Show that $V(F)$ is a subspace to $W.$

Question

Let $F:U\rightarrow W$ be a linear transformation from the vector space $U$ to the vectorspace $W$. Show that the image space to $F$,

$$V(F)=\{w\in W:w=F(u) \ \ \text{for some} \ \ u\in U\},$$

is a subspace of $W$.

Okay, I know that in order for $M$ to be a subspace of a vectorspace $V$ then $M$ has to be

1. non-empty
2. closed under addition with vectors and multiplication with scalars.

So I have to show that $V(F)$ is non empty and closed under addition with vectors and multiplication with scalars.

Can someone break down to me how this is done? I don't really understand what is being stated in the curly brackets and how to apply that to show 1. and 2.

Answer 1

The curly bracket says that $w$ is an element of the set $V(F)$ if and only if $w$ is the result of applying $F$ to some element $u \in U$.

For instance, if $F: \mathbb{R}^2 \to \mathbb{R}^3$ is given by $F(x,y)=(x,0,0)$, then $(1,0,0)$ is in $V(F)$, since $(1,0,0)=F(1,3)$, for example. But $(3,2,0)$ is not in $V(F)$, since if it were, its second coordinate would have to be $0$.

So, suppose that $w_1$ and $w_2$ are in $V(F)$. Then they are the image of some $u_1$ and $u_2$, respectively (i.e., $F(u_1)=w_1$ and $F(u_2)=w_2$).

We want to prove that $w_1+w_2$ is still in $V(F)$. But since $F$ is linear, you have a pretty good candidate for an element $u$ such that $F(u)=w_1+w_2$. Can you tell who it is?

Repeat this train of thought for the rest.

Answer 2

The curly brackets are just defining the image of F. F is a mapping. It takes each element of $U$ to an element of $W$. $V(F)$ is just the set of elements of $W$ that get mapped to by an element of $U$. In other words, take each $x \in U$ and apply $F$. We get $F(u)$ which is an element of $W$. Collect all the elments of W obtained this way and call this set $V(F)$.

Now take two elements of $V(F)$. Call them $w_1$ and $w_2$. This means there is a $u_1$ and $u_2$ in $U$ such that $F(u_1)=w_1$ and $F(u_2)=w_2$. We then have $$w_1 + w_2 = F(u_1) + F(u_2) = F(u_1+u_2)$$ Thus $w_1 + w_2$ is in $V(F)$ since it gets mapped to by $u_1+u_2$.

Proceed similarly to finish the proof that $V(F)$ is a subspace.