Hw3 Problem 9

### Problem 9

Let $H$ be the subset of $M_{2}(\mathbb{R})$ consisting of all matrices of the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ for $a$,$b \in \mathbb{R}$. One can show H is closed under matrix addition and matrix multiplication.

$\hspace{24pt}$ $(a)$ Show that $\langle \mathbb{C} , +\rangle \simeq \langle H,+\rangle$.
$\hspace{24pt}$ $(b)$ Show that $\langle \mathbb{C} , \cdot \rangle \simeq \langle H, \cdot \rangle$.

Solution
In order to prove that $\langle \mathbb{C} , \cdot \rangle \simeq \langle H, \cdot \rangle$, it is sufficient to show that
(i) $\hspace{10pt}$ there exists some function $\phi : \mathbb{C} \rightarrow H$ that provides a one-to-one mapping from $\mathbb{C}$ onto $H$ and
(ii) $\hspace{10pt}$ $\phi$ is a homomorphism

$\hspace{10pt} (a)$
$\hspace{20pt}$ (i)
Let $\phi : \mathbb{C} \rightarrow H$ be defined such that for all $x=(a+b\cdot i)\in \mathbb{C}$, $\phi (x) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$
Suppose that for some $x,y\in \mathbb{C}$, with $x=a+b\cdot i$ and $y=c+d\cdot i, \phi (x) =\phi (y)$.
Then $\begin{bmatrix} a & -b \\ b & a\end{bmatrix} = \begin{bmatrix} c & -d \\ d & c \end{bmatrix}$
Now, by definition of matrix equality, $a = c$ and $b=d$.
Thus, $x = a + b\cdot i = c + d\cdot i = y$
Therefore, $\phi$ is one-to-one.

Let $x\in H$
Then $x$ may be written in the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ with $a,b\in \mathbb{R}$
Now, since by definition of $\mathbb{C}$, $\mathbb{C}$ contains all numbers of the form $q+r \cdot i$ with $q,r \in \mathbb{R}$, there must exist some $y \in \mathbb{C}$ with $y=a + b \cdot i$.
Observe that $\phi (y) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Thus, for any $x \in H$, there exists some $y \in \mathbb{C}$ such that $\phi (y) = x$
Therefore, $\phi$ maps $\mathbb{C}$ onto $H$.
$\hspace{300pt}$ $\clubsuit$

$\hspace{20pt}$ (ii)
let $x,y \in \mathbb{C}$ with $x=a+b\cdot i, y=c+d\cdot i$, and $a,b,c,d \in \mathbb{R}$.
Then $\phi (x) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}, \phi (y) = \begin{bmatrix} c & -d \\ d & c \end{bmatrix}$
Hence, $\phi (x) + \phi (y) = \begin{bmatrix} (a+c) & -(b+d) \\ (b+d) & (a+c) \end{bmatrix}$
Now, $\phi (x+y) = \phi \bigg( (a+c)+(b+d)\cdot i \bigg) = \begin{bmatrix} (a+c) & -(b+d) \\ (b+d) & (a+c) \end{bmatrix}$
Thus, $\phi (x) + \phi (y) = \phi (x+y)$
Therefore the homomorphism property holds.
All conditions hold, so $\langle \mathbb{C} , +\rangle \simeq \langle H,+\rangle$
$\hspace{200pt} \clubsuit$

$\hspace{10pt} (b)$
(i)
By the same argument presented in part $(a)$, $\phi$ is a one-to-one function mapping $\mathbb{C}$ onto $H$.
(ii)
Let $x,y \in \mathbb{C}$ with $x=a+b\cdot i, y=c+d\cdot i$, and $a,b,c,d\in \mathbb{R}$
Then $\phi (x)=\begin{bmatrix} a & -b \\ b & a \end{bmatrix} , \phi (y)=\begin{bmatrix} c & -d \\ d & c \end{bmatrix}$
Hence, $\phi (x)\cdot \phi (y) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \cdot \begin{bmatrix} c & -d \\ d & c \end{bmatrix} = \begin{bmatrix} (a\cdot c-b\cdot d) & (-d\cdot a-c\cdot b) \\ (d\cdot a+c\cdot b) & (a\cdot c-b\cdot d) \end{bmatrix}$
Now, $\phi (x\cdot y) = \phi \bigg((a+b\cdot i)\cdot (c+d\cdot i)\bigg) = \phi (a\cdot c - b\cdot d + a\cdot d\cdot i + b\cdot c\cdot i) = \phi \bigg( (a\cdot c-b\cdot d)+(d\cdot a + c\cdot b) \cdot i \bigg)$
Thus, $\phi (x \cdot y) = \begin{bmatrix} (a\cdot c-b\cdot d) & (-d\cdot a-c\cdot b) \\ (d\cdot a+c\cdot b) & (a\cdot c-b\cdot d) \end{bmatrix}$
Consequently, $\phi (x) \cdot \phi (y) = \phi (x\cdot y)$ for all $x,y\in \mathbb{C}$, so the homomorphism property holds.
Therefore, since all conditions hold, $\langle \mathbb{C}, \cdot \rangle \simeq \langle , H, \cdot \rangle$
$\hspace{100pt} \clubsuit$