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Cauchy-Schwarz inequality of determinants:

for $A_{n\times k}$, $B_{n\times k}$, and $B'B$ non-singular, we have

$|A'B|^2\leq |A'A||B'B|$

I was wondering what's the sufficient and necessary conditions for the equality to hold.

I know a sufficient condition:

when $\exists C_{k\times k}$, such that $A=BC$, then the equality holds.

Is this also a necessary condition?

Thanks.

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Tiny nitpick: You mean Schwarz (as in Hermann Schwarz), not Schwartz (as in Laurent Schwartz). –  J.C. Ottem Jun 11 '11 at 21:48
    
yes, you are right, thanks, edited –  Jeff Jun 11 '11 at 21:51

1 Answer 1

up vote 3 down vote accepted

The standard argument goes as follows: Let $M=B'A(A'A)^{-1}A'B$ and $N=B'(I-A(A'A)^{-1}A')B$. Then it is easy to check that $M$ and $N$ are positive definite and that the Cauchy-Schwarz inequality is equivalent to the following (true) inequality $$ |M+N| \ge |M| $$which has equality if and only if $N$ is the zero matrix, i.e., when $(I-A(A'A)^{-1}A')B=0$, or $B=AC$ where $C=(A'A)^{-1}A'B$. This shows that your inequality is an equality if and only if $B=AC$ where $C$ is invertible.

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