Basic Eigenvalue Question

Question

The rotation matrix $$T=\left[\begin{array}{c c}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]$$ has no eigenvectors as an operator $T:\mathbb{R}^2\to\mathbb{R}^2$. Here $0<\theta<\pi$. But identifiying $\mathbb{C}=\mathbb{R}^2$ and considering $\mathbb{C}$ as a complex inner product space, we have that $$T:\mathbb{C}\to\mathbb{C}\ ,\ \theta\mapsto e^{i\theta}$$ is representing the same map. That is, $T=L_{e^{i\theta}}$. But any complex inner product space, with a normal operator ($T$ is normal) acting on it, admits an orthonormal basis of eigenvectors for that operator. So what am I missing here?

Answer 1

There's two possible interpretations here. Either you mean that this represents a linear map $\mathbb{C} \to \mathbb{C}$ by $z \mapsto e^{-i\theta}z$ in which case the problem is quite simple: without even invoking the algebraic completeness of $\mathbb{C}$, for any field $F$, a "linear map" $T:F \to F$, can be completely decscribed by the data $T(1) = \lambda \in F$, as we have that for any $z \in F$

$$ T(z) = zT(1) = \lambda z$$

Where we just use the fact that in one dimension we can just identify the vector space $F$ with the scalars $F$ and pull anything out. To get to the punchline, any linear map $F \to F$ is all eigenvectors, as it is completely described as just multiplication by $T(1)$. So in our example the eigenvectors are everything (except 0, depending on your definition), and the eigenvalue is $e^{i\theta}$.

Alternatively, the matrix $$T = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$ Can be viewed as a $2 \times 2$ complex matrix specifying a linear map $\mathbb{C}^2 \to \mathbb{C}^2$. To detect eigenvalues compute the characteristic polynomial

$$ \det(\lambda I - T ) = \det \begin{pmatrix} \lambda - \cos\theta & \sin\theta \\ -\sin\theta & \lambda - \cos\theta \end{pmatrix} = \lambda^2 - 2\lambda\cos\theta + 1 $$

So just use the quadratic formula to find eigenvalues.