What do we mean visually by complex eigen values of a matrix? Intuition behind the rotation of space using a 2x2 matrix and eigen values? The rotation matrix on $\Bbb R^2$ , the Euclidean plane given by $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ has two imaginary eigen values $i$ and $-i$. The definition of eigen vectors are those vectors $x$ that are parallel to $x$  [i.e. $Ax= \lambda x$].
Here the definition says $Ax=ix$ or $Ax=-ix$, since multiplying by $ i $ rotates my space by ninety degrees , does that essentially mean that we see two vectors in complex planes that are perpendicular to each other as being parallel to each other at the same time. What intuition am I missing here?
 A: We have an intuition of what "parallel" means in a real inner product space. Algebraically, it means one vector is a scalar multiple of another vector. This can get tricky, though, if our scalars are complex.
A big reason this gets tricky is because not only can there be a complex inner product $\langle u,v\rangle$ (which gives complex values), but there is an induced real inner product $\mathrm{Re}\langle u,v\rangle$. For instance, the standard complex inner product space $\mathbb{C}^n$ may be reinterpreted as the real inner product space $\mathbb{R}^{2n}$. So for instance, this means that within $\mathbb{C}^1$ itself, any two complex numbers are "parallel" because they are all multiples of each other, but interpreted as a real inner product space this is no longer true - e.g. $1$ and $i$ are perpendicular, not parallel.
When you say the matrix $A$ rotates a vector by $90^{\circ}$ and is thus perpendicular, you are describing that from the point of view of the real inner product on $\mathbb{C}^2$ (or, of course, the real inner product subspace $\mathbb{R}^2$), but it is no longer true if we treat $\mathbb{C}^2$ as a complex inner product space, as an eigenvector $x$ and $Ax=ix$ are now "parallel" in the sense that algebraically they are scalar multiples of each other.
A: You may try to think to "i" not as a number, but rather as a funcion.
In complex plane, the multiplication by $i$ is a rotation of 90 degrees. In $\mathbb C^2$, the multiplication by $i$ acts as a rotation on each cohordinate. So we can identify the multiplication by $i$ with a function $R_i:\mathbb C^2\to \mathbb C^2$.
Your matrix $A$ acts on $\mathbb R^2$ which you can think sitting inside $\mathbb C^2$. And in fact $A$ acts on $\mathbb C^2$. So you may ask if the matrix $A$ and the multiplication by $i$ act in the same way on some vector of $\mathbb C^2$. In other words, you may ask if there is $X\in\mathbb C^2$ so that $AX=R_i(X)$. A complex eigenvector of $A$ with eigenvalue $i$, is then a vector $X\in\mathbb C^2$ so that $AX$ gives the same result that the multiplication by $i$.
The equation $AX=iX$ now reads as $AX=R_i(X)$.
Note that if you identify $\mathbb C^2$ with $\mathbb R^4$, then the matrix $A$ becomes $\begin{pmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{pmatrix}$ and the function $R_i$ is given by the multiplication by the matrix $R=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}$
So in $\mathbb C^2=\mathbb R^4$ the problem of searching for eigenvectors with eigenvalue $i$ reduce to solve $AX=RX$ which is equivalent to search eigenvector with eigenvalue $1$ of $R^{-1}A$. That is to say $X$ so that $R^{-1}AX=X$.
