How to prove that det($A^{T}A$) is nonnegative?

Why is the determinant of the product of a matrix and its transpose nonnegative?

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: $$0\le( A\vec{v})\cdot(A\vec{v})=(A\vec{v})^T(A\vec{v})=\vec{v}^TA^TA\vec{v}=\vec{v}^T\lambda\vec{v}=\lambda\vec{v}\cdot\vec{v}$$)

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 $\det A = 0$?