91影视

Since\(A\)is a square matrix, its singular matrix is also square.

Therefore the determinant is as follow:

\(\begin{array}{c}U{U^T} = I\\Det\left( {U{U^T}} \right) = Det\left( I \right)\\ = Det\left( I \right)\\ = 1\end{array}\)

Apply the distributive property.

\(\begin{array}{c}Det\left( U \right)Det\left( {{U^T}} \right) = I\\{\left( {Det\left( U \right)} \right)^2} = 1\\Det\left( U \right) = \pm 1\end{array}\)

The determinant is as follow:

\(\begin{array}{c}V{V^T} = I\\Det\left( {{V^T}V} \right) = Det\left( {V{V^T}} \right)\\ = Det\left( I \right)\\ = 1\end{array}\)

Apply the distributive property.

\(\begin{array}{c}Det\left( V \right)Det\left( {{V^T}} \right) = I\\{\left( {Det\left( V \right)} \right)^2} = 1\\Det\left( V \right) = \pm 1\end{array}\)

As \(A = U\Sigma {V^T}\)

Apply the determinant on both sides of\(A = U\Sigma {V^T}\)and simplify.

\(\begin{array}{c}{\rm{Det}}A = {\rm{Det}}\left( {U\Sigma {V^T}} \right)\\ = {\rm{Det}}\left( U \right){\rm{Det}}\left( \Sigma \right){\rm{Det}}\left( {{V^T}} \right)\\ = \left( { \pm 1} \right){\rm{Det}}\left( \Sigma \right)\left( { \pm 1} \right)\\ = \left( { \pm 1} \right) \cdot {\rm{Product of singular values of }}A\end{array}\)

Thus,\(\left| {{\rm{Det}}A} \right| = {\rm{Product of singular values of }}A\)

Hence Proved.