I am asking myself, if we would weaken the assumptions of the Lax-Milgram Lemma to the finite dimensional case
Lax Milgram Lemma Let ($V$, $(\cdot, \cdot$, $\Vert \cdot \Vert$) be a (real) Hilbert space. Let $\mathcal{A} : V \rightarrow V^*$ be a linear, strongly positive, bounded operator, then it follows that $\mathcal{A} $ is bijective, i.e for all $f \in V^*$ there exists a unique solution $u \in V$ to the problem $\mathcal{A}u = f$ in $V^*$.
how the proof would be easier? Since we are then finite dimensional, we get a linear system of equations. So, we have to show that $A = \mathcal{A}$ is invertible. But how to show that it holds that $\det A \neq 0$?
Thank you already for the responses!
You an observe that your map is injective because, by hypothesis of Lax-Milgram theorem, it satisfies the following property:
For each $v\in V$ $\| \mathcal{A}(v)\|\geq \|v\|$
Then if $v\in \ker(\mathcal{A})$ then
$\|v\|\leq \| \mathcal{A}(v)\|=\|0\|=0$
So $v=0$, that means $\ker(\mathcal{A})=\{0\}$.
Moreover $V$ and $V^*$ have the same dimension, but your map is injective, so it must be also surjective, infact, by nullity+rank theorem, you have
$\dim(V^*)=\dim(V)$
$=\dim(\mathcal{A}(V))+\dim(\ker(\mathcal{A}))=\dim(\mathcal{A}(V) )+0$
So $\dim(\mathcal{A}(V))=\dim(V^*)$ that means
$A(V)=V^*$
So $\mathcal{A}$ is an isomorphism.