Finding a basis such that a linear map can be described through a transformation matrix Let $f: \mathbb{R}^4 \to \mathbb{R}^4 ,  \begin{pmatrix}x\\y\\z\\w\end{pmatrix} \mapsto \begin{pmatrix}y+z+2w\\z+w\\y+z+2w\\y+w\end{pmatrix}$ be a linear transformation. What basis would one have to choose such that $B=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$ is the transformation matrix of $f$.
The vectors of the matrix $B$ are image of the basis vectors, but how can I get $(1,0,0,0)^T$ or $(0,1,0,0)^T$?
 A: Like YiFan observed, this is impossible. On the contrary, if $B$ were the matrix of $f$ with respect to some basis, then $v = \langle 1, 0, 0, 0 \rangle$ is an eigenvector of $f$ corresponding to the eigenvalue $1.$ But clearly, we have that $f(v) = 0$ -- a contradiction. One other way to see the problem (that could be a useful strategy in the case that the problem is actually doable) is the following. We wish to find linearly independent vectors $v_i = \langle x_i, y_i, z_i, w_i \rangle$ such that $f(v_1) = v_1,$ $f(v_2) = v_2,$ $f(v_3) = 0,$ and $f(v_4) = 0.$ Consequently, we obtain four systems of equations. Unfortunately, we see again that the question is patently false. Our first equation in the system is $$\langle x_1, y_1, z_1, w_1 \rangle = v_1 = f(v_1) = \langle y_1 + z_1 + 2w_1, z_1 + w_1, y_1 + z_1 + 2w_1, y_1 + w_1 \rangle.$$ Comparing the entries of the vectors on the left- and right-hand sides, we obtain the following. $$\begin{cases} x_1 = y_1 + z_1 + 2w_1 \\ y_1 = z_1 + w_1 \\ z_1 = y_1 + z_1 + 2w_1 \\ w_1 = y_1 + w_1 \end{cases}$$ By the fourth equation, it follows that $y_1 = 0.$ By the third equation, then, it follows that $w_1 = 0.$ By the second equation, it follows that $z_1 = 0,$ and by the first equation, it follows that $x_1 = 0.$ But this means that the only vector that satisfies $f(v) = v$ is the zero vector -- a contradiction.
