Answer to Question #162398 in Linear Algebra for Nikhil

Question

Let 0<θ<2π, θ ≠ π. Consider the linear transformation T: C^2→C^2 given by matrix

[ cosθ -sinθ](w.r.t standard basis)

[ sinθ cosθ]. Find the vector v1, v2 such that Tv1= e^iθv1, Tv2= e^-iθv2. Is {v1,v2} a basis for C^2? Give reason for your answer.

Answer 1

Matrix of transformation:

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

Let

$v_1=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ , $v_2=\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$

Then:

$T(v_1)=Av_1=\begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$

$T(v_1)=\begin{bmatrix} x_1cos\theta-x_2sin\theta \\ x_1sin\theta+x_2cos\theta \end{bmatrix}=\begin{bmatrix} x_1e^{i\theta} \\ x_2e^{i\theta} \end{bmatrix}$

$x_1cos\theta-x_2sin\theta=x_1e^{i\theta}=x_1cos\theta+ix_1sin\theta$

$x_1sin\theta-x_2cos\theta=x_2e^{i\theta}=x_2cos\theta+ix_2sin\theta$

$-x_2=ix_1\implies x_1=-x_2/i=ix_2$

$x_2cos\theta=0$

$v_1=\begin{bmatrix} x_1 \\ -ix_1 \end{bmatrix}=x_1\begin{bmatrix} 1 \\ -i \end{bmatrix}$

$T(v_2)=Av_2=\begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$

$T(v_2)=\begin{bmatrix} y_1cos\theta-y_2sin\theta \\ y_1sin\theta+y_2cos\theta \end{bmatrix}=\begin{bmatrix} y_1e^{-i\theta} \\ y_2e^{-i\theta} \end{bmatrix}$

$y_1cos\theta-y_2sin\theta=y_1e^{-i\theta}=y_1cos\theta-iy_1sin\theta$

$y_1sin\theta-y_2cos\theta=y_2e^{-i\theta}=y_2cos\theta-iy_2sin\theta$

$y_2=iy_1$

$v_2=\begin{bmatrix} y_1 \\ iy_1 \end{bmatrix}=y_1\begin{bmatrix} 1 \\ i \end{bmatrix}$

The set $\{v_1,v_2\}$ is lineary independent, and dimension of $\{v_1,v_2\}$ is the same as dimension of $C^2$ . So $\{v_1,v_2\}$ is a basis for $C^2.$