A proof that the null linear mapping is the only one whose matrix representation does not depend on the basis [FALSE] I would like to show the fact that the linear mapping
$$
L : E\to E
$$
$$
x\mapsto0_{E}
$$
is the unique linear mapping whose matrix representation does not depend on the choice of the basis.
My attempt : Consider $\mathcal{V}$ and $\mathcal{\hat{V}}$ two basis of $E$. First we show that $A$, the matrix representation of $L$, does not depend on the basis.
To do so just consider the image of any vector of the basis $\mathcal{V}$ :
$$
L(v_i) = 0
$$
clearly its coordinates in the basis $\mathcal{V}$ are all zeros. This gives us the null matrix of size $n\times n$. Now consider the coordinate of $L(v_i)$ in the basis $\mathcal{\hat{V}}$, by linear independance of the $\hat{v}_i$'s its coordinates are also zeros, this yields the null matrix. An analogous reasoning with the base $\mathcal{\hat{V}}$ as starting point allows to conclude.
Now to prove the uniqueness, consider $F$ a linear mapping which is not the null linear mapping and does not depend on the choice of the basis $\mathcal{V}$ and $\mathcal{\hat{V}}$. So what happens if I consider the image $F(v_i)$ and $F(\hat{v_i})$ ? No reason to be the same at first since it is the coordinates of the image $F(v_i)$ and $F(\hat{v}_i)$ which should be the same no matter which base is chosen. But if it does not depend on the choice of the basis, I should have the same matrix wether I consider the basis $\mathcal{V}$ in the space of arrival or the basis $\mathcal{\hat{V}}$, but if I do so and decide to take the same basis on the space of arrival, then if their coordinates coincide, they are the same vector. Thus, we should have
$$
F(v_i) = F(\hat{v}_i),\quad\forall i \in\{1,...,n\}
$$
which implies
$$
F(v_i) - F(\hat{v}_i) = 0_{E}\implies F(v_i-\hat{v}_i) = 0_{E}\implies v_i = \hat{v}_i,\quad\forall i \in\{1,...,n\}
$$
but the bases $\mathcal{V}$ and $\mathcal{\hat{V}}$ are not the same, which yields the contradiction.
I would like to know if the proof is correct please.
Thank you a lot !

EDIT This is false, thanks to Federico Fallucca for his answer.
My mistake is at the last implication.
 A: There are infinitely many linear maps that do not depend on the choice of the basis, if you fix the same basis in the domain and co-domain. They are a multiple of the identity:
Let us fix $\lambda\in \mathbb K$, and consider the linear map $\lambda\cdot Id\colon E\to E$ defined by $v\to \lambda\cdot v$.
If you write the associated matrix $A$ in a fixed basis, then you obtain that $A=\lambda I$. If you change basis, then the new associated matrix will be $A=M^{-1} (\lambda I) M=\lambda I$, so this map does not depend by the choice of the basis.
In particular, for $\lambda=0$ we would get the trivial map.
Remark: You can say much more. The only operators that do not depend on the choice of the basis (when you fix the same basis in the domain and co-domain) are multiples of the identity, so the maps that we have studied above.
This holds because if we consider the associated matrix $A$, then we would get $A=M^{-1}AM$ for each invertible matrix $M$. In other words $A$ commutes with each invertible matrix. Let us take the $(i,j)$ component, $a_{i,j}$, of $A$, let $M_1$ be the invertible matrix that exchanges the $i$-th and $j$-th columns, $i\neq j$ and let $M_2$ be the invertible matrix that exchanges the $i$-th column with minus the $j$-th column. Then
$$M_1A=AM_1 \implies a_{ij}=a_{ji}, \quad a_{ii}=a_{jj}.$$
$$M_2A=AM_2 \implies a_{ij}=-a_{ji}, \quad -a_{ii}=-a_{jj}.$$
Therefore $a_{ij}=0$ and $a_{ii}=a_{jj}$, which means $A=\lambda I$.
(Observe that I have implicitly used above that the characteristic of the field $\mathbb K$ is different from $2$. However, the proof should be generalizable to these kind of fields as well, just by considering  suitable matrices $M_1$ and $M_2$).
Moreover, we can ask ourselves which are the linear maps that does not depend by the choice of the basis in general without fixing the same basis on the domain and co-domain.
Of course, an example is the zero map: $0_E=(M')^{-1}0_EM$.
Is this the unique one? In this case, by what we have said before, we are sure that such kind of maps has to be a multiple of the identity: $A=\lambda \cdot I$. But now if you fix two different basis that gives you two distinct change base matrix $M'$ and $M$, then you would get
$\lambda\cdot I= (M')^{-1}\lambda \cdot I M=\lambda (M')^{-1}M$.
If $\lambda \neq 0$, then we would have $I=(M')^{-1}M \implies M'=M$ that contradicts our hypothesis.
Therefore we have $\lambda=0$, and so the unique linear map independent by the choice of the basis has to be $0_E$.
