Ttykuu

Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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 multiplicacion, i.e., it is a real subalgebra of the algebra $L_2(mathbb C)$;

I have tried to multiply it with this matrix:
beginbmatrix
a & b
\
c & d
endbmatrix

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 simplier way?

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

$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$

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:

$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$

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.

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

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

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