= Positive-definite bilinear form
{title2=$B(v,v)>0\quad(v\ne0)$}
= Positive-definite
{synonym}
A real <symmetric bilinear form> is <positive-definite> when $B(v,v)>0$ for every nonzero <vector>. It defines an <inner product> and the associated <norm> $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.https://www.ma.imperial.ac.uk/~anskor/LINEAR.PDF[Alexei Skorobogatov, Linear Algebra and Matrices, pp. 13–14: positive forms and inner products]{ref}
The <identity matrix> $A=\operatorname{diag}(1,1)$ gives $B(v,w)=v_1w_1+v_2w_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving <norm> $5$. A weighted example, $A=\operatorname{diag}(4,1)$, gives <norm> $\sqrt{4x^2+y^2}$: the first <coordinate> costs twice as much in length. Such <inner products> let geometry assign different weights to different directions.
Every <positive-definite bilinear form> is <nondegenerate>: a <vector> in its <radical of a bilinear form> has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\operatorname{diag}(1,-1)$ shows. Even positivity of the associated <quadratic form> does not force a general <bilinear form> to be symmetric: $A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$ has $v^TAv=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.
The complex <Killing form> is not positive-definite on the entire complex <Cartan subalgebra>; its restriction to the <Euclidean subspace of a Cartan subalgebra> is a real positive-definite form.
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