This has essentially been asked before here but I guess I need 50 reputation to comment. Also, here I have some questions of my own.
My Proof outline: (forward direction/Necessary direction):
Call the symmetric matrix $A$. Write the quadratic form for $A$ as $x^{t}Ax$, where superscript $t$ denotes transpose. $A$ p.d. (positive definite) implies $x^{t}Ax >0 \ \forall x\neq 0$.
if $v$ is an eigenvector of A, then $v^t Av \ =v^t \lambda v \ =\lambda \ >0$ where $\lambda$ is the eigenvalue associated with $v$. $\therefore$ all eigenvalues are positive.
Any hints for the reverse direction? Perhaps I need to write $A$ as $PDP^{-1} $ where D is a diagonal matrix of the eigenvalues of A and the columns of P are eigenvectors?
Also, a more general question but one that is probably important, is that, since the statement does not assume that A is real (in addition to symmetric), does the possibility of complex entries introduce any complications? Do I need to show that the eigenvalues are real?
Thanks.
Best Answer
If $A$ is symmetric and has positive eigenvalues, then, by the spectral theorem for symmetric matrices, there is an orthogonal matrix $Q$ such that $A = Q^\top \Lambda Q$, with $\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n)$. If $x$ is any nonzero vector, then $y := Qx \ne 0$ and
$$ x^\top A x = x^\top (Q^\top \Lambda Q) x = (x^\top Q^\top) \Lambda (Q x) = y^\top \Lambda y = \sum_{i=1}^n \lambda_i y_i^2 > 0 $$
because $y$ is nonzero and $A$ has positive eigenvalues.
Conversely, suppose that $A$ is positive definite and that $Ax = \lambda x$, with $x \ne 0$. WLOG, we may assume that $x^\top x = 1$. Thus, $$0 < x^\top Ax = x^\top (\lambda x) = \lambda x^\top x = \lambda, $$ as desired.