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The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

A Hermitian form is a sesquilinear form $H:V\times V\to \mathbb{C}$ such thatUse the convention that $H$ is linear in its first argument and conjugate-linear in its second. Its matrix in the basis $(v_1,\dots ,v_n)$ isIf $v={\sum }_i{x}_i{v}_i$ and $w={\sum }_j{y}_j{v}_j$, then

Since $w_i={\sum }_j{p}_{ij}{v}_j$, the coordinate row of each new basis vector is a row of $P$. Direct substitution givesso the new matrix is

The subspace $N=V^{\perp }$ is the radical of $H$. In coordinates it is the kernel of $A$, and the rank-nullity theorem gives