Singular value decomposition - determinant

Question

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 $8$?

Answer 1

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 $0$