Q.If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order 2, show that $I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

Concept: Cayley Transform for Rotation Matrices — a skew-symmetric matrix $A$ generates a rotation via $(I-A)^{-1}(I+A)$.

Step 1: Write $I+A$ and $I-A$ explicitly.

Step 2: Compute $(I-A)R$, where $R = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$. Use the double-angle identities:

Step 3: Multiply:

where $t = \tan\frac{\alpha}{2}$. The $(1,1)$ entry: …

The core idea here is beautiful: any rotation matrix can be expressed as a rational function of a skew-symmetric matrix. This is the Cayley transform for rotations. The matrix $A$ is skew-symmetric ($A^T = -A$), and the matrix $\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ is a rotation by angle $\alpha$. The identity $I+A = (I-A)R$ is equivalent to $R = (I-A)^{-1}(I+A)$, which is exactly the Cayley transform formula.

Let’s work through it step by step.

1. Write down the given matrices. We have

Let $t = \tan\frac{\alpha}{2}$ for brevity. Then $A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$.

2. Compute $I+A$ and $I-A$.

3. The goal is to show $I+A = (I-A)R$, where $R = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

   This is equivalent to showing $R = (I-A)^{-1}(I+A)$, provided $I-A$ is invertible. Let’s check: $\det(I-A) = 1\cdot1 - (t)(-t) = 1 + t^2 \neq 0$, so it’s invertible.

4. Find $(I-A)^{-1}$.

   For a $2\times2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

   Here $a=1$, $b=t$, $c=-t$, $d=1$, so $\det = 1(1) - t(-t) = 1+t^2$.

   Thus

5. Compute $(I-A)^{-1}(I+A)$.

Multiply the matrices:

• Top-left: $1\cdot1 + (-t)\cdot t = 1 - t^2$
• Top-right: $1\cdot(-t) + (-t)\cdot1 = -t - t = -2t$
• Bottom-left: $t\cdot1 + 1\cdot t = t + t = 2t$
• Bottom-right: $t\cdot(-t) + 1\cdot1 = -t^2 + 1 = 1 - t^2$ So

6. Now use the double-angle formulas. …

Method: Proving a matrix identity involving an inverse

To prove an identity like $I+A=(I-A)R$, either compute the product on the right directly, or rearrange to $R=(I-A)^{-1}(I+A)$ and evaluate, then simplify entries with the relevant identities.

Steps

Step 1: Write the small matrices explicitly

Form $I+A$ and $I-A$ from the given $A$, and check $\det(I-A)\neq0$ so the inverse exists.

Step 2: Compute the inverse and the product …

Common Mistakes

Mistake 1: Multiplying in the wrong order

Why it's wrong: matrix multiplication is not commutative, so $(I-A)R$ and $R(I-A)$ differ; the identity specifies $(I-A)R$. Correct approach: keep the exact order stated in the problem.

Mistake 2: Using the wrong half-angle conversions …