Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 11

Put $\mu =|b{|}^2>0$ and $\lambda =a\cdot b$. Expanding the nonnegative squared Euclidean norm of $c=\mu a-\lambda b$ givesDividing by $\mu$ and taking nonnegative square roots proves the Cauchy-Schwarz inequality:Equality holds exactly when $c=0$, equivalently when the nonzero vectors $a,b$ are linearly dependent.

Expanding the squared Euclidean norm and applying the Cauchy-Schwarz inequality givesBoth sides are nonnegative, so taking square roots proves the triangle inequality, $\boxed{|a+b|\le |a|+|b|}$. For nonzero vectors, equality requires a positive real proportionality factor.

With conjugation in its first slot, a complex inner product has conjugate symmetry, linearity in its second slot, and positive definiteness. For the proposed Hermitian inner product, direct expansion givesThe last expression vanishes exactly for the zero vector. Conjugate symmetry and linearity also give antilinearity in the first slot: $⟨\alpha z+\beta w,y⟩=\bar{\alpha }⟨z,y⟩+\bar{\beta }⟨w,y⟩$. These verify all three axioms and the associated first-slot consequence.

Using the Hermitian inner product, the scalar product is $1-i+2i^2=-1-i$. The two squared norms are $6$ and $4$. The specified real angle in a complex inner product space therefore hasThe real part is essential: the complex scalar product itself need not be real.

Conjugating the first vector givesTheir squared norms are $3$ and $2$, independently of $x,n$. Thus the specified real angle in a complex inner product space obeysThe complex Cauchy-Schwarz inequality ensures that this real quotient lies in $[-1,1]$.