Let $A,B$ symmetric matrices of $\mathbb R$ and $(\cdot, \cdot)$ be Euclid inner product.
Suppose $(Ax,x)\geqq (Bx,x)\geqq 0$ for all $x\in \mathbb R^n$.
Then, prove rank$A\geqq$ rank$B$.
I don't come to the conclusion by the method below.
Note rank$A\geqq$ rank$B$ $\iff \dim \ker A\leqq \dim \ker B$.
So, it suffices to show $\dim \ker A\leqq \dim \ker B$.
Let $V:=\{x\in \mathbb R^n \mid (Bx,x)=0\}.$
Then, from the condition $(Ax,x)\geqq (Bx,x)\geqq 0$, I get $\ker A\subset V.$
And if I can prove $V$ is subspace of $\mathbb R^n$ (I didn't prove and I don't know whether $V$ is subspace of $\mathbb R^n$), I get $\dim \ker A\leqq \dim V$.
I don't come to conclusion and I don't know whether I'm on right track.
Any idea ? Thanks for your help.
Given : $ A,B $ symmetric matrices of $\Bbb{R}$ and $(⋅,⋅)$ be Euclid inner product and $(Ax,x)\geq (Bx,x)\geqq 0$
To show : $\operatorname{rank}(A)\ge \operatorname{rank}(B) $
It is enough to show that $\ker(A) \subset \ker(B) $
Let $x\in \ker(A) $ then $Ax=0$ implies $0=(Ax,x)\geq (Bx,x)\geq 0$
Hence $(Bx, x) =0$
Let $y\in\Bbb{R}^n$ , then by Cauchy-Schwarz inequality
$|(Bx, y) |^2\le (Bx, x)( By, y) $
implies $(Bx, y) =0$ for all $y\in\Bbb{R}^n$
In particular for $y=Bx$ , we have
$(Bx, Bx)=0$ implies $Bx=0$ i.e $x\in\ker(B) $
Hence $\ker(A) \subset \ker(B) $ implies $\dim \ker A\leq \dim \ker B$
Hence the conclusion.