# A circle and a hyperbola living in one plot

We will see that the 3D plot of \(x^2 + (y+zi)^2 = 1\) contains a circle and a hyperbola, where \(i\) is the imaginary number. Beyond the visuals, this helps us understand the (complex) eigenvalues of real matrices.

If we expand out \(x^2+(y+zi)^2 = 1\) in real and imaginary parts, we get

\[\begin{align*} &\text{Real Part:} &x^2 + y^2 - z^2 &= 1, \\ &\text{Imaginary Part:} &yz &= 0. \end{align*}\]From \(yz=0\) we get two cases.

\[\begin{align*} \text{Case 1: }y&=0 \text{ and so } x^2-z^2 = 1\\ \text{Case 2: }z&=0 \text{ and so } x^2+y^2 = 1 \end{align*}\]Case 1 nets us a hyperbola in the \(xz\)-plane. Case 2 nets us a unit circle in the \(xy\)-plane.

You can check out the plot at desmos.

## Eigenvalues and dynamical systems

Why might we care about this beyond the visuals? This sort of graph often comes up when we’re trying to understand the (complex) eigenvalues of a real matrix that depends on a real parameter. For example, consider the matrix

\[M(\mu) = \begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\]for some real parameter \(\mu\). The eigenvalues \(\lambda\) of \(M\) satisify the equation \(\mu^2 + \lambda^2 = 1.\) Letting \(\mu = x\) and \(\lambda = y+zi\) recovers the plot of \(x^2 + (y+iz)^2 = 1.\)

Just from looking at the plot, we see that

\[\begin{align*} \mu < -1&: M(\mu) \text{ has a pair of complex conjugate eigenvalues} \\ \mu = -1&: M(\mu) \text{ only has one eigenvalue}, \lambda = 0 \\ 0< \mu < 1&: M(\mu) \text{ has one +ve and one -ve eigenvalue} \\ \mu = 1&: M(\mu) \text{ only has one eigenvalue}, \lambda = 0 \\ \mu > 1&: M(\mu) \text{ has a pair of complex conjugate eigenvalues} \end{align*}\]We are sometimes interested in the magnitude of the eigenvalues. We can see that \(\vert \lambda \vert = \sqrt{y^2+z^2}\) is the radius of the \(yz\)-plane. Again just from the plot we can guess / see that

\[\begin{align*} \vert\lambda\vert\leq1 &\iff \mu \in [-\sqrt{2}, \sqrt{2}], \\ \vert\lambda\vert<1 &\iff \mu \in (-\sqrt{2}, \sqrt{2}) \backslash \{0\}. \end{align*}\]Understanding of the eigenvalues of \(M(\mu)\) is generally useful if \(M(\mu)\) shows up in a dynamical system. For example let’s consider the system \(\mathbf{v}_{n+1} = M(\mu)^2 \mathbf{v}_n\), then

\[\begin{align*} \mathbf{v}_{n+1} &= M(\mu)^2\begin{bmatrix} x \\ y \end{bmatrix} \\ &=\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} \\ &=\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} (1+\mu)y \\ (1-\mu)x \end{bmatrix}\\ &= (1-\mu^2) \begin{bmatrix} x \\ y \end{bmatrix} \\ &= (1-\mu^2)\mathbf{v}_n. \end{align*}\]Curiously, we see that the origin is a stable fixed point if and only if \(\vert1-\mu^2\vert \leq 1\), which is equivalent to \(\mu \in [-2,2]\) as well as \(\vert\lambda\vert \leq 1\). Similarly, the origin is an asymptotically stable fixed point if and only if \(\vert 1-\mu^2\vert < 1\) if and only if \(\vert \lambda \vert < 1.\) This hints at a relationship between eigenvalues and stability.

## More examples

Consider the matrix

\[M(\mu) = \begin{bmatrix} 1 & \mu \\ 1 & 1 \end{bmatrix}.\]We can notice a couple of things immediately. When \(\mu=1\), \(M(\mu)\) is degenerate and so has a zero eigenvalue. When \(\mu=0\), \(M(\mu)\) is a shear matrix with only one eigenvalue \(\lambda = 1\). As shear matrices live on the boundary of diagonalisable and non-diagonalisable matrices, we can reasonably further guess that if \(\mu<0\) then \(M(\mu)\) has complex conjugate eigenvalues and if \(\mu>0\) then \((M\mu)\) has real eigenvalues.

Concretely, considering the eigenvalue equation \(\lambda^2 - 2\lambda + (1-\mu) = 0\) and letting \(\mu=x\) and \(\lambda=y+zi\) and re-arranging terms we get

\[\left((y+zi)-1\right)^2=x.\]By splitting into real parts and imaginary parts and re-arranging terms we get

\[\begin{align*} &\text{Real Part:} &(y-1)^2 - z^2 &= x, \\ &\text{Imaginary Part:} &(y-1)z &= 0. \end{align*}\]Again we have two cases

\[\begin{align*} \text{Case 1: }y&=1 \text{ and so } -z^2 = x\\ \text{Case 2: }z&=0 \text{ and so } (y-1)^2 = 0 \end{align*}\]Case 1 nets us a parabola in the \(xz\)-plane while Case 2 nets us a parabola in the \(xy\)-plane. Finally, we arrive at the plot.

You can create these sort of plots for any other 2 by 2 matrix that depends on one real parameter using a desmos calculator I made.