Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Algebra / 2

A real inner product is a symmetric bilinear map $⟨\cdot ,\cdot ⟩:V\times V\to \mathbb{R}$ with $⟨v,v⟩>0$ for $v\ne 0$. It defines $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Set $W=\mathrm{span}\{e_1,\dots ,e_k\}$ and $Pv={\sum }_i⟨v,e_i⟩e_i$. Orthonormality gives $v-Pv\perp W$, hence the Pythagorean identity givesThis proves Bessel's inequality. Equality holds exactly when $v\in W$.

If $v=0$, equality is immediate. If $v\ne 0$, the nonnegative squared normproves the Cauchy-Schwarz inequality. Equality holds precisely when the displayed vector vanishes. Thus equality is equivalent to linear dependence of $u,v$, including the cases with a zero vector.

The sum in question is $⟨Pu,Pv⟩$. Applying Cauchy-Schwarz and then Bessel's inequality givesIf either original vector is zero, equality holds. If both are nonzero, equality forces both projection norms to equal their original norms and also forces dependence of the projected vectors. Thus apart from the zero-vector cases, equality holds exactly when $u,v\in W$ and $u,v$ are linearly dependent.

For the final unnumbered request, apply Cauchy-Schwarz in ${\mathbb{R}}^k$ to $p_i=a^{i/2}$ and $q_i=a^{-i/2}$. Their inner product is $k$, soEquality requires the positive vectors $p,q$ to be proportional, equivalently $a^i$ constant for $i=1,\dots ,k$. For $k\ge 2$ this is $a=1$. If $k=1$, equality holds for every $a>0$.