Let $P,P' $ be proper subspaces of finite dimension vector space $V$ $\Rightarrow$ $\exists$ $S$ , such that $V = S \bigoplus P = S \bigoplus P'$

Question

I would like to know how to prove the following theorem:

Theorem: Let $V$ be a vector space of dimension $n$. If $P$ and $P'$ are subspaces of $V$, such that $\dim(P) = \dim(P') = k$, then exists $S$, subspace of $V$, satisfying $\dim(S) = n - k$ and $\{0\} = P \cap S = P' \cap S.$

Any hints?

Answer 1

Hint:

Make a decreasing induction on $k$.

Prove $P \cup P' \neq V$.

Take $x$ $\notin P \cup P' $. Then let $Q := P + Vect(x)$ and $Q' := P' + Vect(x)$, and use the induction hypothesis on $Q$ and $Q'$.

Answer 2

Hints:

Let $B = \{a_1, \ldots, a_t\}$ be a basis for $P \cap P'$.

Since $P \cap P'$ is a subspace of both $P$ and $P'$, you can extend $B$ to bases $\{a_1, \ldots, a_t, b_1, \ldots, b_r\}$ for $P$ and $\{a_1, \ldots, a_t, c_1, \ldots, c_r\}$ for $P'$.

Show that $$\{a_1, \ldots, a_t, b_1, \ldots, b_r, c_1, \ldots, c_r\}$$ is a basis for $P + P'$.

Extend $\{a_1, \ldots, a_t, b_1, \ldots, b_r, c_1, \ldots, c_r\}$ to a basis $$\{a_1, \ldots, a_t, b_1, \ldots, b_r, c_1, \ldots, c_r, d_1, \ldots, d_q\}$$

for $V$.

Finally, show that the desired subspace $S$ is spanned by $$\{b_1+c_1, b_2 + c_2, \ldots, b_r + c_r, d_1, \ldots, d_q\}$$