 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 ${\sum }_i{a}_i{g}_i$, giving $f(v)=0$. If $v\notin U$, part (b) supplies a functional in $U^0$ with $f(v)\ne 0$, a contradiction. ThereforeFor the dimension statement, take a basis $u_1,\dots ,u_m$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. Relative to the corresponding dual basis, a functional belongs to $U^0$ precisely when its first $m$ coefficients are zero. ThusIf $U=V$, 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 kernels are the two coordinate planesTheir intersection is the stated axis, and $k=2=3-1$.