$Ain mathcal{L}(H)$ and $langle x,Ayrangle =langle Ax,y rangle,forall x,yin H$. Prove $Q(A)(H)=H$

Question

Let $Ain mathcal{L}(H)$, where $mathcal{L}(H)$ is the space of bounded linear operator, $H$ is real Hilbert space and $langle x,Ayrangle =langle Ax,y rangle,forall x,yin H$, $operatorname{ker}A={0}$. Put $Q(t)=sumlimits_{k=0}^na_kt^k$, $a_kin mathbb{R}$ where $Q(t)=0$ has no solutions on $mathbb{R}$. Prove $Q(A)(H)=H$.
I tried use Lax Milgram theorem but no problem yet

Disintegrating By Parts · Answer

The polynomial $Q(t)$ is real, with no real roots. Therefore, it may be written as a non-zero constant times products of terms $(t-t_k)^2+v_k^2$ where $t_k,v_k$ are real. Because $A$ is symmetric, then $(A-t_kI)^2+v_k^2I$ is surjective. It follows that $Q(A)$ is surjective.

Answered by Disintegrating By Parts on February 3, 2021

$Ain mathcal{L}(H)$ and $langle x,Ayrangle =langle Ax,y rangle,forall x,yin H$. Prove $Q(A)(H)=H$

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