PMI in Determinant

Question

In given below picture
How can we interchange the A position to between A^k and B^k+1 . As given condition is only stating AB=BA , NOTHING ABOUT POWERS

Answer 1

A simple induction shows that if $A$ anf $B$ commute, any power of $A$ commutes with any power of $B$.

Indeed, it suffices to show that $A$ commutes with any $B^k$:

• the case $k=1$ is the hypothesis.
• Now suppose that for some $k\ge 1$, we have $AB^k=B^k A$,. Then \begin{align} AB^{k+1}&=A(BB^k)=(AB)B^k=(BA)B^k&\qquad&\text{by hypothesis,}\\ &=B(AB^k)=B(B^kA)&&\text{by the inductive hypothesis,} \\ &=(BB^k)A=B^{k+1}A. \end{align}

Answer 2

This is an unbelievably bad non-proof. Among the errors (besides the one you found):

The statement

So, $P(K)$ is true, whenever $P(K+1)$ is true.

Is logically backwards. If you switched it around to be right, it would be in the wrong place in the argument.

This

$P(K+1 : AB)^{K+1} = \cdots$

makes no sense at all.

And

$P(K+1)$ is true for all $n$ whenever $P(K)$ is true.

is again nonsense.

The best one can say about this is that if you already know how to prove the assertion then you can figure out what the author was trying to say and say it correctly. But then you wouldn't be reading this in the first place.

Do not rely on anything else from this document. If it's from a published source you should write the publisher to complain (or I will if you provide a citation).