# Proof that $\exists f:W_1\to W_2:W=\{v\in V \mid \exists w_1\in W_1:v=w_1+f(w_1)\}$

Let $$V$$ be a finite-dimensional vector space and $$W,W_1,W_2$$ subspaces of $$V$$, such that $$V=W_1\oplus W_2$$, $$W \cap W_2=\{0\}$$ and $$\dim W= \dim W_1$$. Prove that there exists a linear transformation $$f:W_1\to W_2$$, such that $$W=\{v\in V \mid \exists w_1\in W_1:v=w_1+f(w_1)\}$$

My work so far:

Using that $$\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2)$$, for subspaces $$U_1,U_2$$ of $$V$$, we get that $$V=W\oplus W_2$$ and now my aim was to find an explicit linear transformation which has the required properties but I haven't been successful with this so far.

• Do you require $f$ to be a linear transformation? – Aryaman Maithani Jan 2 at 17:46
• @AryamanMaithani Oh yes I do, thank you, I have edited the question. – user Jan 2 at 17:56

Let us construct a function $$f:W_1 \to W_2$$ as follows.
Given any $$w \in W$$, it has a unique representation as $$w = w_1 + w_2,$$ where $$w_1 \in W_1$$ and $$w_2 \in W_2$$.
We define $$f(w_1) = w_2$$. It is clear that this $$f$$ does actually satisfy the condition you want. That is, $$W = \{w_1 + f(w_1) \mid w_1 \in W_1\}$$.
However, what is still left is to show that $$f$$ is indeed a well-defined linear function.

First, we show given any $$w_1 \in W,$$ there do exist vectors $$w \in W$$ and $$w_2 \in W_2$$ such that $$w = w_1 + w_2$$, that is, $$f(w_1)$$ has some value.
This is easy as $$w_1 \in W_1 \subset V = W \oplus W_2.$$ Thus, $$w_1 = w + w_2$$ for some $$w \in W$$ and some $$w_2 \in W_2$$. Rearranging the equation and using the fact that $$-w_2 \in W_2$$ gives us the desired result.

Secondly, we show that there is no ambiguity with this choice of $$f(w_1).$$
Suppose that $$w_1 + w_2 = w \in W \ni \widehat{w} = w_1 + \widehat{w_2}$$ for some $$w_2, \widehat{w_2} \in W_2.$$ We want to show that $$w_2 = \widehat{w_2}.$$
Note that $$w - \widehat{w} = w_2 - \widehat{w_2}.$$
The LHS is an element of $$W$$ and the RHS of $$W_2.$$ Thus, we get that $$w_2 - \widehat{w_2} \in W \cap W_2 = \{0\}$$, that is, $$w_2 = \widehat{w_2}$$, as desired.

Thirdly, we show that $$f$$ is indeed a linear map.
Suppose $$x, y \in W_1$$ and $$\alpha \in \mathbb{F}$$, the field over which $$V$$ is a vector space.
By construction, we have that $$\alpha x + f(\alpha x) = X \in W$$ and $$y + f(y) = Y \in W$$.
Thus, $$\underbrace{\alpha x + y}_{\in W_1} + \underbrace{f(\alpha x) + f(y)}_{\in W_2} = \underbrace{X+Y}_{\in W}.$$
By our definition, we get that $$f(\alpha x + y) = f(\alpha x) + f(y)$$ and hence, $$f$$ is a linear map.

Hint:

From the hypothesis we have $$V=W_1\oplus W_2=W\oplus W_2$$ so for every $$v\in V$$ we can write by uniquely way: $$v=w_1+w_2=w+\omega_2$$ where $$w_1\in W_1, w\in W$$ and $$w_2,\omega_2\in W_2$$. Hence we have

$$w=w_1+ \underbrace{(w_2-\omega_2)}_{\in W_2}$$ Now we construct the function $$f:W_1\to W_2, w_1\mapsto w_2-\omega_2$$ and we prove that $$f$$ is a linear transformation and we are done.

• Thank you very much for the answer but don't we have the problem that we can find $\hat{v}\neq v\in V$, such that $\hat{v}=w_1+\hat{w_2}$ and then $f$ would not be well-defined, would it? – user Jan 2 at 18:58