= Solution
Choose the nonnegative square roots of the <eigenvalues> of $A^TA$ for the diagonal entries of $\Sigma$, in nonincreasing order. Its assumed <matrix inverse> exists, so these entries are positive. The <eigendecomposition> gives
$$
U^TU=\Sigma^{-1}V^TA^TAV\Sigma^{-1}
=\Sigma^{-1}\Sigma^2\Sigma^{-1}=I_n.
$$
Hence $U$ has <orthonormal> columns. Also
$$
\boxed{U\Sigma V^T=AV\Sigma^{-1}\Sigma V^T=A}.
$$
Together with the <orthogonal matrix> $V$ and the nonnegative <diagonal matrix> $\Sigma$, these are precisely the properties of a thin <singular value decomposition>.
Back to article page
