Q.Prove by Mathematical Induction that $(A')^n = (A^n)'$, where $n \in \mathbb{N}$ for any square matrix $A$.

Concept: Matrix Transpose Power — the transpose of a power equals the power of the transpose.

Step 1 – Base case ($n=1$):

$(A')^1 = A'$ and $(A^1)' = A'$, so the statement holds.

Step 2 – Induction hypothesis:

Assume $(A')^k = (A^k)'$ for some $k \in \mathbb{N}$.

Step 3 – Inductive step ($k \to k+1$):

$(A')^{k+1} = (A')^k \cdot A' = (A^k)' \cdot A'$ (by hypothesis). …

Why This Works: The Core Idea

Matrix transposition has a beautiful property: when you transpose a product, the order of multiplication reverses. That is, $(AB)' = B'A'$. This "reversal" is the engine behind the proof.

If we think about $A^n$ as $A$ multiplied by itself $n$ times, then $(A^n)'$ is the transpose of that long product. Using the reversal property repeatedly, each $A$ inside becomes $A'$, but the order flips completely — which, because all factors are the same matrix $A'$, gives back $(A')^n$. Induction just makes this intuition rigorous.

The Proof Step by Step

1. Base case: $n = 1$

For $n = 1$, we have $(A')^1 = A'$ and $(A^1)' = A'$. Both sides are identical, so the statement holds trivially.

2. Inductive hypothesis

Assume that for some $k \in \mathbb{N}$, the statement is true:

3. Inductive step: prove for $n = k+1$

We need to show $(A')^{k+1} = (A^{k+1})'$.

Start with the left-hand side:

By the inductive hypothesis, $(A')^k = (A^k)'$, so:

Now, here's the key move. The product $(A^k)' \cdot A'$ is the transpose of something — but in reverse order. Using the property $(XY)' = Y'X'$ with $X = A^k$ and $Y = A$, we get:

$(A^k)' \cdot A' = (A \cdot A^k)'$ …

Method: Proving a matrix identity for all $n$ by mathematical induction

When a statement must hold for every natural number $n$ (here $(A')^n=(A^n)'$), induction is the standard tool: verify a base case, assume the statement for $k$, then deduce it for $k+1$ using a known matrix rule.

Steps

Step 1: Base case $n=1$.

Check the statement directly for the smallest $n$: $(A')^1=A'$ and $(A^1)'=A'$, so it holds.

Step 2: Inductive hypothesis.

Assume the statement is true for some $k\in\mathbb{N}$: $(A')^k=(A^k)'$.

Step 3: Inductive step — build the $k+1$ case from the $k$ case. …

Common Mistakes

Mistake 1: Applying the reversal law in the wrong order and getting $(A^k A)'=(A^{k+1})'$ but writing $A^{k+1}=A^k\cdot A$ as if it were $A\cdot A^k$ without noticing they match only here.

Why it's wrong: $(XY)'=Y'X'$ forces a specific pairing; sloppy ordering can produce $(A^k)'A'=(A\cdot A^k)'$ only if you set $X=A,\ Y=A^k$ correctly. Correct approach: identify $X$ and $Y$ explicitly before reversing.

Mistake 2: Skipping or mis-stating the base case.

Why it's wrong: induction is invalid without a verified starting point. Correct approach: check $n=1$ explicitly. …