Volume of a k parallelopiped

Question

When we study Manifolds we often encounter the formula for volume of a parallelopiped as

Vol=$\sqrt{det(X.X^{T})}$

My question is: Why is the above determinant($det(X.X^{T})$) always non negative ? Can someone provide an easy proof or intuition ?(It is obvious in the case where $X$ is square,can you prove the other cases)

Answer 1

Two approaches:

First, it suffices to note that if $X$ has linearly-independent rows, then $XX^T$ is positive definite. It follows that the eigenvalues of $XX^T$ are positive, so that $XX^T$ has positive determinant.

Another approach is to use the Cauchy-Binet formula to find that $$ \det(AA^T) = \sum_{S \subset \{1,\dots,n\}, |S| = m} \det(A_S)^2 $$ where $A_S$ denotes the $m \times m$ matrix whose columns are the $i$th column of $A$ for all $i \in S$. Because the above is a sum of squares, it must be non-negative. When $A$ has linearly independent rows, there exists an $S$ such that $\det A_S$ is non-zero, so that the above sum is in fact positive.

Answer 2

Note that $X^TX$ is self-adjoint, hence it is diagonalizable. Also note that $(x,X^TXx)=(Xx,Xx)=||Xx||^2\geq 0$ for any vector $x$. Hence, all eigenvalues are nonnegative.

Hence we can write $X^TX=VDV^*$ where $D$ is a diagonal matrix with nonnegative entries and $V$ is unitary. Hence $det(X^TX)=det(D)\geq 0$.