How to prove: for every two complementary subspaces there exists a projector

Question

In Trefethen and Bau's book, Computational Linear Algebra, in the Projections chapter I've come across the following statement:

Let $S_1$ and $S_2$be two subspaces of $\mathbb{C}^m$ such that $S_1 \cap S_2 = \{0\} $ and $S_1+S_2=\mathbb{C}^m$ (where the sum denotes the span of $S_1$ and $S_2$. Then there exists a projector P such that $range(P)=S_1$ and $null(P)=S_2$

They claimed that an equivalent problem is:

Given $v$, find vectors $v_1 \in S_1$, and $v_2 \in S_2$ such that $v_1+v_2=v$.

and the proof to the problem is:

The projection $Pv$ gives $v_1$ and the complementary projection $(I-P)v$ gives $v2$

1. My Questions: why are the problems equivalent, namely, why does the second problem suggest that there in fact exists a projection $P$?
2. how does their proof suggests that there exists a projection $P$?

Answer 1

Suppose that $S_1$ and $S_2$ are two subspaces such that $S_1 \cap S_2 = \{0\}$ and $S_1 + S_2 = \mathbb{C}^m$.

Suppose that for any $v$ there are $v_1 \in S_1$ and $v_2 \in S_2$ such that $v_1 + v_2 = v$. The crucial thing to observe is that these $v_1$ and $v_2$ are unique with such a property. Indeed, suppose there are also $u_1 \in S_1$ and $u_2 \in S_2$ such that $v = u_1 + u_2$. Then $0 = v - v = (v_1 - u_1) + (v_2 - u_2)$ and because $S_1$ and $S_2$ are subspaces we have $v_1 - u_1 \in S_1$ and $v_2 - u_2 \in S_2$.

From $0 = (v_1 - u_1) + (v_2 - u_2)$ we have $v_1 - u_1 = u_2 - v_2$, thus $v_1 - u_1$ and $u_2 - v_2$ are both in $S_1$ and $S_2$. Hence $v_1 - u_1 = u_2 - v_2 = 0$, since $S_1\cap S_2 = \{0\}$.

Thus define the projection $P$ using this fact. Given $v$ there are unique $v_1 \in S_1$ and $v_2 \in S_2$ such that $v = v_1 + v_2$. Define $Pv = v_1$.

Now one needs to check that this is indeed a linear operator. If $v_1 + v_2$ is the decomposition of $v$ and $u_1 + u_2$ is the decomposition of $u$ then, because $S_1$ and $S_2$ are subspaces we have $(\alpha v_1 + \beta u_1) \in S_1$ and $(\alpha v_2 + \beta u_2) \in S_2$ and so $(\alpha v_1 + \beta u_1) + (\alpha v_2 + \beta u_2)$ is the decomposition of $\alpha v + \beta u$. Thus $P(\alpha v + \beta u) = \alpha P v + \beta P u$ by definition.

Now if $v \in S_1$ then clearly $v = v + 0$ and so $Pv = v$ and by definition we have that $Pv \in S_1$ for any $v \in \mathbb{C}^m$. So $\text{range}(P) = S_1$.

If $Pv = 0$ then $v = 0 + v_2$ for some $v_2 \in S_2$ but clearly $v_2 = v$ hence $\text{null}(P) = S_2$.

Answer 2

I don't think their proof is complete. It seems like they assume that there already exists a (unique) decomposition of any $v$ into $v_1+v_2$ (once given this decomposition though, their proof is fine (other than skipping a few steps) - perhaps earlier in the book they proved such a decomposition exists).

Why is their proof fine if we assume the (unique) decomposition? Let's fill out the details.

First we prove such a projection exists: Define an operator $P$ such that $P(v)=v_1$. Then $P$ is certainly a projection because $P(P(v))=P(v_1)=v_1$, and so $P(P(v))=P(v)$ for every $v\in\mathbb{C}^m$.

Now we just need to prove that $\operatorname{range}(P)=S_1$ and $\operatorname{null}(P)=S_2$.

Proof of $\operatorname{range}(P)=S_1$: Note that $P(v)=v_1\in S_1$, so $\operatorname{range}(P)\subset S_1$. However, for every $w\in S_1$, $P(w)=w$ and so we also have $S_1\subset\operatorname{range}(P)$. Hence $\operatorname{range}(P)=S_1$.

Proof of $\operatorname{null}(P)=S_2$: First suppose $P(v)=0$. Then $v=0+v_2=v_2$ where $v_2\in S_2$. Hence $\operatorname{null}(P)\subset S_2$. However, for every $w\in S_2$, we have $w=0+w$ and $P(w)=0$, and so we also have $S_2\subset\operatorname{null}(P)$. Hence $\operatorname{null}(P)=S_2$.

In short we have shown that if we are given the existence of such a decomposition, then it implies the existence of the required projection $P$.