= Solution
Every $g_i\in U^0$ vanishes on $U$, so $U\subseteq\bigcap_i\ker g_i$. Conversely, suppose $v\in\bigcap_i\ker g_i$. Every $f\in U^0$ is a <linear combination> $\sum_i a_ig_i$, giving $f(v)=0$. If $v\notin U$, part (b) supplies a functional in $U^0$ with $f(v)\neq0$, a contradiction. Therefore
$$
\boxed{U=\bigcap_{i=1}^k\ker g_i.}
$$
For the dimension statement, take a <basis> $u_1,\ldots,u_m$ of $U$ and extend it to a <basis> $u_1,\ldots,u_n$ of $V$. Relative to the corresponding <dual basis>, a functional belongs to $U^0$ precisely when its first $m$ coefficients are zero. Thus
$$
\boxed{k=\dim U^0=n-m=\dim V-\dim U.}
$$
If $U=V$, this says $k=0$ and the empty intersection is understood to be $V$.
In the coordinate example choose the <linear functionals> $g_1(x,y,z)=y$ and $g_2(x,y,z)=z$. Their <kernel of a linear map>[kernels] are the two coordinate planes
$$
\boxed{W_1=\{(x,0,z):x,z\in\mathbb R\},\qquad
W_2=\{(x,y,0):x,y\in\mathbb R\}.}
$$
Their intersection is the stated axis, and $k=2=3-1$.
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