Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 12H / Solution

Both candidate sets contain zero. The sum of subspaces is closed under linear combinations because $a(u_1+w_1)+b(u_2+w_2)=(a{u}_1+b{u}_2)+(a{w}_1+b{w}_2)$, with the two terms in $U$ and $W$. The intersection is closed because any linear combination of vectors lying in both subspaces still lies in both. Thus each is a vector subspace.

Take a basis $s_1,\dots ,s_r$ of the intersection, extend it by $u_1,\dots ,u_a$ to a basis of $U$, and by $w_1,\dots ,w_b$ to a basis of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the basis of $U$ makes all $u$ coefficients zero; independence of the basis of $W$ then makes all remaining coefficients zero. Hence $\dim (U+W)=r+a+b$, giving the dimension formula for a sum of subspaces

For the concrete kernels, the equations $Ax=0$ giveApplying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has basis $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. ThenIndeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.