How do you show that linear transformation $T$ does not have eigenvector?

Question

I came across this question:

Consider the plane as the real linear space $R^2$, and let $T$ be a rotation of $R^2$ through an angle of $\pi/2$ radians about the origin. Although $T$ has no eigenvectors, prove that every nonzero vector is an eigenvector for $T^2$ for $T^2$.

1). My main question is how to do you show that $T$ has no eigenvectors?

2). Second question: does this suffice as a proof of the above statement.

Let $T^2 = \left(\begin{array}{cc}-1 & 0 \\0 & -1\end{array}\right)$. You get this by matrix multiplying the rotation matrix for $R^2$ twice. Let $a,b \in$ Reals, and $x = (a,b)^T$. $T^2[x] = -1x$. Thus, for every non-zero vector $x$, it is an eigenvector (with eigenvalue of -1).

Answer 1

Your second proof should suffice, although I don't know what $(a, b)$ and $\mathbf{x}$ are both doing in there -- are $a, b$ the elements of the vector $\mathbf{x}$? You could just note that $T^2 = -I$, hence $T^2\mathbf{x} = -I\mathbf{x} = -\mathbf{x}$.

You can show that $T$ has no eigenvectors by showing that it has no (real) eigenvalues. Consider the eigenvalue equation $$\det{(T - \lambda I)} = 0$$

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

we have $$T - \lambda I = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix} - \begin{bmatrix}\lambda & 0\\0 & \lambda\end{bmatrix} = \begin{bmatrix}-\lambda & -1\\1 & -\lambda\end{bmatrix}$$

we can see that $\det({T - \lambda I}) = \lambda^2 + 1$. The solutions of this polynomial in $\lambda$ are the eigenvalues of $T$. What are the (real) solutions? None. Clearly the parabola $\lambda^2 + 1$ never touches the $x$-axis!

Answer 2

For (1), you might either write down a formula for $T(x, y)$ and solve $T(x, y) = \lambda(x, y)$ "manually", or find the standard matrix of $T$ and compute the characteristic polynomial.

Your answer for (2) is correct.

Answer 3

the rotation matrix $A = \pmatrix{0&-1\\1&0}$ has no real eigenvalues and no eigenvectors with all real components; however, it has complex eigenvalues $\pm i$ and the corresponding eigenvectors are $\pmatrix{\pm i\\1}.$

one reason that $A$ has no real eigenvalue is $Au$ is never parallel to $u$ for any $u \neq 0$ because $A$ rotates $u$ by $90^\circ$ counter clockwise.