Linear Algebra - Proving a projection onto a subspace is a linear transformation

Question

How do I prove that a projection onto a subspace is a linear transformation ?

Given that V is a vector space, and M is a subspace of V.

I know these two facts:

i) There exists a subspace $N$ such that every vector $v \in V$ can be written uniquely as $v = x + y$ for some $x \in M$ and $y \in N$.

ii) $P$ is given by $P(x + y) = x$, for all $x \in M$ and $y \in N$

Answer 1

If $v=x+y$ and $w=x'+y'$, with $x,x'\in M$ and $y,y'\in N$, then\begin{align}P(v+w)&=P(x+y+x'+y')\\&=P(x+x'+y+y')\\&=x+x'\\&=P(v)+P(w).\end{align}Can you prove now that if $\lambda$ is a scalar, then $P(\lambda v)=\lambda P(v)$?

Answer 2

You need to show $P(v_1+v_2)=P(v_1)+P(v_2)$ for $v_1,v_2\in W$, and $P(\lambda v)=\lambda P(v)$ for $v\in V$, $\lambda\in F$.

Write $v_1=x_1+y_1$, $v_2=x_2+y_2$ with $x_i\in M$, $y_i\in N$, so that $P(v_i)=x_i$. Let $x=x_1+x_2\in M$, $y=y_1+y_2\in N$. Then $v_1+v_2=x+y$, and by uniqueness of this decomposition, $P(v_1+v_2)=x=P(v_1)+P(v_2)$, as desired.

Similarly, if $v=x+y$ with $x\in M$, $y\in N$, then $\lambda v=\lambda x+\lambda y$ with $\lambda x\in M$, $\lambda y\in N$. We conclude that $P(\lambda v)\lambda x=\lambda P(v)$.