Question

Consider a linear transformation T in <4 is defined by T(x1, x2, x3, x4) =(−x2, x1, −x4, x3). Show that it does not have real eigen-values.

Answer 1

T(x1, x2, x3, x4) =(−x2, x1, −x4, x3)

We should write in matrix firstly:

$T= \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} ;$

In order to show that T matrix does not have real eigen values, should calculate $(\lambda\times I-A)$ matrix:

$(\lambda\times I-A)= \lambda \times \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} -$ $\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} =$

$\begin{bmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \\ \end{bmatrix} - \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} =$ $\begin{bmatrix} \lambda & 1 & 0 & 0 \\ -1 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & -1 & \lambda \\ \end{bmatrix};$

And find det of $(\lambda\times I-A)$ matrix:

$\lambda \times (-1)^{1+1} \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & -1 & \lambda \\ \end{bmatrix}+ 1 \times (-1)^{1+2} \times$ $\begin{bmatrix} -1 & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & -1 & \lambda \\ \end{bmatrix}=$

$\lambda \times (\lambda^3+ \lambda) -(-\lambda^2-1)=$ $\lambda \times (\lambda^3+ \lambda) -(-\lambda^2-1)= \lambda^4+\lambda^2+\lambda^2+1=$ $\lambda^4+2\lambda^2+1;$

And then will equate the polynomial formed by the eigen values to 0:

$\lambda^4+2\lambda^2+1=0$

$\lambda^2$ is greater than or equal to 0. The left-hand side of the equation is greater than or equal to 1, but the right-hand side of the equation is 0. The equality cannot be true. So this equation doesn't have real solutions. Thus,the matrix T does not have real eigen values.