Thanks for reading.
I'm trying to better understand how rotation matrices work on complex-valued vectors.
Say we have the rotation matrix:
$R_\theta =\begin{bmatrix} \cos(\theta) & \ -sin(\theta) \\ \sin(\theta)& \cos(\theta) \end{bmatrix}$
This rotates a real-valued vector on the cartesian plane by $\theta$ degrees.
So, if we have a vector $\vec{v}$, and we compute $R_\theta \vec{v}$, the output is $\vec{v}$ but rotated by $\theta $ degrees couunter-clockwise .
However, now say $\vec{v}$ is allowed to have complex components.
If $\vec{v}$ is in $\mathbb{C}^2$, then we can picture it as really existing in $4D$ space, where there's a real $X$ direction, a real $Y$ direction, an imaginary $X$ direction and an imaginary $Y$ direction.
Let's say $\vec{v}=\begin{bmatrix} 2\\ 3 \end{bmatrix}$, or something like that. The point is that its components are only real - it exists entirely on the real $(X,Y)$ plane.
When we rotate $\vec{v}$ by any amount, it'll stay on the real $(X,Y)$ plane. It won't ever rotate into any of the imaginary dimensions.
Similarly, say $\vec{v}=\begin{bmatrix} 2i\\ 3i \end{bmatrix}$.
Now once more, when we rotate $\vec{v}$ by any amount by multiplying it by $R_\theta$ on the left, it'll stay on the imaginary $(X,Y)$ plane. It won't ever rotate into any of the real dimensions.
But, now say $\vec{v}=\begin{bmatrix} (1+i)\\ (3) \end{bmatrix}$.
...or something like that. The point is that it's $x$ and $y$ components have both real and imaginary components themselves.
Would a rotation (multiplication by $R_\theta$ on the left) still keep $\vec{v}$ on some plane, with the axes of that plane pointing into some combination of the imaginary and the real directions?
If that's the case, how could we find what that plane is?
What I've done thus far:
I'm tempted to answer yes to my question, and here's why.
When we have a vector
$\vec{v}=\begin{bmatrix} a\\ b \end{bmatrix}$
and we rotate it by $90^0$, we end up with $\vec{v}=\begin{bmatrix} -b\\ a \end{bmatrix}$
$\vec{v}=\begin{bmatrix} -b\\ a \end{bmatrix}$ and $\vec{v}=\begin{bmatrix} a\\ b \end{bmatrix}$
are orthogonal directions, which very clearly could be made to define the axes of some plane.
If $a$ and $b$ were both real, then this quite obviously was the real $(X,Y)$ plane, and rotation by ANY angle would've kept the vector on this plane.
Now, say we have... $\vec{v}=\begin{bmatrix} (2+3i)\\ (1+4i) \end{bmatrix}$.
...or something similar, where its components have both imaginary and real parts, and we multiply $\vec{v}$ by the rotation matrix that corresponds to rotating by $90^0$:
$\begin{bmatrix} \cos(90) & \ -sin(90) \\ \sin(90)& \cos(90) \end{bmatrix} \begin{bmatrix} (2+3i)\\ (1+4i) \end{bmatrix}=\begin{bmatrix} -(1+4i)\\ (2+3i) \end{bmatrix}$
Those orthogonal directions, $\begin{bmatrix} -(1+4i)\\ (2+3i) \end{bmatrix}$ and $\begin{bmatrix} (2+3i)\\ (1+4i) \end{bmatrix}$ could also define some plane.
However, it's not clear to me that rotation by an angle less than $90^0$ would've kept $\vec{v}$ on that same plane...
Would it have?
Thanks!
(If my question is unclear, please leave a comment!)
I guess by plane you mean $2$-dimensional real subspace?
View $\mathbb C^2$ as an $\mathbb R$ vector space. Any two linearly independent vectors of any real vector space determine a plane. If they are linearly dependent, then there are many such planes to choose from. In either way, the transformation "keeps $v$ and $R_\theta(v)$ in a plane."
If you're trying to ask if the image of $R_\theta$ acting on $\mathbb C^2$ is $2$-real-dimensional, then of course not. It is a nonsingular transformation acting on a $4$-real-dimensional space. Its image will also be $4$ dimensional.