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).
Best Answer
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!