Q.Show that if $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$, then $A^{n} = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for every positive integer $n$.

Let $P(n): A^{n} = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$.

Base: $P(1)$ is just the given $A$, so it holds.

Step: Assume $P(k)$. Then

using $\cos(k\theta+\theta)$ and $\sin(k\theta+\theta)$. So $P(k)\Rightarrow P(k+1)$, and by induction $P(n)$ holds for all $n$.

Let $P(n)$ be the statement

Step 1 — Base case $n = 1$.

which is exactly $P(1)$. So $P(1)$ is true.

Step 2 — Inductive hypothesis.

Assume $P(k)$ is true for some positive integer $k$, i.e.

Step 3 — Inductive step: show $P(k+1)$.

Using $A^{k+1} = A^{k}\cdot A$ and the hypothesis,

Multiplying the matrices entry by entry,

Step 4 — Apply the compound-angle identities.

Using $\cos(k\theta+\theta) = \cos k\theta\cos\theta - \sin k\theta\sin\theta$ and $\sin(k\theta+\theta) = \sin k\theta\cos\theta + \cos k\theta\sin\theta$,

This is precisely $P(k+1)$, so $P(k)$ true $\Rightarrow P(k+1)$ true.

Step 5 — Conclude.

Since $P(1)$ holds and $P(k) \Rightarrow P(k+1)$, by the principle of mathematical induction $P(n)$ is true for every positive integer $n \ge 1$.

Method: Proving a Matrix-Power Formula by Mathematical Induction

This method applies whenever you must prove a formula for $A^n$ (a matrix raised to the power $n$) holds for every positive integer $n$.

Steps

Step 1: State the statement $P(n)$ to be proved

Write out explicitly what $P(n)$ claims $A^n$ equals, in terms of $n$.

Step 2: Verify the base case $P(1)$

Check that $A^1 = A$ matches the given matrix $A$ exactly as stated by the formula — this is usually immediate since $A^1$ is just $A$ itself.

Step 3: Assume $P(k)$ (the inductive hypothesis)

Assume the formula holds for some positive integer $k$; this gives you an explicit matrix expression to use for $A^k$.

Step 4: Prove $P(k+1)$ using $A^{k+1} = A^k \cdot A$

Multiply the assumed expression for $A^k$ by $A$ using ordinary matrix multiplication (row-by-column, entry by entry). This produces four trigonometric expressions (one per matrix entry).

Step 5: Simplify with the compound-angle identities

Recognise that each entry matches the expansion of $\cos(k\theta+\theta)$ or $\sin(k\theta+\theta)$, i.e.

Apply these to collapse each entry to $\cos(k+1)\theta$ or $\sin(k+1)\theta$, matching $P(k+1)$.

Step 6: Conclude by the principle of mathematical induction

Since $P(1)$ holds and $P(k) \Rightarrow P(k+1)$ for every $k$, state explicitly that $P(n)$ holds for all positive integers $n$.

Common Mistakes

Mistake 1: Treating $A^n$ as entrywise powers

Why it's wrong: $A^n$ means multiplying the matrix $A$ by itself $n$ times using full matrix multiplication, not raising each individual entry (like $\cos\theta$) to the power $n$. This is a fundamentally different operation and gives a completely wrong result. Correct approach: always compute $A^{k+1}=A^k \cdot A$ using row-by-column matrix multiplication.

Mistake 2: Forgetting to verify the base case

Why it's wrong: The inductive step ("if $P(k)$ then $P(k+1)$") only chains correctly if there is a confirmed starting point; without checking $P(1)$, the whole induction has no foundation, and technically nothing has been proved. Correct approach: always explicitly verify $P(1)$ (or whatever the smallest case is) before moving to the inductive step.

Mistake 3: Arithmetic slips multiplying the two matrices

Why it's wrong: Expanding $A^k \cdot A$ involves four separate dot products of rows and columns; a sign error in one entry (especially with the negative signs already present in the rotation matrix) silently breaks the pattern and makes the compound-angle identity not apply cleanly. Correct approach: compute each of the four entries separately and carefully, tracking signs, before trying to match them to the angle-sum identities.