Inequality of quadratic form and generalized inverse

Question

The original question is asked here by myself. But seems like nobody give an answer. So I decide to ask this question in mathoverflow. Sorry for duplication.

I want to prove the following results,

Given two non-negative definite matrices $A$ and $B$, suppose vector $a \in C(A)\cap C(B)$ , where $C(A)$ and $C(B)$ are column spaces of $A$ and $B$ separately.

Then

$(a' A^- a)(a' B^- a) \geq (a' (A+B)^- a) ( a' A^- a+ a' B^- a)$

where $A^-, B^-, (A+B)^-$ are generalized inverses, i.e. , $A A^-A=A$.

I start from the simplest case $A=B$ etc. But for general case, I can't prove it. Any hints or ideas?

Thanks in advance!

Answer 1

Yes, this is true. I write $(x,y)=x'y=y'x$ for an inner product of (column) vectors $x,y$. We have $a=Ax=By$ for certain $x,y$ and also $a=(A+B)z$ (since the image of the operator $A+B$ is an orthogonal complement to the kernel of $A+B$ and thus contains the image of $A$.) We have $(Ax,x)(Az,z)\geqslant (Ax,z)^2=(a,z)^2$ (that is Cauchy-Bunyakovsky-Schwarz for the vectors $\sqrt{A}x,\sqrt{A}z$), therefore either $(a,x)=(a,z)=0$ or $(Az,z)\geqslant (a,z)^2/(a,x)$.Analogously $(Bz,z)\geqslant (a,z)^2/(a,y)$. Summing up we get $(a,z)\geqslant (a,z)^2(1/(a,x)+1/(a,y))$, $(a,x)(a,y)\geqslant (a,z)((a,x)+(a,y))$ (if $(a,x)=(a,z)=0$ this is also so.)

But $(A^-a,a)=(A^-a,Ax)=(AA^-a,x)=(AA^-Ax,x)=(Ax,x)=(a,x)$ and so on, thus we have proved exactly what you ask for.