If $U$ is a finite-dimensional real vector space with inner product $\langle\cdot, \cdot\rangle$, prove that the linear map $\phi: U \rightarrow U^{*}$ given by $\phi(u)\left(u^{\prime}\right)=\left\langle u, u^{\prime}\right\rangle$ is an isomorphism. [You do not need to show that it is linear.]

If $V$ and $W$ are inner product spaces and $\alpha: V \rightarrow W$ is a linear map, what is meant by the adjoint $\alpha^{*}$ of $\alpha$ ? If $\left\{e_{1}, e_{2}, \ldots, e_{n}\right\}$ is an orthonormal basis for $V,\left\{f_{1}, f_{2}, \ldots, f_{m}\right\}$ is an orthonormal basis for $W$, and $A$ is the matrix representing $\alpha$ in these bases, derive a formula for the matrix representing $\alpha^{*}$ in these bases.

Prove that $\operatorname{Im}(\alpha)=\operatorname{Ker}\left(\alpha^{*}\right)^{\perp}$.

If $w_{0} \notin \operatorname{Im}(\alpha)$ then the linear equation $\alpha(v)=w_{0}$ has no solution, but we may instead search for a $v_{0} \in V$ minimising $\left\|\alpha(v)-w_{0}\right\|^{2}$, known as a least-squares solution. Show that $v_{0}$ is such a least-squares solution if and only if it satisfies $\alpha^{*} \alpha\left(v_{0}\right)=\alpha^{*}\left(w_{0}\right)$. Hence find a least-squares solution to the linear equation

$$\left(\begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right)$$