- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
30.$\quad$ Let $S$ and $T$ be in $\mathscr{L}(V, V)$ and assume that $ST - TS = I.$ Prove that $ST^n - T^n S = n T^{n-1}$ for all $n \geq 1.$
Proof. $\quad$ If $n = 1,$ we have $ST^n - T^nS = ST - TS = I,$ and the relation holds trivially (where by definition, $T^0 = I$). If $n = 2,$ we have \begin{align*} ST^n - T^nS &= ST^2 - T^2S \\ &= ST(T) - (T)TS \\ &= (I + TS)T + T(I - ST) \\ &= 2T \\ &= nT^{n-1} \end{align*} Thus, we have proven the relation for $n = 1$ and $n = 2.$ Now, assume $ST^k - T^kS = kT^{k-1}$ for some $k \geq 2.$ We will show that it also holds for $k + 1.$ Let $n = k + 1.$ By definition we have \begin{align*} ST^{k+1} - T^{k+1}S &= ST(T^k) - (T^k)TS \\ &= (I + TS)T^k + T^k(I - ST) \\ &= (T^k + TST^k) + (T^k - T^kST) \\ &= 2T^k + T(ST^k - T^{k-1}ST) \\ &= 2T^k + T(ST^{k-1} - T^{k-1}S)T \end{align*} But by hypothesis, we know that $ST^{k-1} - T^{k-1}S = (k-1)T^{k-2},$ and because $T$ is a linear transformation, we get \begin{align*} ST^{k+1} - T^{k+1}S &= 2T^k + T[(k-1)T^{k-2}]T \\ &= 2T^k + (k-1)T(T^{k-2})T \\ &= (k+1)T^{k} \end{align*} And by induction, since this is true for $k + 1,$ it is true for all $n \geq 1. \quad \blacksquare$