Exercise 1.9.7

When $n=2$, the matrix $A$ has maximum rank of $k=2$, so it makes sense to discuss the case when $k=1$, otherwise if $k=2$, the best approximation of rank-2 is the $A$ matrix itself.

So we have $\mathit{CR}=(m\times 1)(1\times 2)$, so $D=C^{T}C$ is a 1 by 1 matrix, i.e. a number. From the second derivative, we have $A^{T}A{R}^T=R^{T}D$, so it’s clear that $D$ is the eigenvalue of $A^{T}A$, and $R^T$ is the eigenvector. From the first derivative, we have $A{R}^T=C$, so $A{A}^{T}C=A{R}^{T}D=\mathit{CD}$, so $C$ is the eigenvector of $A{A}^T$.