I know that when expressing a matrix $A$ in SVD, the determinant of $A$ can be calculated by finding the product of the singular values $\sigma_1 *… *\sigma_n $. My question is, does this hold true only for non-zero values of $\sigma_i$? So for instance, I am given the singular matrix given below:
$\Sigma$ = $$
\begin{bmatrix}
4 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
Would the determinant in this case be 0? Or would we just multiply the non-zero singular values to get a determinant of $8$?
Best Answer
Consider a matrix $A$ which has an SVD as $A = U \Sigma V^{T}$
$$ \det(A) = \det(U \Sigma V^{T}) \\ = \det(U)\det(\Sigma) \det(V^{T}) \\ = \det(U)\det(\Sigma) \det(V) $$
now the determinant of an orthogonal matrix like $U,V$ is $\pm 1$
$$ \det(U) , \det(V) = \pm 1 \implies \\ |\det(A)| = \det(\Sigma) $$
Additionally the determinant of a diagonal matrix is the product of the diagonal
$$ \det(D) = \prod_{i} \textrm{diag}(D)_{i} $$
So the determinant of $A$ is
$$ |\det(A)| = \prod_{i} \textrm{diag}(\Sigma)_{i} $$
now if any singular value $\sigma_{i} = 0$ then the entire product will be $0$