Prove that a positive definite matrix is nonsingular.
Definitions:
A is positive definite if it is symmetric and $x^TAx>0$ for all x.
A square matrix that is full rank is nonsingular.
Attempt 1:
$x^TAx>0\rightarrow x^T\frac 12 (A+A^T)x>0\rightarrow \frac 12 x^TAx+\frac 12 x^TA^T x>0$ stuck.
Attempt 2:
$x^TAx=x^T\sum x_i\textbf a_i=\sum x_i\langle x,\textbf a_i\rangle$ where $x_i$ is the i-th entry of $x$ and $\textbf a_i$ is the i-th column of A. This seems closer but still does not show that $\sum x_i\textbf a_i\ne0_n$ unless $x_i$ are all $0$. Any ideas?
Prove the contrapositive: if $A$ is singular, then it cannot be positive definite.
If $A$ is singular, then there exists $x$ such that $Ax=0$. Then, ...