If $A = B^{T}B $, then all the eigenvalues of $A$ are non-negative How can I prove that if $A$ & $B$ are symmetric matrices, and $A = B^{T} B$ , then all the eigenvalues of $A$ are non-negative? I understand that I can take the transpose manipulation which yields:
$$ X(AX) = X\left(\left(B^TB\right)X\right) = X\left(B^T\left(BX\right)\right) = (BX)(BX) = (||BX||)^2 .$$
Because $(||{BX}||)^2$ cannot be negative, $X(AX) \geq 0$ no matter what $X$ is.
But I'm struggling to move forward from here to explain why all the eigenvalues of $A = B^{T} B$ are non-negative?

HINT: Consider your calculation in the event that $AX = \lambda X$.
