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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 $8$?

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  • $\begingroup$ 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. $\endgroup$ May 11, 2019 at 23:47
  • $\begingroup$ @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? $\endgroup$ May 11, 2019 at 23:55
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    $\begingroup$ @user3424575 Yes the determinant is 0 and that is also the product of the singular values. $\endgroup$
    – Dzoooks
    May 11, 2019 at 23:56
  • $\begingroup$ @Dzoooks thank you! $\endgroup$ May 11, 2019 at 23:56

1 Answer 1

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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 $0$

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    $\begingroup$ Thank you. It's always helpful to see how thee formulas are derived. $\endgroup$ May 12, 2019 at 0:21
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    $\begingroup$ Changed to $| \det(A) | $ $\endgroup$
    – user3417
    May 12, 2019 at 1:02

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