Why is the determinant of a square matrix equal to the determinant of the diagonal of an SVD?

Question

If I have a square matrix $A$, where $ A = UDV^T $ is the SVD of $A$,
how can $\det A = \pm \det D$ ?

Answer 1

The determinant has a multiplicative property:

$$\det(UDV^t) = \det(U)\det(D)\det(V^t)$$

Now you just need to use a certain property of $U$ and $V$.

Answer 2

The matrix $D$ has on the diagonal the nonnegative square roots of the eigenvalues of $A^TA$ (counted with their multiplicity).

Note that the product of the eigenvalues (counted with their multiplicity) equals the determinant of the matrix. Thus $$ (\det(D))^2=\det(A^TA)=\det(A^T)\det(A)=(\det(A))^2 $$