Show that the projection is $\in End(V)$

Question

Consider a direct sum $V=U\oplus W$ where $U,W \subseteq V$. In my lecture notes I have been given the definition of the projection on U along W: $E\in End(V), E(v)=u$ with $v\in V$ and $u\in U$. I have then been given the task to prove that $E$ is indeed in $End(V)$, however, to me it seems that that simply follows from the definition, since $u\in U$ and $U\subseteq V$. How else would I prove this statement?

Answer 1

Each element of $V$ can be uniquely written as $v=u+w$ for some $u\in U$ and $v\in V$.

Thus the mapping $f:V\rightarrow U: v=u+w\mapsto u$ is well-defined.

Let $v_i=u_i+w_i\in V$ as above, $i=1,2$.

Then $f(v_1+v_2) = f((u_1+w_1)+(u_2+w_2)) = f((u_1+u_2) + (w_1+w_2)) = u_1+u_2$

and

$f(v_1)+f(v_2) = u_1 + u_2$.

Let $v=u+w\in V$ as above and $k\in K$ (scalar).

Then $f(kv) = f(ku+kw) = ku$

and

$k\cdot f(v) = ku$.

Hence $f$ is a linear mapping as claimed.

Answer 2

Hint

What you're requested to do is to prove that $E$ is linear.

Do do so, use the fact that by definition of $V=U\oplus W$, the decomposition for any vector exists and is unique. Then prove that for the sum of two vectors the decomposition is the sum of the decompositions by unicity. Same for the multiplication by a scalar.

Answer 3

You have to prove that for every vector $v\in V$ and for every scalar $\lambda$, you have $$E(\lambda v)=\lambda E(v)$$ and that for every vectors $v,v'\in V$, you have $$E(v+v')=E(v)+E(v)$$ If you want, you can perform both checks at once by proving that $E$ preserves linear combinations of vectors, that is $$E(\lambda v+\mu v')=\lambda E(v)+\mu E(v')$$ It is not difficult to see that $E$ is indeed linear, given the direct decomposition of $V$ as $U\oplus W$, which allows you to write each vector $v\in V$ as a suitable sum of two vectors, in a unique way.