How to prove that det($A^{T}A$) is nonnegative? Why is the determinant of the product of a matrix and its transpose nonnegative? 
 A: I assume that $A$ is real.
$A^TA$ is symmetric, so it is (orthogonally) diagonalizable. So its determinant is the product of its eigenvalues. Let $\lambda$ be an eigenvalue for $\vec{v}$, where $\vec{v}$ is an eigenvector of $A^TA$.
Using inner product notation:
$$0\le\langle A\vec{v},A\vec{v}\rangle=(A\vec{v})^T(A\vec{v})=\vec{v}^TA^TA\vec{v}=\vec{v}^T\lambda\vec{v}=\lambda\langle\vec{v},\vec{v}\rangle$$
This implies that $\lambda$ is nonnegative, since $\langle\vec{v},\vec{v}\rangle>0$. So the determinant is a product of nonnegative real numbers, and therefore a nonnegative real number. Note that this shows something much more specific about $A^TA$ than merely having positive determinant.
(If you prefer dot product notation:
$$0\le( A\vec{v})\cdot(A\vec{v})=(A\vec{v})^T(A\vec{v})=\vec{v}^TA^TA\vec{v}=\vec{v}^T\lambda\vec{v}=\lambda\vec{v}\cdot\vec{v}$$)
A: Assuming $A$ is square, (hence $\det (A)$ is defined):
Recall $$\det(A) = \det(A^T)$$
$$\det(A^TA) = \det(A^T)\det(A)$$ 
What does this imply about $\det(A^TA)$ if 


*

*If $\det A > 0?$ 

*If $\det A < 0$?

*If $\det A = 0$?

A: Consider $A=Q\cdot R$ where $R$ is a square matrix and the columns of $Q$  consist of an orthonormal basis of the column space of $A$. Now $\det(A^T A)=\det((QR)^T(QR))=\det(R^TQ^TQR)=\det(R^T)\det(Q^TQ)\det(R)$, and since $Q^TQ$ is a unit matrix and its determinant is 1, the final product is $\det(R^T)\det(R)=\det(R)\det(R)=\det(R)^2$, which is non-negative (see also amWhy's argument above).
