Proof that determinant of inner product of linearly independant vectors is always positive

Question

If $x,y \: \in \:\mathbb{R}^{3}$ are two linearly independent vectors in three dimensional space, and $\langle x,y\rangle$ represents the dot product of the two vectors, then prove that \begin{align*} \begin{vmatrix} \langle x,x \rangle & \langle x,y\rangle \\ \langle y,x \rangle & \langle y,y\rangle \end{vmatrix} >0 \end{align*}

Answer 1

For two vectors this is equivalent to the Cauchy-Schwarz inequality. For more $k$ vectors, note that replacing them by linear combinations of themselves according to a $k\times k$ matrix $P$ (in coordinates $(xy\ldots z):=(xy\ldots z)P$) changes the Gram matrix (which is what you have before taking the determinant) $M$ into $P^\top MP$, and therefore multiplies its determinant by $\det(P)^2$. In particular if $P$ is upper uni-triangular (each vector only gets modified by a combination of vectors coming before it) then the determinant does not change. This allows performing Gram-Schmidt on the vectors, after which the Gram matrix has become diagonal. Since the diagonal entries are squared norms of the vectors after Gram-Schmidt, the determinant is obviously positive.

Since we are only interested in the sign of the Gram determinant here, we didn't really need to worry about the factor $\det(P)^2$ (as long a $P$ is invertible), and may replace the vectors by any basis whatsoever of the space they span. So we could have done without Gram-Schmidt and just choose any orthonormal basis of the span right away.

The second argument incidentally suggests an easy way to prove the Cauchy-Schwarz inequality.