Question

Which one of the following statements is false?
(I) The linear transformation \(L:\mathbb{R}^2 \to \mathbb{R}^2\) that reflects a vector across the line \(y=x\) is both self-adjoint and normal.
(II) The linear transformation \(L:\mathbb{R}^2 \to \mathbb{R}^2\) which rotates a vector by an angle \(\pi/4\) in the anti-clockwise direction is both self-adjoint and normal.

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Check whether each transformation's matrix is symmetric (self-adjoint) and orthogonal (normal).

Answer 1

The Correct Option is B

Step 1: The reflection across \(y=x\) has matrix \(A = \begin{pmatrix}0&1\\1&0\end{pmatrix}\) with respect to the standard basis. Since \(A^T = A\), this matrix is symmetric, so \(L\) is self-adjoint. Every self-adjoint operator is automatically normal (\(AA^T = A^2 = A^TA\)), so statement (I) is true.

Step 2: The rotation by \(\pi/4\) anticlockwise has matrix \(B = \begin{pmatrix}\cos(\pi/4)&-\sin(\pi/4)\\\sin(\pi/4)&\cos(\pi/4)\end{pmatrix} = \begin{pmatrix}\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\\\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\end{pmatrix}\).

Step 3: Check self-adjointness: \(B^T = \begin{pmatrix}\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\end{pmatrix} \ne B\), since the off-diagonal entries differ in sign. So \(B\) is not symmetric and \(L\) is not self-adjoint.

Step 4: Check normality: rotation matrices are orthogonal, so \(B^TB = BB^T = I\), which means \(L\) is normal even though it is not self-adjoint. So the claim in (II) that it is both self-adjoint and normal is false, since self-adjointness fails.

Step 5: So (I) is true and (II) is false, meaning only statement (II) is false.

\[\boxed{\text{Only (II)}}\]