# (Basic Linear Algebra) Eigenvalues of SPD Matrix

I think I may have missed some basic Linear Algebra knowledge at some point, so I want to clarify the relationship between two things I (think I) know about Matrices:

1. A matrix $A$ is described as positive-definite if: $\;\;x^\top A x >0 \;\;\;\forall x \in \mathcal{R^n}$

2. For a symmetric positive-definite matrix, all its eigenvalues are $\;>0$

What is the relationship between $x^\top A x$ and the eigenvalues of $A$, if any?

As an aside, is there a name for the operation $x^\top A x$ ? The term appears a lot in many contexts (seems like a quadratic, also similar to a decomposition?) and I'd probably find it easier to learn about the relevant properties if I knew a name to look for.

• Note that if $\lambda$ is an eigenvalue of a positive matrix $A$ and $0\neq x$ a corresponding eigenvector then $$0<x^{T}Ax=\lambda x^Tx$$ and since $x^Tx>0$ this implies $\lambda>0$. – Peter Melech May 11 '18 at 12:48

$x^\top A x$ is the quadratic form associated with matrix $A$ where $x$ is any non-zero vector. The definiteness of $A$ is defined according to its sign. (positive definite if $>0$ ; semi-positive definite if $\geq 0$ and so on...)
If $A$ is a positive definite matrix, i.e. $x^\top A x>0$ for any non-zero vector $x$, then its eigenvalue is positive. (why?) Same result for semi-definite matrix, just replace $>$ with $\geq$.