Find the matrix of $f:(x,y,z) \rightarrow (y+z,x-y, -x+y+2z)$ in a new basis Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ such that $$(x,y,z) \mapsto (y+z,x-z,-x+y+2z) $$
Find its matrix, then show that this function is a projection parallel to a vector subspace $G$ onto a vector subspace $F$. Find the basis $B_F$ for $F$ and the basis $B_G$ for $G$ and then find the matrix of $f$ in the basis $B= B_F \cup B_G$.
So the matrix $A$ of $f$ looks as follows: $$
\begin{bmatrix}
0 & 1 &1 \\ 
1 & 0 & -1\\ 
-1 & 1 & 2
\end{bmatrix}
 $$ 
To show that it's a projection, I simply notice that $A^2 =A$. Now, I know that the $F,G$ vector subspaces are equal to $F= \text{Im(f)}$ and
$G= \text{Ker(f)}$.
Now to find the basis for $\text{Ker(P)}$ I simply have to solve the equation $AX= 0$, which gives me the following results: $A (x,y,z)^T =0 \iff x=z,y=-z,z=z$. So a basis for $\text{Ker(f)}$ would be: $(1,0,1),(0,-1,1)$ and as it is identical to the first two columns of $A$, I can simply choose the basis of $\text{Im(p)}$ as the last column (am I right? I am unsure about this).
Now the hard part comes, finding the matrix of $A$ in the new basis. The new basis is : $(1,0,1),(0,1,-1),(1,-1,2)$. So if we call the vectors of the new basis this way: $\epsilon_1 =(1,0,1), \epsilon_2 = (0,1,-1), \epsilon_3 =(1,-1,2) $ we have that 
$$\epsilon_1= 1e_1 + 1e_3$$ $$ \epsilon_2=1e_2 -1e_3 $$ $$\epsilon_3= 1e_1 -1e_2 + 2e_3 $$
So the matrix $P$ for the change of basis is equal to:
$$\begin{bmatrix}
1 & 0 &1 \\ 
0 & 1 & -1\\ 
1 & -1 & 2
\end{bmatrix}
 $$
So the matrix of $f$ in the new basis would be $P^{-1}AP$, is this correct?
 A: $P$ is the matrix of the identity linear transformation  
$$I: \mathbb{R}^3 \to \mathbb{R}^3$$
In other words, suposse the canonical base $\{ e_1, e_2, e_3\}$ and $\{ \epsilon_1, \epsilon_2, \epsilon_3\}$ the new base. Then, $P(e_1)= \epsilon_1$, $P(e_2) = \epsilon_2$ and $P(e_3)  = \epsilon_3$. 
So to use $P$, you have to use a vector $v$ in the canonical base and its output is the same vector in the other base. In other wise, $P^{-1}$ take a vector in the new base and its output is the same vector in the canonical base. Remark that $A$ take vector $v$ in the canonical base and the output $Av$ is too in the canonical base . So to find the matrix of $f$ in the new base, you have to do  $$PAP^{-1}$$
UPDATE
I made a mistake
In fact, the matrix $P$ can be considered as the identity transformation matrix that carries vectors on the noncanonical basis in the canonical base. As you wrote above
$$\epsilon_1 = (1,0,1); \epsilon_2 = (0,1,-1); \epsilon_3 = (1,-1,2)$$
where $(1,0,1),(0,1,-1),(1,-1,2)$ are vector on canonical basis. So, to use $P$, you have to use a vector on the noncanonical basis and its output is the same vector in the canonical base. So
$$P^{-1} A P$$
is your target.
A: The last part of the problem is actually the easiest: with a little bit of thought about the properties of a projection, you can write down the matrix with almost no work. The restriction of any projection to its image is the identity map and it of course vanishes on its kernel. Now recall that the columns of a transformation matrix are the images of the basis vectors. So, in any ordered basis of $\mathbb R^n$ that consists of a basis for the image of a projection concatenated with a basis for its kernel, the matrix of the projection is diagonal with a number of 1’s along the main diagonal equal to $f$’s rank and the rest of them $0$. (This is a special case of a more general theorem.)  
Going back to $\ker(f)$, the equations that you’ve derived are correct, but you’ve drawn the wrong conclusion from them. Indeed, if you compute $f(1,0,1)^T$ and $f(0,-1,1)^T$ you’ll find that neither is zero, so these two vectors aren’t even elements of the kernel, let alone a basis for it. Even if they were, choosing the last column of $A$ to complete a basis for $\mathbb R^3$ would be wrong, anyway: $A$ doesn’t have full rank, so its columns are not linearly independent.  
Your system of equations in only has one free variable, not two. All elements of the kernel are of the form $(z,-z,z)^T=z(1,-1,1)^T$. Knowing this, you also know based on the first paragraph above, that the matrix of $f$ in the basis $B$ is $\operatorname{diag}(1,1,0)$. You still have to find a basis for $F$, but that’s a fairly standard computation. Indeed, you could find bases for both $G$ and $F$ at the same time by row-reducing $A$. It would be a good idea to then check your work by applying the resulting change of basis to $A$ to see that you indeed get the described diagonal matrix.
