Over the reals, non-negative and positive operators correspond exactly to positive semidefinite and positive definite symmetric forms

For a linear map $T$, the form $B_T(u,v)=⟨Tu,v⟩$ is bilinear. Conversely, let $B$ be a bilinear form on $V$. By [L1], the assignment $u\mapsto B(u,-)$ is a linear map $V\to V^{\ast }$. For each $u\in V$, [L2] gives a unique vector $Tu\in V$ such that $B(u,v)=⟨Tu,v⟩$ for every $v\in V$. If $a,b\in \mathbb{R}$ and $u,u^{\mathrm{\prime }}\in V$, then for every $v$ one has $⟨T(au+b{u}^{\mathrm{\prime }}),v⟩=B(au+b{u}^{\mathrm{\prime }},v)=aB(u,v)+bB(u^{\mathrm{\prime }},v)=⟨aTu+bT{u}^{\mathrm{\prime }},v⟩,$ so uniqueness in [L2] gives $T(au+b{u}^{\mathrm{\prime }})=aTu+bT{u}^{\mathrm{\prime }}$. Thus $T$ is linear, and the correspondence $T\mapsto B_T$ is bijective on all linear maps and bilinear forms. In the real case, $T$ is self-adjoint exactly when $⟨Tu,v⟩=⟨u,Tv⟩$ for all $u,v$, and symmetry of the inner product makes this exactly the condition $B_T(u,v)=B_T(v,u)$. Thus self-adjoint endomorphisms correspond exactly to symmetric bilinear forms.

For every $v\in V$, one has $B_T(v,v)=⟨Tv,v⟩$. Therefore the weak inequality in Non-negative and positive operators is exactly the positive-semidefinite condition in [L3], and the strict inequality on nonzero vectors is exactly the positive-definite condition in [L3].