Show that for a square matrix $A$, $|\det A| = \det|A|$, where $|A| = \sqrt{A^tA}$ is denoted as the modulus of A.
May I get a hint as to where to start this proof?
Show that |detA|= det|A|
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linear-algebra
2 Answers
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The modulus of $A$ is a positive semidefinite matrix $|A|$ such that $|A|^2=A^TA$ (or the hermitian transpose, if $A$ has complex entries). Then $$ (\det |A|)^2=\det(|A|^2)=\det(A^TA)=(\det A)^2 $$ Since $|A|$ is positive semidefinite, its determinant is $\ge0$.
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0I interepreted it as modulus as in [this](https://math.stackexchange.com/questions/214423/what-is-the-modulus-of-a-matrix) question, meaning the unique matrix $X$ such that $X^2 = A^tA$. – 2017-02-22
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0@Mark Thanks for noting – 2017-02-22
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$|det(A)|^2 = det(A^TA) = det(|A|^2) = (det|A|)^2 \Rightarrow |det(A)| = det|A|$