# Singular value decomposition - determinant

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 $$8$$?

• This recipe calculates the absolute value of the determinant. Consider the SVD of the $1\times1$ matrix with a $-1$ in it. Put another way, this recipe implies no matrix has a negative determinant. May 11 '19 at 23:47
• @kimchilover gotcha - I didn't realise it was the absolute value of the determinant we were calculating here. So 0 is a valid determinant in this case? May 11 '19 at 23:55
• @user3424575 Yes the determinant is 0 and that is also the product of the singular values. May 11 '19 at 23:56
• @Dzoooks thank you! May 11 '19 at 23:56

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 $$0$$

• Thank you. It's always helpful to see how thee formulas are derived. May 12 '19 at 0:21
• Changed to $| \det(A) |$
– user3417
May 12 '19 at 1:02