Let $S:\mathbb R^n \to \mathbb R^n$ be given by $S(v) = \alpha (v)$, for a fixed $\alpha \in \mathbb R, \alpha \neq 0$. Let $T:\mathbb R^n \to \mathbb R^n$ be a linear operator such that $B = (v_1, v_2, ..., v_n)$ is a set of linearly independent eigen vectors of $T$. Then prove or disprove that the matrix of $T-S$ with respect to $B$ is diagonal.
Clearly, $B$ forms a basis for $\mathbb R^n$. This implies that $T$ is diagonalizable and hence the matrix representation of $T$ with respect to $B$ (i.e. $[T]_B$) is diagonal, with corresponding eigen values of $T$ (say $\lambda_1, \lambda_2, ..., \lambda_n$) as the diagonal entries.
Also, $S$ is another linear operator on $\mathbb R^n$, by its definition in the problem. With respect to $B$, the matrix representation of $S$ (i.e. $[S]_B$) is also diagonal, with $\alpha$ as the diagonal entries.
Therefore, the matrix representation of $T-S$ with respect to $B$ is again diagonal, with diagonal entries $\lambda_1-\alpha, \lambda_2-\alpha, ..., \lambda_n-\alpha$ because $(T-S)(v_i) = T(v_i) - S(v_i) = (\lambda_i - \alpha)(v_i)$, $v_i \in B, 1 \leq i \leq n$.
Is my conclusion correct?
Your answer is correct.
A different way to formulate it is by using the following properties of basis representation:
Let $B$ be any basis of $\Bbb R^n$:
If $\mathcal L,\mathcal L'\colon \Bbb R^n \to \Bbb R^n$ are linear mappings and $r,s\in\Bbb R$ , then $[r\mathcal L+s\mathcal L']_B=r[\mathcal L]_B+s[\mathcal L']_B$.
If $\mathcal{I}(x)=x$ for all $x\in\Bbb R^n$, and $B'$ is a basis of $\Bbb R^n$ then $[\mathcal{I}]_B=[\mathcal{I}]_{B'}$.
So in your case, if we let $\mathcal{L}=T$, $\mathcal{L'}=S$, $r=1$, $s=\alpha$ and $B'$ the canonical basis, then
$$[T-S]_B = [T]_{B}-[S]_B=[T]_{B}-\alpha [\mathcal{I}]_{B'}$$
which is a diagonal matrix as $[T]_{B}$ is diagonal because $B$ contains the eigenvectors of $T$ and $[\mathcal{I}]_{B'} =I_n$ is the identity matrix.