Exercise 10.A.3

Proof. Let $A$ denote the matrix of $T$, which is the same with respect to every basis. Let $v_1,\dots ,v_n$ be basis of $V$. Then

for $k=1,\dots ,n$. We’ll asumme $n\ge 2$ here (if $n=1$ then every operator on $V$ already is a multiple of the identity). The list $v_1-v_2,v_2,\dots ,v_n$ is also a basis of $V$. The matrix of $T$ with respect to this basis is the same, therefore we have

Calculating $T{v}_1-T{v}_2$ using $(1)$ we get

Because $T(v_1-v_2)$ can be uniquely written as combination of basis vectors, equating $(2)$ and $(3)$ shows that all entries in second column of $A$ equal $0$, except possibly $A_{2,2}$ which equals $A_{1,1}$. This same argument can be repeated with every pair of the basis vectors. Hence all nondiagonal entries of $A$ are $0$ and all diagonal entries are equal. Thus $A$ is a scalar multiple of the identity, which implies that so is $T$. □