Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 8E (Linear Algebra)

A Hermitian form is a sesquilinear form $H:V\times V\to \mathbb{C}$ such thatUse the convention that $H$ is linear in its first argument and conjugate-linear in its second. Its matrix in the basis $(v_1,\dots ,v_n)$ isIf $v={\sum }_i{x}_i{v}_i$ and $w={\sum }_j{y}_j{v}_j$, then

Since $w_i={\sum }_j{p}_{ij}{v}_j$, the coordinate row of each new basis vector is a row of $P$. Direct substitution givesso the new matrix is

The subspace $N=V^{\perp }$ is the radical of $H$. In coordinates it is the kernel of $A$, and the rank-nullity theorem gives

The symmetric matrix of the quadratic form in the standard basis isbecause the off-diagonal entries contribute twice to $x^{T}Ax$.

Apply Gram-Schmidt orthogonalization for a symmetric bilinear form toThese vectors are pairwise orthogonal for $B(u,v)=u^{T}Av$, andThus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrixIt follows that its rank is $3$ and its signature is

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.