Q.If $AB = BA$ for any two square matrices, prove by mathematical induction that $(AB)^n = A^n B^n$.

Claim: if $AB=BA$, then $(AB)^{n}=A^{n}B^{n}$ for every positive integer $n$. Prove by induction on $n$.

Base $n=1$: $(AB)^{1}=AB=A^{1}B^{1}$. True.

Hypothesis: assume $(AB)^{k}=A^{k}B^{k}$ for some $k\ge1$.

Step $k\to k+1$:

Since $A$ and $B$ commute, so do $A$ and $B^{k}$, i.e. $B^{k}A=AB^{k}$. Hence …

The idea

Matrix multiplication normally cannot be reordered. But if $A$ and $B$ commute, we may slide $A$ past $B$ freely, which lets us collect all the $A$'s on the left and all the $B$'s on the right inside $(AB)^{n}$. We make this rigorous with induction.

A helper fact: $B^{k}A=AB^{k}$

Before the main proof, note that if $A$ commutes with $B$ it also commutes with every power $B^{k}$. Quick induction on $k$:

• $k=1$: $B^{1}A=BA=AB=AB^{1}$ (given).
• Assume $B^{k}A=AB^{k}$. Then

So $B^{k}A=AB^{k}$ for all $k\ge1$.

Main proof by induction on $n$

Base case $n=1$. $(AB)^{1}=AB$ and $A^{1}B^{1}=AB$, so the statement holds.

Inductive hypothesis. Suppose $(AB)^{k}=A^{k}B^{k}$ for some $k\ge1$.

Inductive step. Then

By associativity, regroup as $A^{k}(B^{k}A)B$. Apply the helper fact $B^{k}A=AB^{k}$:

$A^{k}(B^{k}A)B=A^{k}(AB^{k})B=(A^{k}A)(B^{k}B)=A^{k+1}B^{k+1}.$ …

Method: Proving a power identity by induction using commutativity

To prove $(AB)^n=A^nB^n$ for all $n$ when $AB=BA$, use mathematical induction, and at the key step use commutativity to move an $A$ past a power of $B$.

Steps

Step 1: Base case $n=1$.

$(AB)^1=AB=A^1B^1$, which holds trivially.

Step 2: Inductive hypothesis.

Assume $(AB)^k=A^kB^k$ for some $k\ge1$.

Step 3: Inductive step.

Step 4: Use commutativity to swap $B^kA$. …

Common Mistakes

Mistake 1: Assuming $B^kA=AB^k$ without justification.

Why it's wrong: this step is exactly where the hypothesis $AB=BA$ is used; $A$ commutes with $B^k$ only because it commutes with $B$. Correct approach: state that $AB=BA$ implies $A$ commutes with every power of $B$.

Mistake 2: Rearranging $ABAB\ldots$ freely as if all factors commute. …