Why is $H = \left\{ \left(\begin{array}{rl} z_1&z_2\\ -\bar z_2&\bar z_1 \end{array}\right) \bigg\vert \ z_1, z_2 \in \mathbb C \right\}$ closed?

Question

Consider the set of matrices $$H = \left\{ \left(\begin{array}{rl} z_1&z_2\\ -\bar z_2&\bar z_1 \end{array}\right) \bigg\vert \ z_1, z_2 \in \mathbb C \right\}.$$ It is a four-dimensional real subspace of the vector space $L_2(\mathbb C)$, and enjoys the following remarkable properties:

$1)$ $H$ is closed under multiplication, i.e., it is a real subalgebra of the algebra $L_2(\mathbb C)$;

I have tried to multiply it with this matrix: \begin{bmatrix} a & b \\ c & d \end{bmatrix}

where $a$, $b$, $c$, and $d$ are complex numbers but I got a very big formula that I do not know how this formula still is in $H$. Is there any suggestions for proving this in a simpler way?

Answer 1

So, for a set $S$ of matrices (or any sort of element) to be closed under an operation $\ast$ on it, we require that, for all $a,b \in S, a \ast b \in S$.

As I noted in the comments, your issue lied in multiplying a matrix of $H$ by a generic matrix of complex elements, which is too general to have closure. You have to take two generic matrices of the set. So, let $a,b,c,d \in \Bbb C$ and then consider the multiplication

$$\begin{bmatrix} a & b\\ -\bar{b} & \bar{a} \end{bmatrix} \begin{bmatrix} c & d\\ -\bar{d} & \bar{c} \end{bmatrix} =\begin{bmatrix} ac - b \bar{d} & ad+b\bar{c}\\ -\bar{a} \bar{d} - \bar{b}c & \bar{a} \bar{c}-\bar{b}d \end{bmatrix} $$

You can see immediately the left two matrices are of the form of matrices in $H$; on the right is their product. You can verify that it, too, matches by noting a couple of properties of the complex conjugate:

$$\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2} \;\;\;\;\; \text{and} \;\;\;\;\; \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \;\;\;\;\; \text{and} \;\;\;\;\; \overline{\overline{z_1}} = z_1$$

where $z_1,z_2 \in \Bbb C$. So if...

• ...the bottom-left entry is the negative of the conjugate of the top-right
• ...the bottom-right entry is the conjugate of the top-left

...then the product is in the form for a matrix in $H$. It does happen to hold, and thus $H$ is closed under matrix multiplication.

Answer 2

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.

Hint Denote $$J := \pmatrix{\cdot&-1\\1&\cdot}.$$ It follows immediately from the definition that $$\{X \in M(2, \Bbb C) : \textrm{$X$ satisfies $X^\dagger J = J X^\top$}\} .$$

So, for $X, Y \in H$, $$(X Y)^\dagger J = Y^\dagger X^\dagger J = Y^\dagger JX^\top = J Y^\top X^\top = J (XY)^\top .$$