Explain proof that any positive definite matrix is invertible

Note that if $Ax=0=0\cdot x$ for some $x\ne 0$ then by definition of eigenvalues, $x$ is an eigenvector with eigenvalue $\lambda = 0$, contradicting that $0$ is not an eigenvalue of $A$. $$Ax=\lambda x$$

$$\det A = \prod_{j=1}^n \lambda_j \implies \det A = 0 \Leftrightarrow \exists\ i \in \{1,2,\ldots, n\}:\lambda_i = 0$$

In other words, the determinant is the product of the eigenvalues, and can only be zero if at least one eigenvalue is zero.

Because if $Ax=0$ for some nonzero $x$, then $0$ would be an eigenvalue.