= Positive definiteness
{title2=$q(v)>0\quad(v\ne0)$}
The strict positivity of a <quadratic form> on every nonzero <vector>. A real symmetric <matrix> has positive definiteness exactly when its associated <quadratic form> does, making it a <positive-definite matrix>. A <Hermitian form> is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.
Think of a <positive-definite quadratic form> as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real <symmetric matrix>
$$
A=\begin{pmatrix}3&1\\1&3\end{pmatrix},\qquad q(x,y)=3x^2+2xy+3y^2=3\left(x+\frac y3\right)^2+\frac83y^2
$$
is a <positive-definite matrix> because the two squares can vanish together only at $(0,0)$. The mixed term couples the <coordinates>, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive <level sets> are <ellipses> with axes along the <eigenvectors>, rather than along the original <coordinates>.
\Image[/positive-quadratic-energy-elliptical-levels.png]
{title=Positive quadratic energy bowl and elliptical level curves}
The real and complex settings require different symmetry assumptions. For a real <symmetric matrix> use $q(x)=x^TAx$ with $x\in\mathbb R^n$; for a complex <Hermitian matrix> use $q(z)=z^*Az$ with $z\in\mathbb C^n$, where $z^*$ is the <conjugate transpose>. The latter expression is real, so strict positivity is meaningful. A complex <quadratic form> such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A <Hermitian form> uses <complex conjugation>, which changes this example to $|z_1|^2+|z_2|^2$.
For a real <matrix> $M$ without a symmetry hypothesis, its <quadratic form> depends only on its <symmetric part of a matrix>:
$$
x^TMx=x^T\frac{M+M^T}{2}x.
$$
Positive <eigenvalues> of $M$ alone do not establish <positive definiteness> of this <quadratic form>. For example, $M=\begin{pmatrix}1&4\\0&1\end{pmatrix}$ has both <eigenvalues> equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the <symmetric part of a matrix>, not the <spectrum> of an arbitrary <matrix>.<Bilinear Forms by Keith Conrad>[Theorem 1.7, equation (1.3), p. 3: symmetric and alternating parts]{ref}
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