Trying to find the matrix by a given linear transformation and eigenvalue / eigenvector the question I'm trying to resolve is the following: 
Given f: R3->R3 and f(1,1,0) = (1,0,0), f(0,1,1) = (1,1,0) and where 3 is an eigenvalue of the eigenvector (1,0,0).
a) Find the associated matrix to f on canonical base R3
Any help would be much appreciated! 
 A: You know how $f$ acts on some basis, namely $\{(1,1,0),(0,1,1),(1,0,0)\}$, so represent that canonical basis in this basis, determine how $f$ acts on the canonical basis by linearity, and use this to construct a matrix in the standard way, where the $j$th column is the output of the linear transformation acting on the $j$th basis vector.
A: Denote $B= \{(1,1,0), (0,1,1), (1,0,0)\}$ a basis given. Let $C = \{e_1,e_2,e_3\}$ the canonical basis. 
Note that $e_1 = 0\cdot (1,1,0) + 0\cdot(0,0,1) + 1(1,0,0), e_2 = 1 \cdot (1,1,0) + 0 \cdot(0,1,1)+ (-1)(1,0,0)$ and $e_3 = (-1)(1,1,0) + 1 \cdot (0,1,1) + 1 \cdot(1,0,0).$
Now, the matrix representation A of the linear map with respect to the canonical basis  is given by $A = [f(e_1) \hspace{0.5cm}f(e_2) \hspace{0.5cm} f(e_3)]$.
Note that $3$ is eigevalue of $(1,0,0)$, we have $f(1,0,0) = (3,0,0).$ This is the first column.
So, we have:
$f(e_1) = f(0\cdot (1,1,0) + 0\cdot(0,0,1) + 1(1,0,0)) = 0\cdot f(1,1,0) + 0\cdot f(0,0,1) + 1f(1,0,0) = f(1,0,0) = (3,0,0).$
$f(e_2) = f(1 \cdot (1,1,0) + 0 \cdot(0,1,1)+ (-1)(1,0,0)) = 1 \cdot f(1,1,0) + 0 \cdot f(0,1,1)+ (-1) f(1,0,0)= 1 \cdot (1,0,0) + (-1)(3,0,0) = (-2,0,0).$
$f(e_3) = f((-1)(1,1,0) + 1 \cdot (0,1,1) + 1 \cdot(1,0,0)) = (-1) f(1,1,0) + 1 \cdot f(0,1,1) + 1 \cdot f(1,0,0) = (-1) \cdot(1,0,0) + 1 \cdot (1,1,0) + 1 \cdot (3,0,0) = (3,1,0).$
So $A$ is given by:
$$A = \left(\begin{array}{cc}
3 & -2 & 3\\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right)
$$ 
