Prove that there's no ordered basis $E$ in which $T{x\choose y}={0\choose y}$ can be represented as $1\ 2\choose 2\ 4$ Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that:  
$$T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix}0\\ y\end{pmatrix}$$
Show that there's no ordered-basis, $E$ such that:  
$$[T]_E = \begin{pmatrix}
   1 & 2\\
   2 & 4\\ 
 \end{pmatrix}$$
I didn't get how should I prove it. I tried some ways but end up with nothing. I guess I'm missing something.
 A: The trace is unvariant under change of basis. In the canonical basis, $T$ has for matrix $\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}$ and then $\mathrm{trace}T = 1$. The second matrix has trace $5 \neq 1$. Therefore, there is no basis in wich this is the matrix of $T$.
A: With
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ y \end{pmatrix}, \tag 1$
we have
$T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \tag 2$
and
$T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 3$
which shows the eigenvalues of $T$ are $0$ and $1$; on the other hand, the eigevalues of the matrix
$[T]_E = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \tag 4$
are the roots of its characteristic polynomial
$\det ([T]_E - \lambda I) = \begin{pmatrix} 1 - \lambda & 2 \\ 2 & 4 - \lambda \end{pmatrix}$
$= (1 - \lambda)(4 - \lambda) - 4 = 4 - 5\lambda + \lambda^2 - 4 = \lambda^2 - 5 \lambda, \tag 5$
which are $0$ and $5$; since the eigenvalues of the transformation $T$ as in (1) are $0$ and $1$, it cannot be represented by (4) in any basis $E$.
