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Assume $A$ is an arbitrary $m\times n$ real matrix.

Is $\det(AA^T)$ always positive? Is it non-negative or it can have any value?

Edit:

It seems I have to emphasis that $m \ne n$ i.e. matrix is non-squared.

Community
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Pouya
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2 Answers2

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If $A$ has a singular value decomposition $$ A = VDU^{-1} $$ (note since $A$ is rectangular, $V \neq U$ but $V^T = V^{-1}$ and $U^T = U^{-1}$ and finally, $D^T=D$ since $D$ is diagonal) then $$ A^T = (U^T)^{-1}D V^T = UDV^T $$ and now $$ AA^T = VD^2 V^T $$ so $\det(AA^T) = \det D^2$ which must be real but cuold be negative is one of the singular values of $A$ is complex.

gt6989b
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  • Can you make your final sentence a little more clearer. I'm afraid I'm not following your point. – Pouya Mar 25 '14 at 17:01
  • @Pouya just like a square matrix can have complex eigenvalues, the rectangular matrix can have complex singular values, which would make the final result task but negative – gt6989b Mar 25 '14 at 17:03
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In the case n=m (square matrix)

$$det(AB)=det(A)det(B)$$

and

$$det(A^T)=det(A)$$

yes, it is positive

T_O
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    Well, $\det(A)$ is not exactly defined for non-square $A$, and here $A$ is definitely rectangular, you cannot assume $m \neq n$ – gt6989b Mar 25 '14 at 16:40
  • you are right, my bad – T_O Mar 25 '14 at 16:41
  • @gt6989b Can we say that T_O is right in the case of square matrix? – nam Mar 25 '14 at 16:41
  • @nam Yes, for square matrices this holds. For rectangular, you have to use similar value decomposition – gt6989b Mar 25 '14 at 16:42
  • @gt6989b By singular value decomposition, $A=U \Sigma V^{T}$. And so, $ A^{T} = V \Sigma^{T} U^T$.

    Then, $AA^{T}=U \Sigma V^{T} V \Sigma^{T} U^T = U \Sigma \Sigma^{T} U^T$

    But, I do not know how to proceed.

     – nam Mar 25 '14 at 16:46
  • @gt6989b Thz for your answer. I know that I missed the fact that $\Sigma^{T} = \Sigma$ – nam Mar 25 '14 at 16:53