We say $A$ is a positive definite matrix if and only if $x^T A x > 0$ for all nonzero vectors $x$. Then why does every positive definite matrix have strictly positive eigenvalues?
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$\begingroup$Suppose our matrix $A$ has eigenvalue $\lambda$.
If $\lambda = 0$, then there is some eigenvector $x$ so that $Ax = 0$. But then $x^T A x = 0$, and so $A$ is not positive definite.
If $\lambda < 0$, then there is some eigenvector $x$ so that $Ax = \lambda x$. But then $x^T A x = \lambda \lvert x \rvert^2$, which is negative since $\lvert x \rvert^2 > 0$ and $\lambda < 0$. Thus $A$ is not positive definite.
And so if $A$ is positive definite, it only has positive eigenvalues.
$\endgroup$ 4 $\begingroup$Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.
$\endgroup$ $\begingroup$As you said, we say $A \in \mathbb{R}^{n\times n}$ is a positive definite matrix if and only if $x^T A x > 0$ for all nonzero vectors $x \in \mathbb{R}^n$.
Consider the eigenvector $v \neq 0$ and its associated eigenvalue $\lambda$. It means $Av = \lambda v$.
Since $A$ is positive definite, it means $v^T A v > 0$, wich also means $\lambda v^Tv > 0$. We have that:
$$v^Tv = \sum\limits_{i=1}^{n} v_i^2$$
which is always strictly positive. So $\lambda$ must be strictly positive.
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