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Let $A$ be a $n \times m$ matirx and $B$ a $m \times m$ matrix as they are all real-valued. Then does it hold $$ \det ( ( A^{T}A ) ( B^T B ) ) = \det ( (AB)^T (AB) ) $$ in general?

Do I prove this by mere use of transposition and the property of determinants? If there is a major trick in the proof of it, could you let me know?

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Ah! After just writing out, I've found it too easy.. –  julypraise Oct 11 '12 at 22:38
    
Is $M$ the same as $B$? Is $M$ suppose to be $m \times n$? –  hardmath Oct 11 '12 at 22:54
    
@hardmath oh sorry yes you are right! I will edit –  julypraise Oct 11 '12 at 22:54
    
So it is a key part of your question (and answer) that $B$ is a square matrix, here $m \times m$. –  hardmath Oct 12 '12 at 0:19
    
@hardmath Yep, by your note I've solved the question. Thanks! –  julypraise Oct 12 '12 at 0:48
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1 Answer

up vote 3 down vote accepted

$$ |A^{T}A||B^{T}B| = |A^{T}A||B^{T}||B| = |B^{T}||A^{T}A||B| $$ which equals $$ |B^{T}(A^{T}A)B| = |(B^{T}A^{T})(AB)| = |(AB)^{T}AB|. $$

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