Two-dimensional Hermitian matrices YouTube

Eigenvalues

Recall that for Hermitian matrices, the diagonal terms are real and the cross-diagonal elements are complex conjugates of their transposes.

Thus, all two-dimensional Hermitian matrices can be written as $$M = \begin{bmatrix} x & z \\ z^* & y \end{bmatrix}$$ where $x, y \in \mathbb{R}$ and $z \in \mathbb{C}$.

Notice that the determinant is real $$\mathrm{Det}(M) = xy - z^* z \in \mathbb{R}$$

Write $z = re^{\mathrm{i}\theta}$. Then, the eigenvalues of the two-dimensional Hermitian matrix, $$M = \begin{bmatrix} x & z \\ z^* & y \end{bmatrix}$$ are given by $$\lambda_{\pm } = \frac{x + y}{2} \pm \sqrt{r^2 + \left(\frac{x - y}{2} \right)^2}$$ where $r^2 = z^* z$.

To see why this is true, we expand out the characteristic equation, $$\begin{align*}0 &=\mathrm{Det}(M - \lambda \mathbf{I})\\&= \left | \begin{matrix}x - \lambda & z\\z^* & y - \lambda\end{matrix} \right |\\&= \left( x - \lambda \right) \left( y - \lambda \right) - z^* z\\&= xy - \left( x + y \right)\lambda + \lambda^2 - r^2\\&= \lambda^2 - \left( x + y \right)\lambda + xy - r^2\\&= \left( \lambda - \frac{x + y}{2} \right)^2 + xy - \left( \frac{x + y}{2} \right)^2 - r^2 & \text{ completing the square }\\&= \left( \lambda - \frac{x + y}{2} \right)^2 - r^2 - \left( \frac{x - y}{2} \right)^2\end{align*}$$

The result follows from re-arranging and taking the square root.

Non-trivial eigenvalues

Non-trivial two-dimensional Hermitian matrices have two distinct real eigenvalues

Any Hermitian matrix, $$M = \begin{bmatrix} x & z \\ z^* & y \end{bmatrix}$$ where $x, y \in \mathbb{R}$ and $z =r e^{\mathrm{i} \theta}\in \mathbb{C}$, has two distinct real eigenvalues unless, trivially, $$r = 0 \text{ and } x = y$$ in which case it has two equal (real) eigenvalues.

To see why, putting $r \neq 0$ or $x \neq y$ means that the term under the square root sign is positive, $$r^2 + \left(\frac{x - y}{2} \right)^2 > 0$$ and so there will always be two distinct solutions, and they will be real.

But if we put $r=0$ and $x = y$ into the formulae for the eigenvalues, $$\begin{align*}\lambda_{\pm } &= \frac{x + y}{2} \pm \sqrt{r^2 + \left(\frac{x - y}{2} \right)^2}\\&= \frac{x + x}{2} \pm \sqrt{(0)^2 + \left(\frac{x - x}{2} \right)^2}\\&= x\end{align*}$$ meaning we get two equal eigenvalues. In this case, the matrix is a (real) multiple of the identity, $$M = x \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = x \mathbf{I}$$

We won't be interested in matrices with two equal eigenvalues.

Unit eigenvalues

For any (non-trivial) two-dimensional Hermitian matrices, if $$M^2 = \mathbf{I}$$ then the eigenvalues satisfy $$\lambda_\pm = \pm 1$$

To see why, expand out the product, $$\begin{align*}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} &= \begin{bmatrix} x & z \\ z^* & y \end{bmatrix} \begin{bmatrix} x & z \\ z^* & y \end{bmatrix}\\&= \begin{bmatrix} x^2 + z z^* & (x + y)z \\ z^*(x + y) & z^* z + y^2 \end{bmatrix}\\&= \begin{bmatrix} x^2 + r^2 & (x + y)re^{\mathrm{i}\theta} \\ (x + y)re^{-\mathrm{i}\theta} & y^2 + r^2 \end{bmatrix}\end{align*}$$ then we get four equations to solve, $$\begin{align*}x^2 + r^2 &= 1\\y^2 + r^2 &= 1\\(x + y)re^{\mathrm{i}\theta} &= 0\\(x + y)re^{-\mathrm{i}\theta} &= 0\end{align*}$$

The first two imply that $$y = \pm x$$ and the second two imply that $$y = - x \text{ or } r = 0$$

Thus, if we set $y = +x$ then that would immediately imply that $r = 0$ and we are back to two equal eigenvalues, which we don't want. So, we choose $$y = -x$$ which means that $$\begin{align*}\lambda_{\pm } &= \frac{x + y}{2} \pm \sqrt{r^2 + \left(\frac{x - y}{2} \right)^2}\\&= \tfrac{0}{2} \pm \sqrt{r^2 + \left(\frac{2x}{2} \right)^2}\\&= \pm \sqrt{r^2 + x^2}\\&= \pm \sqrt{1}\\&= \pm 1\end{align*}$$

So, the Hermitian matrices that we will be considering are of the form $$M = \begin{bmatrix} x & z \\ z^* & -x \end{bmatrix}$$