Proving projection map is a linear map Let $V$ be a finite-dimensional inner product space and $W$ be a subspace of $V$. Recall that any $v \in V$ can be written as $x+y$ with $x\in W$ and $y \in W^\perp$; and that the projection  $\mathrm{proj}_{W}v$ is equal to $x$ by definition.  
Prove that the function $T:V\rightarrow V$ defined by $T(v)=\mathrm{proj}_Wv$ is a linear transformation.  
The algebra is supposed to be really easy, but I just can't seem to get it.
I did:
Let $v = v_1 + v_2$ with $v_1 \in W, v_2 \in W^\perp$;
$w = w_1 + w_2$ with $w_1\in W, w_2 \in W^\perp$.  
$T(v+w)=T(v_1 + v_2 + w_1 + w_2) = T(\mathrm{proj}_Wv + v_2 + \mathrm{proj}_Ww + w_2) $ and I'm stuck..
 A: If $\langle\cdot,\cdot\rangle$ is your inner product and if $(e_1,\ldots,e_k)$ is an orthonormal basis of $W$, then$$(\forall v\in V):T(v)=\sum_{j=1}^k\langle v,e_j\rangle e_j.$$This expresses $T$ as a sum of linear maps. Therefore, $T$ is linear.
A: If you have that $V=W\oplus W^{\perp}$ you can always write down $v=w_1+w_2, \forall v \in V, w_1 \in W, w_2 \in W^{\perp}$.
Now, for definition, your projection is given by the direct sum you have. 
Definition: $\forall v \in V, \text{proj}_W(v)=w_1$
$\forall v \in V, \text{proj}_{W^{\perp}}(v)=w_2$.
So now you can easily verifiy that.
A: By definition of the projection onto $W$ we have
$$
\DeclareMathOperator{proj}{proj}
\proj_W v = \proj_W (x +  y) = x
$$
For two arbitrary scalars $\lambda_i$ and two vectors $v_i \in V$ we have individual
splits $v_i = x_i + y_i$ with $x_i \in W$ and $y_i \in W^\perp$.
We can combine them into
$$
\lambda_1 v_1 + \lambda_2 v_2 = \lambda_1 (x_1 + y_1) + \lambda_2(x_2 + y_2)
= (\lambda_1 x_1 + \lambda_2 x_2) + (\lambda_1 y_1 + \lambda_2 y_2)
$$
due to the calculation rules in $V$.
Because of the closure of $W$ and $W^\perp$ each (a vector space property) we have
$$
\lambda_1 x_1 + \lambda_2 x_2 \in W \\
\lambda_1 y_1 + \lambda_2 y_2 \in W^\perp
$$
which implies
$$
\proj_W(\lambda_1 v_1 + \lambda_2 v_2) = \lambda_1 x_1 + \lambda_2 x_2 
$$
On the other hand we have
$$
\lambda_i \proj_W(v_i) = \lambda_i x_i 
$$
so we get
$$
\proj_W(\lambda_1 v_1 + \lambda_2 v_2) 
= \lambda_1 x_1 + \lambda_2 x_2
= \lambda_1 \proj_W(v_1) + \lambda_2 \proj_W(v_2)
$$
This means $\proj_W$ is linear on $V$.
A: $$T(v+w)=T(v_1+v_2+w_1+w_2)=T(v_1+w_1+v_2+w_2)=v_1+w_1=T(v)+T(w)$$
Edit: Let $v \in V$. Then $v=x+y$ where $x \in W,y \in W^\perp$. Then $T(v) = x$.
Now to show it is linear. $T(v)=T(v_1+v_2)=v_1$. Similarly, $T(w)=w_1$.
Also, $T(v+w)=T(v_1+v_2+w_1+w_2)$. Let $x=v_1+w_1,y=v_2+w_2$.
Then $x \in W$ and $y \in W^\perp$. Therefore, $$T(v+w)=T(x+y)=x=v_1+w_1=T(v)+T(w)$$
