Missing last step to define a suitable linear transformation. Consider the vector space $\Bbb R^4$ and the subspaces $$\Bbb S=\{\rm x\in\Bbb R^4:3x_1+x_2-x_3+4x_4=0\}$$ $$\Bbb T=\langle (1,1,0,-1);(-1,0,1,0)\rangle$$
Find a linear transformation $f$ such that $${\rm Im} (f)=\Bbb T$$
$${\rm Ker} (f)\subset \Bbb S$$
and with eigenvalues $3,-2,0$. Note: $\subset$ is strict inclusion.
Now, one can readily check that $$\Bbb S=\langle(0,1,1,0),(1,0,3,0),(0,4,0,-1)\rangle$$
and that the set is linearly independent so that $\dim \Bbb S=3$. By the dimension theorem, since $\dim \Bbb T=2$, we must have $\dim \;{\rm Ker} (f)=2$. We show $$\mathscr B=\{(0,1,1,0),(1,0,3,0),(1,1,0,-1),(-1,0,1,0)\}$$ is a basis for $\Bbb R^4$. By expanding throughout the last row $$\begin{vmatrix}0&1&1&-1\\1&0&1&0\\1&3&0&1\\0&0&-1&0\end{vmatrix}=(-1)\cdot(-1)^{3+4}\begin{vmatrix}0&1&-1\\1&0&0\\1&3&1\end{vmatrix}$$
$$=1\cdot(-1)^{2+1}\begin{vmatrix}1&-1\\3&1\end{vmatrix}=-4\neq0$$
Thus, $$\mathscr B=\{(0,1,1,0),(1,0,3,0),(1,1,0,-1),(-1,0,1,0)\}$$ is a basis.
It suffices to define $f$ for the basis vectors. Since ${\rm Ker} (f)\subset \Bbb S$ and $(0,1,1,0),(1,0,3,0)\in\Bbb S$, I set $$f(0,1,1,0)=\bar 0$$
$$f(1,0,3,0)=\bar 0$$
and I get the first eigenvalue, $0$. Now, I'm only missing $$f(1,1,0,-1)=\lambda (1,1,0,-1)$$ $$f(-1,0,1,0)=\mu (-1,0,1,0)$$
which will ensure the other two eigenvalues and ${\rm Im}(f)=\Bbb T$. How can I correctly choose wether $\mu=3$ and $\lambda =-2$ or $\lambda =3$, $\mu=-2$? I have to be careful, since I need $\dim{\rm Ker}(f)=2$. 
 A: Instead of constructing your linear map in pieces, it is easier to do it by diagonalization. Once you have your basis $\mathscr{B}$, simply pick these to be your eigenvectors corresponding to eigenvalues $0,0,3,-2$, and form the associating matrix $V$, where the columns of $V$ are precisely the vectors $\mathscr{B}$. Then your linear map has as its matrix representation in the standard basis of $\mathbb{R}^4$ the decomposition $VDV^{-1}$, where $D = \mathrm{diag}(0,0,3,-2)$.
To answer your question, however, it does not matter whether you pick $\mu$ or $\lambda$ to have the values $3,-2$, or the other way around. This is because the kernel will precisely be the eigenspace of dimension 2 corresponding to eigenvalue $\lambda = 0$ that you already set up.
A: It doesn't matter whether you choose $\mu = 3, \lambda=-2$ or vice versa. Your basis vectors are $e_1 = (0,1,1,0)$, $e_2 = (1,0,3,0)$, $e_3 = (1,1,0,-1)$, $e_4 = (-1,0,1,0)$ and you have $\mathbb{S} = \langle e_1, e_2, e_3 \rangle$ and $\mathbb{T} = \langle e_3, e_4 \rangle$.
So if you define $$f(\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \alpha_4 e_4) = \mu \alpha_3 e_3 + \lambda \alpha_4 e_4,$$
you have $\mathrm{Im}(f) = \mathbb{T}$ and $\mathrm{Ker}(f) = \langle e_1, e_2 \rangle \subset \mathbb{S}$, because $\mu, \lambda \neq 0$.
