Question

Let A be a rotation matrix on a plane, please use the concept of linear transformation to explain it doesn't have real eigenvector.

Answer 1

There is a kind of intuitive way to view the eigenvalues and eigenvectors, and it ties in with geometric ideas as well (without resorting to four dimensions!).

The matrix, is unitary (more specifically, it is real so it is called orthogonal) and so there is an orthogonal basis of eigenvectors. Here, as you noted, it is (1i)(1i) and (1−i)(1−i), let us call them v1v1and v2v2, that form a basis of C2C2, and so we can write any element of R2R2 in terms of v1v1 and v2v2 as well, since R2R2 is a subset of C2C2. (And we normally think of rotations as occurring in R2R2! Please note that C2C2 is a two-dimensional vector space with components in CC and need not be considered as four-dimensional, with components in RR.)

We can then represent any vector in R2R2uniquely as a linear combination of these two vectors x=λ1v1+λ2v2x=λ1v1+λ2v2, with λi∈Cλi∈C. So if we call the linear map that the matrix represents RR

R(x)=R(λ1v1+λ2v2)=λ1R(v1)+λ2R(v2)=eiθλ1(v1)+e−iθλ2(v2)R(x)=R(λ1v1+λ2v2)=λ1R(v1)+λ2R(v2)=eiθλ1(v1)+e−iθλ2(v2)

In other words, when working in the basis v1,v2v1,v2:

R(λ1λ2)=(eiθλ1e−iθλ2)R(λ1λ2)=(eiθλ1e−iθλ2)

And we know that multiplying a complex number by eiθeiθ is an anticlockwise rotation by theta. So the rotation of a vector when represented by the basis v1,v2v1,v2 is the same as just rotating the individual components of the vector in the complex plane!