Whitening a square ICA model gives orthogonal mixing

By [F1], $C=\mathrm{Cov}(X)=A{I}_p{A}^T=A{A}^T$. For every nonzero $v\in {\mathbb{R}}^p$, $v^{T}Cv=\parallel A^{T}v{\parallel }^2>0$ because $A$ is invertible; hence $C$ is symmetric positive definite.

By [F2], write $A=U\mathrm{\Sigma }{V}^T$ with $U,V$ orthogonal. Since $A$ is invertible, every diagonal entry of $\mathrm{\Sigma }$ is positive. Thus $C=U{\mathrm{\Sigma }}^2{U}^T$, and $B=U{\mathrm{\Sigma }}^{-1}{U}^T$ is symmetric positive definite with $B^2=C^{-1}$.

If $K$ is another symmetric positive-definite matrix with $K^2=C^{-1}$, then $K$ commutes with $C^{-1}$ and preserves each eigenspace $E_{\lambda }$ of $C^{-1}$. On each such subspace, [F3] gives an orthonormal eigenbasis for $K{\mathrm{\mid }}_{E_{\lambda }}$. Its eigenvalues $\mu$ are positive and satisfy ${\mu }^2=\lambda$, so $K{\mathrm{\mid }}_{E_{\lambda }}=\sqrt{\lambda }I$. This determines $K$ uniquely and agrees with $B$.

Therefore $C^{-1\mathrm{/}2}=B$ and $C^{-1\mathrm{/}2}A=U{\mathrm{\Sigma }}^{-1}{U}^{T}U\mathrm{\Sigma }{V}^T=U{V}^T$. Since $(U{V}^T{)}^T(U{V}^T)=I_p$, the whitened mixing is orthogonal and .