Modulus question, determinant

Question

Show that for a square matrix $A, | \det A| = \det |A|$

For this question, $|A|$ is the modulus of $A$, so $|A|$= $(A^*A)^{\frac{1}{2}}$. Writing det $|A|$ makes sense, but I am unsure how to wrote $| \det A|$. I just am unsure how to apply these properties, if anyone could help.

So $Det|A|= Det|A^*A|$

Would $| \det A| = \det(A)^* \det(A)=\det(A^*)\det(A)= Det|A^*A| = Det |A|?$ That just seems sort of odd but it is what I am speculating.

Answer 1

We have $$ |\det A|^2 = (\det A)(\overline{\det A}) = (\det A)(\det A^\ast) = \det A^\ast A = \det |A|^2.\tag{1} $$ Since $A^\ast A$ is positive semidefinite, so is $|A|=(A^\ast A)^{1/2}$. Hence $\det|A|\ge0$ and by taking square root on both sides of $(1)$, we get $|\det A|=\det|A|$.

Answer 2

We have $|A|^2=A^{\star}A$, hence

$(\det(|A|))^2=\det(|A|^2)=\det(A^{\star}A)=\det(A^{\star})*\det(A)=(\det(A))^2$.

Your turn !