How to interpret the concept of real vs complex vectors in the context of eigenvectors? For simplicity, let's use vectors of two elements. In $\mathbb{R}^2$, a vector $\langle x, y\rangle$ is, on the surface, indistinguishable from a similar vector in $\mathbb{C}^2$. A big difference is that if in the complex domain, it would represent a number $x+yi$.
I am studying some basic linear algebra and the idea of eigenvectors and eigenvalues has brought me to where I am.
A real eigenvector implies that in the real space, there are lines in which any vector that lies on it is only scaled.
For the case of, say, a rotation matrix
$$
\left(\begin{matrix} 
0 & -1\\
1 & 0
\end{matrix}\right) 
$$
it has complex eigenvalues $\pm i$ and complex eigenvectors $\langle i, 1\rangle$ and $\langle -i, 1\rangle$.
I'm confused about how to interpret this.
I can apply a matrix of real elements (such as the above rotation matrix) to both real and complex vectors. However, depending on the matrix, there are either real or complex eigenvectors/values. I understand for the real case the concept of an eigenvector being vectors that are scaled. Similarly, with the example above, a complex eigenvector implies a complex vector being scaled (thus all of the real plane rotates).
Where I'm confused is understanding how the real and complex spaces come together.
In the real space, what does it mean to have a vector with a complex component? Similarly, what about in the complex plane? From my understanding, vectors in both spaces have real components. How can I intuit this?
 A: If $V$ is an $n$-dimensional real vector space and if $\mathcal B=\{e_1,e_2,\ldots,e_n\}$ is a basis of $V$, then any $v\in V$ can be written as $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n$, with $\alpha_1,\alpha_2,\ldots,\alpha_n\in\Bbb R$. So, if what you call the components of a vector are the coefficients of the expression of $v$ in the basis $\mathcal B$, then they are always real numbers.
In the case in which $V=\Bbb R^2$ and you have the linear map $T\colon\Bbb R^2\longrightarrow\Bbb R^2$ defined by $T(x,y)=(-y,x)$, then this linear map has no real eigenvalues, since its characteristic polynomial is $\lambda^2+1$. The fact that it has two complex (non-real) eigenvalues, which are $\pm i$ means that if we see $T$ as a map from $\Bbb C^2$ into $\Bbb C^2$ (I will denote it by $T_{\Bbb C}$), then now it does have eigenvalues. It turns out that $T_{\Bbb C}(i,1)=(-1,i)=i(i,1)$ and that $T_{\Bbb C}(-i,1)=(-1,-i)=-i(-i,1)$.
Is it surprising that $T$ has no eigenvectors whereas $T_\Bbb C$ has them? No, since $T_{\Bbb C}$ is an extension of $T$; it is not the same map.
