How does one calculate the projections onto sub-spaces of $C^3$? Given the following two sub-spaces of  $C^3$:
$W = \operatorname{span}[(1,0,0)] \,, U = \operatorname{span}[(1,1,0),(0,1,1)].$
I want to find the linear operators $P_u , P_w$ which represent the projections onto U & W according to: $C^3 = U⊕W$
I generally understand the idea of projections, but I can't seem to find a way to actually calculate these projections.
Could anyone explain this to me?
 A: You know that $P_W(\vec x)=\vec x$ for every $\vec x\in W$ and $P_W(\vec x)=\vec0$ for every $\vec x\in U$. In particular, $P_W(1,0,0)=(1,0,0)$ and $P_W(1,1,0)=P_W(0,1,1)=(0,0,0)$.
Since the three given vectors are a basis of $\mathbb C^3$, this uniquely determines a linear transformation $\mathbb C^3\to\mathbb C^3$. And you can use the standard method to find the matrix of a linear transformation - see, for example, my answer here: Find the standard matrix for a linear transformation.
$$\left(\begin{array}{ccc|ccc}
  1 & 0 & 0 & 1 & 0 & 0 \\
  1 & 1 & 0 & 0 & 0 & 0 \\
  0 & 1 & 1 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
  1 & 0 & 0 & 1 & 0 & 0 \\
  0 & 1 & 0 &-1 & 0 & 0 \\
  0 & 1 & 1 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
  1 & 0 & 0 & 1 & 0 & 0 \\
  0 & 1 & 0 &-1 & 0 & 0 \\
  0 & 0 & 1 & 1 & 0 & 0
\end{array}\right)$$
So you get that the matrix of the projection onto $W$ is
$$P=
\begin{pmatrix}
  1 & 0 & 0 \\
 -1 & 0 & 0 \\
  1 & 0 & 0 \\
\end{pmatrix}.
$$
You can find projection onto $U$ as $P'=I-P$.
A: For $(x,y,z) \in \mathbb{C}^3$, there is a unique solution $(a,b,c)$ for the system
\begin{align}
(x,y,z) = a(1,0,0) + b(1,1,0) + c(0,0,1) 
\end{align}
because $\mathbb{C}^3 = W \oplus U$ is a direct sum. As $a(1,0,0) \in W$ and $b(1,1,0) + c(0,1,1)\in U$, they are the projections of $(x,y,z)$ onto $W$ and $U$ respectively.
The question reduces to : solve this system of linear equations.
A: The restriction of any projection onto its image is the identity map. In a suitable basis, its matrix will be of the form $\operatorname{diag}(1,\dots,1,0,\dots,0)$, where the number of $1$s is equal to the dimension of the image. In addition, since we want $\mathbb C^3=U\oplus W$, we need $\ker(P_U)=W$ and $\ker(P_W)=U$. We already have bases for both $U$ and $W$, so the standard-basis matrices of the two projections can be constructed via a straightforward change of basis: $$P_W = B\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}B^{-1} \\ P_U = B\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}B^{-1}$$ with $$B = \begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}.$$
