# Do complex vectors remain on a plane when getting rotated?

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!

Would a rotation (multiplication by 𝑅𝜃 on the left) still keep 𝑣⃗ on some plane, with the axes of that plane pointing into some combination of the imaginary and the real directions?

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.

• I believe the question is: for a given vector $v\in\mathbb{R}^4\cong\mathbb{R}^2\oplus \mathbb{R}^2$, does the image of $\theta \mapsto (R_{\theta} v_1, R_{\theta} v_2)$ lie in a two-dimensional linear subspace? – user7530 Jul 17 at 15:57
• @user7530 The user writes: "$\begin{bmatrix} \cos(90) & \ -sin(90) \\ \sin(90)& \cos(90) \end{bmatrix} \begin{bmatrix} (2+3i)\\ (1+4i) \end{bmatrix}=\begin{bmatrix} -(1+4i)\\ (2+3i) \end{bmatrix}$" which is not what you are conjecturing they mean, (which I believe is "apply the rotation to the two complex entries individually.) The appear to be simply treating the matrix as a $\mathbb C$-linear operator on $\mathbb C^2$. – rschwieb Jul 17 at 16:00
• @user7530 that's exactly the question...if I understand the notation correctly. For a specific vector (with possibly complex components), does its multiplication by any rotation matrix $R_\theta$, regardless of what $\theta$ is, give us a vector that lies on the same plane? The same two-dimensional linear subspace? – Joshua Ronis Jul 17 at 16:00
• @rschwieb you may be correct as well on what I mean...I'm not very good at understanding all the notation :( Also, thanks for your answer, +1, but I'm still confused since I don't know if you understood my question... – Joshua Ronis Jul 17 at 16:02
• @JoshuaRonis What is "complex notation"? It makes no difference if you are using $\mathbb R$ or $\mathbb C$, the matrix algebra is all the same... If you meant "complex" in the literal english sense, then never mind :) – rschwieb Jul 17 at 16:03