=S_x$. Thus $S$ is smooth near every point, and the unique local roots agree on overlaps. In rank zero the unique empty endomorphism supplies the result.