In the book of Linear Algebra by Werner Greub at page 95, question 2,
Assume that $\phi$ is a linear transformation $E\to E$ having the same
matrix relative to every basis $x_v$.Prove that $\phi = \lambda i$,
where $\lambda $ is scalar, and $i$ is the identity map.
Let $A$ be the matrix representation of $\phi$ respect to basis $x_v$ and $B$ respect to the basis $y_v$, and C be the basis transformation $x_v \to y_v$. I have derived that
$$AC = CA = CB = BC$$, but after that I stuck.
Actually, as a method I don't know how to show the result, so I tried things to get some feeling what is going on, but, as I have said, it didn't go nowhere.
So how can we show this result ? I would appreciated if you give some hint, but if you directly give the answer, it is OK too.
Edit:
We are working on a give $\phi$ such that its matrix representation $M(\phi; x_v, x_u)$ is the same for any basis $x_v$.
Best Answer
The following proof is valid over any field $K$ that has at least $3$ elements.
Assume that $\phi$ is not a scalar function. Then there is $x$ s.t. for every $\alpha\in K\setminus \{0\}$
$x,\alpha\phi(x),e_3,\cdots,e_n$ is a basis of $E$.
In such a basis, the matrix of $\phi$ has as first column: $[0,\dfrac{1}{\alpha},0,\cdots,0]^T$, a contradiction.
EDIT 1. A solution valid over any field $K$.
Let $A=[a_{i,j}]$ be a representative of $\phi$ and let $(E_{i,j})$ be the canonical basis of $M_n(K)$. As Pierre-Yves Gaillard wrote, for every $P\in GL_n(K)$, $P^{-1}AP=A$, that is $PA=AP$.
Method 1. In particular, for every $k\not= l$, $A(I_n+E_{k,l})=(I_n+E_{k,l})A$, that implies for every $k\not= l$, $a_{l,k}=0,a_{k,k}=a_{l,l}$. Finally, $A$ is a scalar matrix.
EDIT 2. Method 2. We can also use the fact that (over any field) any matrix is the sum of two invertible matrices. cf. the user1551's answer in
Real square matrix as a sum of two invertible matrices