Q.If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, then prove that $A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}$, $n \in N$.

Concept: Matrix Rotation Power — the given matrix $A$ is a rotation matrix (by $-\theta$), and its $n$-th power corresponds to $n$ successive rotations, i.e., rotation by $-n\theta$.

Step 1: Check base case $n=1$:

$A^1 = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, which matches the given form.

Step 2: Assume true for $n=k$:

$A^k = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix}$.

Step 3: Multiply $A^{k+1} = A^k \cdot A$:

Compute entry $(1,1)$: $\cos k\theta \cos\theta - \sin k\theta \sin\theta = \cos(k\theta + \theta) = \cos((k+1)\theta)$.

Entry $(1,2)$: $\cos k\theta \sin\theta + \sin k\theta \cos\theta = \sin(k\theta + \theta) = \sin((k+1)\theta)$.

Entry $(2,1)$: $-\sin k\theta \cos\theta - \cos k\theta \sin\theta = -\sin(k\theta + \theta) = -\sin((k+1)\theta)$.

Entry $(2,2)$: $-\sin k\theta \sin\theta + \cos k\theta \cos\theta = \cos(k\theta + \theta) = \cos((k+1)\theta)$.

Thus $A^{k+1}$ matches the required form. By induction, the statement holds for all $n \in \mathbb{N}$.

Concept and Intuition: Matrix Rotation Power

The matrix

is the standard rotation matrix that rotates a vector in the plane by angle $\theta$ clockwise (notice the minus sign on the lower-left entry).

If you rotate once by $\theta$, then again by $\theta$, the net effect is a rotation by $2\theta$. Repeating this $n$ times gives a rotation by $n\theta$.

So the product $A^n$ should be exactly the rotation matrix for angle $n\theta$:

We prove this formally using mathematical induction, which is the cleanest way for statements about natural numbers $n$.

Step-by-step Proof

1. Base case: $n = 1$

For $n = 1$, the statement says

This is exactly the definition of $A$, so the base case holds trivially.

2. Induction hypothesis

Assume that for some $k \in \mathbb{N}$,

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

We need to show

Multiply $A^k$ (from the hypothesis) by $A$:

4. Perform the matrix multiplication

• Top-left entry:

  $(\cos k\theta)(\cos\theta) + (\sin k\theta)(-\sin\theta) = \cos k\theta \cos\theta - \sin k\theta \sin\theta$.

• Top-right entry:

  $(\cos k\theta)(\sin\theta) + (\sin k\theta)(\cos\theta) = \cos k\theta \sin\theta + \sin k\theta \cos\theta$.

• Bottom-left entry:

  $(-\sin k\theta)(\cos\theta) + (\cos k\theta)(-\sin\theta) = -\sin k\theta \cos\theta - \cos k\theta \sin\theta$.

• Bottom-right entry:

  $(-\sin k\theta)(\sin\theta) + (\cos k\theta)(\cos\theta) = -\sin k\theta \sin\theta + \cos k\theta \cos\theta$.

5. Apply trigonometric sum formulas

Recall the standard identities:

Using these:

• Top-left becomes $\cos(k\theta + \theta) = \cos(k+1)\theta$.
• Top-right becomes $\sin(k\theta + \theta) = \sin(k+1)\theta$.
• Bottom-left becomes $-\big[\sin(k\theta + \theta)\big] = -\sin(k+1)\theta$.
• Bottom-right becomes $\cos(k\theta + \theta) = \cos(k+1)\theta$.

Thus

6. Conclusion of induction

The base case holds, and the induction step is valid. Therefore, by the principle of mathematical induction, the statement is true for all $n \in \mathbb{N}$.

Method: Proving a matrix-power formula by mathematical induction

Use this for statements of the form "$A^n = (\text{closed form})$ for all $n \in \mathbb{N}$", such as a rotation matrix raised to the $n$-th power.

Steps

Step 1: Base case.

Verify the formula for $n = 1$ — it should reduce to $A$ itself.

Step 2: Inductive hypothesis.

Assume the formula holds for $n = k$, i.e. $A^k$ has the stated closed form.

Step 3: Inductive step.

Compute $A^{k+1} = A^k \cdot A$ using the hypothesis, then simplify with the relevant identities (for rotations, the angle-sum formulas $\cos(k\theta+\theta)$ and $\sin(k\theta+\theta)$) to reach the $n=k+1$ form. Conclude by induction.

Common Mistakes

Mistake 1: Skipping or mis-stating the base case.

Why it's wrong: induction needs a verified starting point ($n=1$); without it the chain is unfounded. Correct approach: confirm $A^1$ matches the formula first.

Mistake 2: Sign errors when applying the angle-sum identities.

Why it's wrong: the $(2,1)$ entry must stay $-\sin(k+1)\theta$; a dropped minus breaks the rotation form. Correct approach: use $\cos(k\theta+\theta)=\cos k\theta\cos\theta-\sin k\theta\sin\theta$ and $\sin(k\theta+\theta)=\sin k\theta\cos\theta+\cos k\theta\sin\theta$ carefully.

Mistake 3: Raising each entry to the power $n$.

Why it's wrong: $A^n$ is repeated matrix multiplication, not entrywise powering. Correct approach: use $A^{k+1}=A^k\cdot A$ in the inductive step.