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Given $S$ and $T$, linear transformations such that $ST-TS$ commutes with $S$, I am supposed to show that $S^kT-TS^k=kS^{k-1} (ST-TS)$ for every positive integer $k$. As a matter of fact I do not know how to go about the solution. Can I have a hint? Thank you.

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up vote 4 down vote accepted

Let $A=ST-TS$, so $A$ commutes with $S$. Note that $XSTY=XTSY+XAY$. You can use this to move $T$ to the left by one, and since $A$ commutes with $S$, you can move the resulting $A$s to the right without producing any further terms.

[More detail:]

Your problem can be rephrased as follows: If I know the effect of moving $T$ past $S$ to the left once, what's the effect of moving $T$ past $S$ to the left $k$ times? The effect of moving $T$ past $S$ to the left once is to produce an additional term involving the commutator $A$: $XSTY=XTSY+XAY$. So if you start out with $S^kT$, you can apply this once to get $S^{k-1}TS+S^{k-1}A$. Now you've moved $T$ to the left by one and produced one of the $k$ terms $S^{k-1}A$ in the desired result. If you continue moving $T$ to the left by one in each step, you'll continue producing additional terms with $A$, and since $A$ commutes with $S$, you can move all these $A$s to the right, so all the $k$ terms you generate before $T$ ends up on the left will be the same as the first. Then you just have to subtract $TS^k$ to get the desired result.

Alternatively, you could use induction. The result is clearly correct for $k=1$. Assuming it's correct for $k$, you can write

$$ \begin{align} S^{k+1}T-TS^{k+1} &=SS^kT-TSS^k\\ &=S(kS^{k-1}(ST-TS)+TS^k)-TSS^k\\ &=kS^k(ST-TS)+STS^k-TSS^k\\ &=kS^k(ST-TS)+(ST-TS)S^k\\ &=kS^k(ST-TS)+S^k(ST-TS)\\ &=(k+1)S^k(ST-TS)\;.\\ \end{align} $$

The basic idea is the same; each step of moving $T$ to the left produces one term with the commutator; the difference is just that in the inductive proof you encapsulate $k$ of the steps in the induction hypothesis and only perform the $(k+1)$-th explicitly.

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Can you please give a little more detail? I am not completely clear on what to do. Thank you. –  smanoos Oct 10 '11 at 0:19
    
This is perfect. You are great. Thanks a lot. –  smanoos Oct 10 '11 at 0:55
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