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I asked a question on an matrix inequality several days ago, see the link Matrix Identity. Is this inequality still true for this strengthened version?

Let $A$ be a $ n \times n $ positive definite matrix, $P, Q $ be $n\times m$ and $n\times l$ real matrix with full-rank orthongonal column vector($m+l \leq n$), which is to say $$P^{T}P=I_{m \times m}, \quad Q^{T}Q = I_{l \times l}, \quad P^{T}Q=O_{m \times l}. $$ I'm wondering whether we have the following inequalty $$ P( P^{T} A P) ^{-1} P^{T} + Q( Q^{T} A Q) ^{-1} Q^{T} \preceq A ^{-1}.$$

And the equality holds when $ m+l =n$?

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mewmew
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1 Answers1

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With the same argument that Wild Chan gave in the other answer, we can at least show that if $A=I$ , we would want to show that $I - PP^\dagger - QQ^\dagger$ is positive semidefinite. But $PP^\dagger + QQ^\dagger$ is just an orthogonal projection onto the direct sum of the column spaces of $P,Q$ as they have orthogonal column spaces, and thus it is true.

So let us transfer to the other problem. Letting $A = LL^T$, we have (after multipling on the left by $L^T$ and on the right by $L$,

$L^TP(P^T L L^TP)^{-1}P^TL + L^TQ(Q^T L L^T Q)^{-1} Q^TL = R(R^TR)^{-1}R^T + S(S^T S)^{-1}S^T \preceq I$ where $R = L^TP$ and $S = L^TQ$.

As per comments, the same argument would hold if $P^T A Q = R^T S = 0$ but it may not be true.

We have a negative result if $R$ and $S$ are not orthogonal: For a counterexample we would need to find an $x$ for which $|x_R|^2 + |x_S|^2 > |x|^2$ where $x_R$ is the orthogonal projection to the column space of $R$ and likewise for $x_S$. If we take $x$ to be a column of $L^TP=R$ which is not orthogonal to $L^TQ=S$, then $x_R = x$ but $x_S \neq 0$, and thus we have strict inequality which violates the claim.

I suppose it remains to determine when $P^T A Q = 0$. One such condition is if the column space of $Q$ is $A$-invariant or the column space of $P$ is $A^T$-invariant.

Evan
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  • The main problem I'm not very sure is that when $A \neq I$, $$ P^{T} A Q \neq O$$, how can you use orthogonal projection arguments, can you explain it in detail? Thanks! – mewmew Aug 25 '13 at 07:20
  • @mewmew Did you do the same trick of writing $A=LL^T$ and multiplying the whole equation by $L^T$ on left and $L$ on right? I don't see the P^TAQ appear anywhere? – Evan Aug 25 '13 at 15:20
  • Ok I see now. But you are taking inner products of a linear combination of $P$s with linear combination of $Q$'s so they are still orthogonal. – Evan Aug 25 '13 at 16:46
  • So you mean, if $P^{T}Q = O$, then $P^{T} A Q =O$ for any symmetric matrix $A$? – mewmew Aug 25 '13 at 16:52
  • Yes, the proof is fairly straight forward , any linear combination of the $P$ will be orthogonal to any linear combination of the $Q$ just by bilinearity of the inner product. – Evan Aug 25 '13 at 17:22
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    Please see this simple counterexample: $$e_{1}=(1,0)^{T}, \quad,e_{2}=(0,1)^{T}, \quad A= \begin{pmatrix} 1 & x \ x & 1 \end{pmatrix},$$ then $e_{1}^{T}e_{2}=0$ and $e_{1}^{T}Ae_{2}=x.$ If $x \neq 0$, it is not true. – mewmew Aug 25 '13 at 23:51
  • @mewmew Sorry for not being very careful, thanks for pointing the counterexample out. – Evan Aug 26 '13 at 01:02