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.
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 $0_E$.