Write out linear mapping of $L:\mathbb{R}^3 \to \mathbb{R}^3$ 
Write out the matrix $A$ of linear mapping of $L:\mathbb{R}^3 \to \mathbb{R}^3$ that projects $\mathbb{R}^3$ onto its linear subspace of columns with $x^2=x^3$ and parallel to the column $\left[0,1,-1 \right]^T$ that is  
i) $L\left[0,1,-1\right]^T=0$,
ii) $L$ is identical on the subspace $x^2=x^3$.

I have
$$A=\begin{bmatrix}
a_1^1 &a_2^1  &a_3^1 \\ 
 a_1^2&a_2^2  &a_3^2 \\ 
 a_1^3&a_2^3  &a_3^3 
\end{bmatrix}$$
The first condition is
$$\begin{bmatrix}
a_1^1 &a_2^1  &a_3^1 \\ 
 a_1^2&a_2^2  &a_3^2 \\ 
 a_1^3&a_2^3  &a_3^3 
\end{bmatrix}\begin{bmatrix}
0\\ 
1\\ 
-1
\end{bmatrix}=\begin{bmatrix}
0\\ 
0\\ 
0
\end{bmatrix}$$
now for (say) $L[1,-1,0]^T$
$$\begin{bmatrix}
a_1^1 &a_2^1  &a_3^1 \\ 
 a_1^2&a_2^2  &a_3^2 \\ 
 a_1^3&a_2^3  &a_3^3 
\end{bmatrix}\begin{bmatrix}
1\\ 
-1\\ 
0
\end{bmatrix}$$
Here the second condition says
$$\begin{bmatrix}
a_1^1 &a_2^1  &a_3^1 \\ 
 a_1^2&a_2^2  &a_3^2 \\ 
 a_1^3&a_2^3  &a_3^3 
\end{bmatrix}\begin{bmatrix}
x^1\\ 
x^2\\ 
x^2
\end{bmatrix}=\begin{bmatrix}
x^1\\ 
x^2\\ 
x^2
\end{bmatrix}$$
What does the second condition say for the case $L[0,1,-1]$? Further, determine $A$ completely?
 A: There are several ways to attack this problem, some of which I’ll describe here. First, notice that the subspace $W$ onto which we are projecting is the orthogonal complement of the span of $v=[0,1,-1]^T$. Since we are projecting parallel to this vector, the projection is thus orthogonal. This makes things a little easier.  
One way is to find an orthogonal/orthonormal basis of $W$, find the orthogonal projection matrix for each of these basis vectors and add them together to produce the matrix for projection onto $W$. This is a rather tedious exercise. A simpler approach is to compute $I-\pi_v=I-{v v^T\over v^Tv}$, the orthogonal rejection relative to $v$, instead since this involves computing only one individual projection matrix.  
Another approach is to find a basis $(w_1,w_2)$ for $W$, then extend it to a basis of $\mathbb R^3$ by adding $v$. Relative to this basis, the matrix of the projection is simply $\operatorname{diag}(1,1,0)$, and a change of basis back to the standard basis gives the required matrix.  
Related to this method, observe that $\operatorname{span}(v)$ is the eigenspace of $0$ and that $W$ is the eigenspace of $1$. I.e., $Av=0$ and $Aw=w$ for all $w\in W$. We can combine these equations into the matrix equation $$A\begin{bmatrix}|&|&|\\w_1&w_2&v\\|&|&|\end{bmatrix}=\begin{bmatrix}|&|&|\\w_1&w_2&0\\|&|&|\end{bmatrix}$$ where $(w_1,w_2)$ is a basis of $W$. It should be clear how to solve this equation for $A$. Note that right-multiplying a $3\times3$ matrix by $\operatorname{diag}(1,1,0)$ zeroes out its last column, so this is effectively the same as the previous method.  
Finally, let $B$ be a matrix with a basis of $W$ as its columns. The orthogonal projection matrix onto $W$ is given by the well-known formula $B(B^TB)^{-1}B^T$.
