Exercise 3.F.23

Proof. Use the notation from Theorem 1 in Chapter 2 notes.

Extend $v_1,\dots ,v_n,u_1,\dots ,u_m,w_1,\dots ,w_p$ to a basis $v_1,\dots ,v_n,u_1,\dots ,u_m,w_1,\dots ,w_p,s_1,\dots ,s_q$ of $V$. Let ${\phi }_1,\dots ,{\phi }_n,{\psi }_1,\dots ,{\psi }_m,{\omega }_1,\dots ,{\omega }_p,{\sigma }_1,\dots ,{\sigma }_q$ be its dual basis.

We will prove $U^0=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->({\omega }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\omega }_p,{\sigma }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\sigma }_q)$.

We have

Where the first equation follows from 3.106. Therefore ${\omega }_1,\dots ,{\omega }_p,{\sigma }_1,\dots ,{\sigma }_q$ is a linearly independent list of length $\dim <!--\; FUNCTION\; APPLICATION\; -->U^0$, in other words, a basis of $U^0$. Similarly we can prove $W^0=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->({\psi }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\psi }_m,{\sigma }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\sigma }_q)$ and ${(U\cap W)}^0=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->({\psi }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\psi }_m,{\omega }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\omega }_p,{\sigma }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\sigma }_q)$ and now we can see that ${(U\cap W)}^0=U^0+W^0$ by the definition of sum of subspaces. □