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

A second way is to test the definition $\langle Lx,y\rangle = \langle x,Ly\rangle$ directly on coordinates instead of comparing matrices. Reflection: $L(a,b) = (b,a)$. For $x=(x_1,x_2)$, $y=(y_1,y_2)$: $\langle Lx,y\rangle = x_2y_1+x_1y_2$ and $\langle x,Ly\rangle = x_1y_2+x_2y_1$. These agree for all $x,y$, so $L$ is self-adjoint. Then $L^*=L$, so trivially $LL^*=L^2=L^*L$, and $L$ is also normal. Statement (I) is true. Rotation: $L(a,b) = \left(\frac{a-b}{\sqrt2}, \frac{a+b}{\sqrt2}\right)$. Test on the standard basis $e_1=(1,0)$, $e_2=(0,1)$: $Le_1 = \left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$ and $Le_2 = \left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$. Self-adjointness would require $\langle Le_1,e_2\rangle = \langle e_1,Le_2\rangle$. Here $\langle Le_1,e_2\rangle = \frac{1}{\sqrt2}$ while $\langle e_1,Le_2\rangle = -\frac{1}{\sqrt2}$. Since $\frac{1}{\sqrt2} \ne -\frac{1}{\sqrt2}$, $L$ is not self-adjoint. For normality, $L^*$ is rotation by $-\pi/4$ (the inverse of an orthogonal transformation equals its adjoint), and rotations by different angles always commute, so $LL^* = L^*L = I$; hence $L$ is normal despite not being self-adjoint. So statement (II), which claims both properties, is false. \[\boxed{\text{Only (II)}}\]