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Let $A$ be a $ n \times n $ positive definite matrix, $P$ be a $n\times m$ real matrix with full column rank, I'm wondering whether we have the following inequalty $$ P( P^{T} A P) ^{-1} P^{T} \preceq A ^{-1} ,$$ And in which sufficient conditions the the equality holds.

mewmew
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  • If $A$ is semi-positive, $A^{-1}$ might not be defined. Also, what do you mean by $\preceq$? – Martin Argerami Aug 20 '13 at 02:22
  • There are something wrong with your statement. First, $P$ has to be $n\times m$ real matrix. Second, $A$ should be positive defined. And @Martin, $A\preceq B$ means $B-A$ is semi-positive defined. – Wei Zhan Aug 20 '13 at 02:32
  • @wild Chan： Thanks for your pointing out the false in my statement！ I've corrected my question。 – mewmew Aug 20 '13 at 03:01
  • @mewmew I suppose that $P$ has to be column full rank, otherwise you cannot ensure $P^TAP$ is invertible. – Wei Zhan Aug 20 '13 at 03:07
  • @Wild Chan: You're right! P is column full rank. – mewmew Aug 20 '13 at 03:33

1 Answers1

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First we consider a easier version when $A=I$. We have to prove the matrix below is semi-positive defined: $$I-P(P^TP)^{-1}P^T=I-PP^+$$ Here $P^+$ refers to Moore–Penrose pseudoinverse. Because $P$ is column full rank, we have $P^+=(P^TP)^{-1}P^T$.

A wellknown result for $PP^+$ is that it's an orthogonal projection onto $R(P)$, the range of $P$. We can prove it within three steps:

  1. $PP^+$ is idempotent.
  2. $PP^+$ is Hermite.
  3. $R(PP^+)=R(P)$.

And then $I-PP^+$ is also an orthogonal projection, onto the $N(P^+)=\mathrm{Ker}P^+$. It is a simple deduction from those information that $I-PP^+$ is semi-positive defined, and it's $\mathbf{0}$ iff $\mathrm{rank}(P)=n$.

Now let's get to the general case. Because $A$ is positive defined, we have a invertible $n\times n$ matrix $Q$ so that $A=Q^TQ$. Notice $$P(P^TQ^TQP)^{-1}P^T\preceq (Q^TQ)^{-1}\Leftrightarrow S(S^TS)^{-1}S^T\preceq I$$ where $S=QP$ is also full column rank. That's because pre-multiply by $Q$ and post-multiply by $Q^T$ cannot change the property of semi-positive defined. And now we can apply the result we've already obtained above.

The conclusion is: the inequality holds, and equality holds iff $\mathrm{rank}(P)=n$.

Wei Zhan
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