Determining a $4\times4$ matrix knowing $3$ of its $4$ eigenvectors and eigenvalues The precise question given goes as follows;  
The matrix 'A' $\in \Bbb R^{4 \times4}$ has eigenvectors $u_1, u_2, u_3, u_4$ where 
$u_1 = \begin{pmatrix}
    1 \\
    -1 \\
    1 \\
    1 \\
    \end{pmatrix}
 , u_2 = \begin{pmatrix}
    0 \\
    2 \\
    1 \\
    -1 \\
    \end{pmatrix}, u_3 = \begin{pmatrix}
    3 \\
    -1 \\
    1 \\
    2 \\
    \end{pmatrix}$ satisfy:  
$Au_1 = 2u_1,\; Au_2 = 14u_2,\; Au_3 = 18u_3$ 
Calculate A$w$ where $w = \begin{pmatrix}
    49 \\
    13 \\
    47 \\
    18 \\
    \end{pmatrix}$.
Usually I'd approach a question like this by using the relationship of the similar matrices $A = PA'P^{-1}$ where $A$ is the matrix , $P$ has the matrix $A$'s eigenvectors for columns, and $A'$ has the matrix $A$'s eigenvalues along its diagonal and $0$ elsewhere. And from there just do the calculation.
However only 3 of the 4 eigenvectors/values are given, so I dont know if this method is still applicable, and am having no success, or if I'm going down a rabbit hole and missing an obvious alternate. 
Any suggestions? Thanks.

The answer could be arbitrary unless $w$ is a combination of the given eigenvectors (if it's not then an arbitrary value of $Aw$ defines completely the linear map since you know the images of a linear basis).
One approach will be to find this combination -- it is 
$$
 w = 16 u_1 + 20 u_2 + 11 u_3.
$$
From this you compute $A w$ easily.
Another approach is to complete $(u_1, u_2, u_3)$ by adding any (independent) fourth vector. This defines the matrix $P$ in your method. $A'$ is still diagonal, taking for fourth eigenvalue whatever you like since it will not change the result for $A w$ (though it does change $A$ of course).
I'll give the problem to my students, I like it :) thanks!
