Exercise from Linear Algebra Done Right.
We are supposing that $V$ is finite dimensional and $T$ $\in$ $L(V)$. We need to prove that if $T$ has the same matrix with respect to every basis of $V$, then $T$ is a scalar multiple of the identity operator.
To prove this, I decided to show that the matrix of $T$ is a scalar multiple of the identity matrix.
I have managed to be able to prove that each matrix of $T$ has $0$ in every entry except the diagonals. This is how I proved this:
Assume that $T$ has the same matrix with respect to every basis of $V$. Let $v_1, ..., v_n$ be a basis of $V$. Let $k \in$ {$1, ..., n$}. It can be proven that the list $v_1+v_k, ... , v_k, ..., v_n$ is also a basis of $V$. Now, since the matrix of $T$ is the same for every basis, then column k of the matrix of $T$ has the same entries for any basis. As such,
$Tv_k$ = $a_1v_1 + ... + a_kv_k + ... + a_nv_n$ and
$Tv_k$ = $a_1(v_1 + v_k) + ... + a_kv_k + ... + a_nv_n$
where $a_1, ...,a_n \in F$.
If we subtract the right side of the top equation by the right side of the bottom equation, we find that
$a_1v_k = 0$.
This implies that $a_1 = 0$. By choosing other bases of V similar to $v_1+v_k, ... , v_k, ..., v_n$, we can find that
$a_1 = ... = a_{k-1} = a_{k+1} = ... = a_n = 0$.
Hence, $Tv_k = a_kv_k$.
We have effectively shown that the matrix of $T$ has $0$ in each entry except the diagonals. Q.E.D.
But here is where I am not sure how I can prove that the diagonals are all equal to each other. Proving this will show that the matrix of $T$ is a scalar multiple of the identity matrix. And this will prove that $T$ is a scalar multiple of the identity operator.
I would appreciate help that continues my method.
