= Solution
For $u\in U$, define $\phi(u)\in U^*$ by $\phi(u)(u')=\langle u,u'\rangle$. If $\phi(u)=0$, then in particular
$$
0=\phi(u)(u)=\langle u,u\rangle=\lVert u\rVert^2,
$$
so $u=0$. Thus $\phi$ is <injective>. Since a finite-dimensional vector space and its <dual space> have the same <dimension of a vector space>[dimension], $\phi$ is also <surjective function>[surjective] and hence an <isomorphism>. This is the finite-dimensional real case of the <Riesz representation theorem>.
The <adjoint operator> of $\alpha:V\to W$ is the unique linear map $\alpha^*:W\to V$ satisfying
$$
\langle\alpha v,w\rangle_W=\langle v,\alpha^*w\rangle_V
$$
for every $v\in V$ and $w\in W$. In the stated <orthonormal basis>[orthonormal bases], if $v$ and $w$ denote coordinate columns, then
$$
\langle Av,w\rangle=(Av)^Tw=v^TA^Tw.
$$
Therefore the matrix of $\alpha^*$ is the <matrix transpose> $A^T$.
For $w\in W$,
$$
w\in\ker\alpha^*
\iff \langle v,\alpha^*w\rangle=0\text{ for every }v
\iff \langle\alpha v,w\rangle=0\text{ for every }v
\iff w\in(\operatorname{im}\alpha)^\perp.
$$
Hence $\ker\alpha^*=(\operatorname{im}\alpha)^\perp$. Taking orthogonal complements in the finite-dimensional space $W$ proves the <image-kernel orthogonality for an adjoint>
$$
\boxed{\operatorname{im}\alpha=(\ker\alpha^*)^\perp}.
$$
Put $r=\alpha(v_0)-w_0$. For any $h\in V$,
$$
\lVert\alpha(v_0+h)-w_0\rVert^2
=\lVert r\rVert^2+2\langle r,\alpha h\rangle+\lVert\alpha h\rVert^2.
$$
This is minimized at $h=0$ exactly when $r\perp\operatorname{im}\alpha$, equivalently when $\alpha^*r=0$. Thus the <normal equation for a linear inverse problem> is
$$
\boxed{\alpha^*\alpha(v_0)=\alpha^*(w_0)}.
$$
For the given <linear least-squares problem>,
$$
A^TA=\begin{pmatrix}2&1\\1&2\end{pmatrix},
\qquad
A^Tb=\begin{pmatrix}3\\5\end{pmatrix}.
$$
The normal equations $2x+y=3$ and $x+2y=5$ have the unique solution
$$
\boxed{x=\frac13,\qquad y=\frac73}.
$$
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