With square matrices, prove by mathematical induction that $(ABA^{-1})^k=AB^kA^{-1}$

Question

Assume that matrix $A$ and $B$ are square with same dimensions.

prove by mathematical induction that $(ABA^{-1})^k=AB^kA^{-1}$

if I let $k=2$ then I think I can do

$$ (ABA^{-1})^k=AB^kA^{-1}\\ (ABA^{-1})^2=AB^2A^{-1}\\ (ABA^{-1})(ABA^{-1})=ABBA^{-1}\\ ABBA^{-1}=ABBA^{-1}\\ $$

From here I can see a repeating pattern where I can manually make $k$ any positive integer and do the above but I'm not sure how to do this formally for the general case.

Answer 1

You have already formed a basis for induction.

Assuming that $n=k$ is true, we have $$(ABA^{-1})^k=AB^kA^{-1}.$$ Hence if we consider $n=k+1$, we get $$(ABA^{-1})^{k+1}=(ABA^{-1})^kABA^{-1}=AB^kA^{-1}ABA^{-1}=AB^{k+1}A^{-1}$$ which completes the induction step.

Answer 2

Hint:

The inductive set is

Inductive hypothesis: for some $k\ge2$, $(ABA^{-1})^k=AB^kA^{-1}$ is true.

You then have to prove that $\;(ABA^{-1})^{k+1}=AB^{k+1}A^{-1}$ is true.

Just write $$(ABA^{-1})^{k+1}=(ABA^{-1})^k(ABA^{-1}),$$ use the inductive hypothesis and play with associativity of multiplication.

Answer 3

Base case: $k = 1$

$(ABA^{-1})^1 = AB^1A^{-1}$

Hypothesis: Suppose for some $k, (ABA^{-1})^k = AB^kA^{-1}$

Inductive step: We must show that $(ABA^{-1})^{k+1} = AB^{k+1}A^{-1}$

$(ABA^{-1})^{k+1}\\ (ABA^{-1})^{k}(ABA^{-1})\\ (AB^{k}A^{-1})(ABA^{-1}) \text { by the inductive hypothesis}\\ AB^{k}BA^{-1}\\ AB^{k+1}A^{-1}\\ $