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 $n$.

> [!TLDR] > Prove the formula by the **principle of mathematical induction** on $n$: verify $n = 1$, assume it for $n = k$, then derive it for $n = k+1$ using $A^{k+1} = A^{k}\cdot A$. Let $P(n)$ be the statement $$P(n): \ A^{n} = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}.$$ **Step 1 — Base case $n = 1$.** $$A^{1} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix},$$ 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. $$A^{k} = \begin{pmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{pmatrix}.$$ **Step 3 — Inductive step: show $P(k+1)$.** Using $A^{k+1} = A^{k}\cdot A$ and the hypothesis, $$A^{k+1} = \begin{pmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}.$$ Multiplying the matrices entry by entry, $$A^{k+1} = \begin{pmatrix} \cos k\theta\cos\theta - \sin k\theta\sin\theta & \cos k\theta\sin\theta + \sin k\theta\cos\theta \\ -\sin k\theta\cos\theta - \cos k\theta\sin\theta & -\sin k\theta\sin\theta + \cos k\theta\cos\theta \end{pmatrix}.$$ **Step 4 — Apply the compound-angle identities.** Using $\cos(k\theta+\theta) = \cos k\theta\cos\theta - \sin k\theta\sin\theta$ and $\sin(k\theta+\theta) = \sin k\theta\cos\theta + \cos k\theta\sin\theta$, $$A^{k+1} = \begin{pmatrix} \cos (k+1)\theta & \sin (k+1)\theta \\ -\sin (k+1)\theta & \cos (k+1)\theta \end{pmatrix}.$$ 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 $n \ge 1$. > [!ANSWER] > $A^{n} = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for all $n \in \mathbf{N}$.

Show that if A = cosθ sinθ \-sinθ cosθ , then Aⁿ =

Examples A.1 · Example 3

A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}

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Concept understanding — Matrix Rotation Power

Matrix Rotation Power

A rotation matrix turns every vector in the plane through a fixed angle. So what happens when you apply it again and again? Applying a rotation of $\theta$ twice is just a rotation of $2\theta$ ; three times, $3\theta$ ; and so on. Matrix power is exactly this idea written algebraically: $A^n$ means "apply the transformation $A$ a total of $n$ times."

The intuition

Multiplying a vector by a matrix $A$ transforms it once. Multiplying by $A$ again transforms the result once more. Hence

$A^n = \underbrace{A \cdot A \cdots A}_{n \text{ times}},$

with the conventions $A^1 = A$ and $A^0 = I$ (the identity), just as $x^0 = 1$ for numbers.

Watch out

$A^n$ is not raising each entry to the power $n$ . You must carry out full matrix multiplication. For example, with $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , $B^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ — the top-right entry becomes $2$ , not $1^2$ .

The rotation case

The cleanest example is the rotation matrix through angle $\theta$ (counterclockwise):

$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$

Because stacking two rotations adds their angles,

$R_\theta^{\,n} = R_{n\theta} = \begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{pmatrix}.$

Proving it by induction

This is a classic exam result, proved by mathematical induction on $n$ .

Base case ( $n = 1$ ): $R_\theta^{\,1} = R_\theta = R_{1\cdot\theta}$ , true.
Inductive step: assume $R_\theta^{\,k} = R_{k\theta}$ . Then $R_\theta^{\,k+1} = R_\theta^{\,k}\, R_\theta = R_{k\theta}\, R_\theta.$ Multiplying the two matrices and using the addition formulas $\cos k\theta\cos\theta - \sin k\theta\sin\theta = \cos(k+1)\theta, \qquad \sin k\theta\cos\theta + \cos k\theta\sin\theta = \sin(k+1)\theta,$ gives $R_\theta^{\,k+1} = R_{(k+1)\theta}$ . By induction the formula holds for all positive integers $n$ .

Tip

This is why a rotation matrix is easy to raise to a high power — you never actually multiply $n$ matrices. You just read off the answer: replace $\theta$ by $n\theta$ .

Takeaway: $A^n$ means applying the linear map $A$ repeatedly, done by genuine matrix multiplication. For the rotation matrix this collapses to the neat rule $R_\theta^{\,n} = R_{n\theta}$ , which you can establish rigorously by induction using the angle-sum identities.

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Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

Multiplying a vector by a matrix $A$ transforms it once. Multiplying by $A$ again transforms the result once more. Hence

$A^n = \underbrace{A \cdot A \cdots A}_{n \text{ times}},$

with the conventions $A^1 = A$ and $A^0 = I$ (the identity), just as $x^0 = 1$ for numbers.

Watch out

The rotation case

The cleanest example is the rotation matrix through angle $\theta$ (counterclockwise):

$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$

Because stacking two rotations adds their angles,

$R_\theta^{\,n} = R_{n\theta} = \begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{pmatrix}.$

Proving it by induction

This is a classic exam result, proved by mathematical induction on $n$ .

Base case ( $n = 1$ ): $R_\theta^{\,1} = R_\theta = R_{1\cdot\theta}$ , true.
Inductive step: assume $R_\theta^{\,k} = R_{k\theta}$ . Then $R_\theta^{\,k+1} = R_\theta^{\,k}\, R_\theta = R_{k\theta}\, R_\theta.$ Multiplying the two matrices and using the addition formulas $\cos k\theta\cos\theta - \sin k\theta\sin\theta = \cos(k+1)\theta, \qquad \sin k\theta\cos\theta + \cos k\theta\sin\theta = \sin(k+1)\theta,$ gives $R_\theta^{\,k+1} = R_{(k+1)\theta}$ . By induction the formula holds for all positive integers $n$ .

Tip

This is why a rotation matrix is easy to raise to a high power — you never actually multiply $n$ matrices. You just read off the answer: replace $\theta$ by $n\theta$ .

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 $n$ .

Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

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 $n$ .

Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

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Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

🔒Unlock the whole platform

Unlock the whole platform

Unlock the whole platform