Suppose, following is true for a square matrix $A$:
$|AA^T| = |A^TA| = |A|^2$
Then, is it some special kind of matrix which has a name?
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5It's true for all square matrices. – Mar 30 '14 at 06:21
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2Yes, the name is any square matrix! – Robert Lewis Mar 30 '14 at 06:40
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1If you had $|A|=1$, now that is special! – user76568 Mar 30 '14 at 06:51
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Since for all square matrices: $|AB|=|A||B|= |B||A|$, and $|A|=|A^T|$: $$|AA^T| = |A||A^T| = |A^T||A|= |A^TA| = |A|^2$$ Holds For all $n \times n$ matrices
user76568
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It is true for any two square matrices $A,B$. That is: $$\det(AB)=\det(A)\cdot\det(B)=\det(B)\cdot\det(A)=\det(BA).$$
Now, the part $\det(AA^T)=\det(A^2)$ follows from $\det(A)=\det(A^T)$ and the multiplicativity of the determinant (that was also used above).
DKal
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1Multiplicativity of the determinant is like the mean-value theorem of linear algebra ;-). When in doubt, use it. – user76568 Mar 30 '14 at 06:33
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1@RobertLewis didn't have time to touch them past few weeks.. My desk at work is about to break though. Hope It's going to make it :-O – user76568 Mar 30 '14 at 06:46