LinArg assignment 8

exercise

Fitting a parabola

a)

To show that $A$ has full column rank, we can prove that the columns of $A$ are linear independent. Assume $Ax=0$ for some $x=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ Then we must have $x_1{t}_i^2+x_2{t}_i+x_3=0$ for every $i\in [m]$ Since $t_i$ are all distinct to each other and $m\ge 3$, this quadratic equation must have at least three distinct root. Therefore, $x_1,x_2,x_3$ must all be zero Since $x$ is the only possible solution to $Ax=0$ ,$A$ must be linear independent.

Conversely: Counterexample Let $A$ be $\left[\begin{array}{ccc}0& 0& 1\\ 1& 1& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right]$ where $t_1=0,t_2=1,t_3=1,t_4=2$ The matrix $A$ has full rank 3 but $t_2=t_3$, which contradicts the claim

b)

$$\left[\begin{array}{ccc}4& -2& 1\\ 1& -1& 1\\ 0& 0& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right]\left[\begin{array}{c}{\alpha }_2\\ {\alpha }_1\\ {\alpha }_0\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ 1\\ 4\\ 5\end{array}\right]$$

$\left[\begin{array}{c}{\alpha }_2\\ {\alpha }_1\\ {\alpha }_0\end{array}\right]=(A^{\top }A{)}^{-1}{A}^{\top }b=\left[\begin{array}{c}\frac{4}{7}\\ \frac{3}{5}\\ \frac{13}{7}\end{array}\right]$

Uniqueness: yes. Because $A$ has full column rank, $A^{\top }A$ is invertible, so it has exactly one result, hence the least-squares minimizer is unique.