prove that a Projection is a linear transfor

Question

given $V$ be a vector space over $\mathbb{F}$. let $P:V\longrightarrow V$ be a function $\negthickspace$ as $V=U_{1}\oplus U_{2}$, and for every $v$, $v=u_{1}+u_{2}$ as $u_{1}\in U_{1}$ , $\!u_{2}\in U_{2}$.

$P(v)=p(u_{1}+u_{2})=u_{1}$. which means $P$ is a Projection.

I need to prove that p is a linear transformation.

I know that you need to start with:

let $v_{i}$,$v_{j}\in V$ as $v_{i}=u_{1i}+u_{2i}$ , $v_{j}=u_{1j}+u_{2j}$
$P(v_{i}+v_{j})=P(u_{1i}+u_{1j}+u_{2i}+u_{2j})$ and P$(v_{i})=u_{1i}$, $P(v_{j})=u_{1j} \Longrightarrow P(v_{i})+P(v_{j})=u_{1i}+u_{1j}$

but i'm missing a part how to connect between the two.

Answer 1

Hint.

The key fact is that, if $x=x_1+x_2$ with $x_1 \in U_1$ and $x_2 \in U_2$, and $y=y_1+y_2$ with $y_1 \in U_1$ and $y_2 \in U_2$ than we have: $$ x+y= (x_1+x_2)+(y_1+y_2)=(x_1+y_1) + (x_2+y_2) $$

with: $$ x_1+y_1 \in U_1 \quad and \quad x_2+y_2 \in U_2 $$

Use this in the definition of $P(x+y)$

Answer 2

Let's clean up the notation a little bit: Given a a vector space $V$ over $\mathbb{K}$ which decomposes as the direct sum $V = V_1 \oplus V_2$. Then every element $v \in V$ can be written as $v = v_1 + v_2$ with $v_i \in V_i$.

The projection $P: V \rightarrow V, v \mapsto v_1$ is a linear transformation because: $$ P(a + b) = P( (a+b)_1 + (a + b)_2 ) = (a+b)_1 = a_1 + b_1 = P(a) + P(b)$$ and likewise $$ P(\lambda a) = (\lambda a) _1 = \lambda a_1 = \lambda P(a) $$

for any elements $a,b \in V$ and $\lambda \in \mathbb{K}$.