Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 4 (Gram matrices and determinant bounds)

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

The two-vector Gram matrix isThe upper bound follows because the subtracted square is nonnegative. If either vector is zero, the determinant is zero. Otherwise apply the non-strict version of the calculation in i: $\Vert v_1-t{v}_2{\Vert }^2\ge 0$ with $t=⟨v_1,v_2⟩/\Vert v_2{\Vert }^2$ gives $⟨v_1,v_2{⟩}^2\le \Vert v_1{\Vert }^2\Vert v_2{\Vert }^2$. ThereforeEquality on the left holds exactly when the two vectors are linearly dependent; equality on the right holds exactly when they are orthogonal. The Gram determinant measures the squared area of their parallelogram: in the nonzero case, $\det A=\Vert v_2{\Vert }^2\Vert v_1-t{v}_2{\Vert }^2$, the square of base times perpendicular height.

Bilinearity of the real inner product gives, for the specified linear combinations,The last equality is the entrywise rule for matrix multiplication, with $x^T$ the transpose of the column vector $x$.

Symmetry of the real inner product gives $A^T=A$. Moreover, i with $y=x$ shows thatso the Gram matrix is a positive semidefinite matrix. To also check reality of every eigenvalue, allow a nonzero vector $z=a+ib\in {\mathbb{C}}^n$ with $Az=\lambda z$. Since $A$ is a real symmetric matrix,Here $z^{\ast }$ is the conjugate transpose, and $a^{T}Ab=b^{T}Aa$ cancels the imaginary part. Thusbecause $z^{\ast }z>0$. This proves both that every eigenvalue is real and that it is nonnegative.

By the spectral theorem for real symmetric matrices, there is an orthogonal matrix $Q$ such thatFor clarity, the diagonalization can also be justified recursively. The characteristic polynomial has a complex root of a polynomial by the fundamental theorem of algebra. By ii its eigenvalue is real; taking a nonzero real or imaginary part of an associated eigenvector in ${\mathbb{C}}^n$ gives a real eigenvector, which can be normalized. Its orthogonal complement is invariant, because $⟨Ax,z⟩=⟨x,Az⟩=\lambda ⟨x,z⟩=0$. Restriction to this linear subspace is again symmetric, and induction on dimension gives an orthonormal basis of eigenvectors.

All ${\lambda }_i\ge 0$ by ii. The permitted invariance of determinant and matrix trace under change of basis, applied to the diagonal matrix, yields

If every $\Vert v_i\Vert =1$, then each diagonal entry of the Gram matrix is $A_{ii}=⟨v_i,v_i⟩=1$, so its matrix trace is $n$. Using iii and the permitted arithmetic-geometric mean inequality,The bound is attained by any orthonormal family, whose Gram matrix is the identity matrix.