Complex number isomorphic to certain $2\times 2$ matrices?

Question

I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.

$$\phi:a+bi \rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$

This is what I have so far:

Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.

$$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$

So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.

Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$.

Surjective: I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.

I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated!

Answer 1

• For injectivity you need to show that if $\phi(z_1)=\phi(z_2)$ then $z_1=z_2$. So assume that $$\phi(a+bi)=\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a'&-b'\\b'&a'\end{pmatrix}=\phi(a'+b'i)$$ then $$\begin{pmatrix}a-a'&-b+b'\\b-b'&a-a'\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$ so $a=a'$, and $b=b'$ which means that $a+bi=a'+b'i$.

• For surjectivity all the matrices are on the form $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$ with $a,b\in\mathbb{R}$, and the element $a+bi$ maps to it.

Answer 2

You could prove injectivity simply by showing $$z\ne w \Rightarrow \phi(z) \ne \phi(w)$$ wich is next to obvious.
Regarding surjectivity: A function is always surjective on its range. All you need to show here is that any Matrix $\pmatrix{a&-b\\b&a}$ is in the range of $\phi$, explicitly $a+bi$ is the required argument (sounds like nothing to show, because it's simply the definition of $\phi$ wich guarantees this)

Answer 3

For injectivity assume that for some $(a,b), (a',b') \in \mathbb C$:

$$ \phi (a,b) = \left (\begin{array}{cc} a & -b \\ b & a \end{array}\right ) = \left (\begin{array}{cc} a' & -b' \\ b' & a' \end{array}\right ) = \phi(a',b')$$

Then since two matrices are equal if and only if each entry us equal it follows that $a=a'$ and $b=b'$ hence $\phi$ is injective.

Your argument for surjectivity is good: Let $ \left (\begin{array}{cc} a & -b \\ b & a \end{array}\right ) \in M_2(\mathbb R)$. Then $\phi$ maps $(a,b)$ to it.

Answer 4

The map is bijective by the argument you give; the trick is showing that it respects addition and (especially) multiplciation. Here's another way of doing so: Consider $\mathbb{C}$ as a $2$-dimensional real vector space with basis $\{1, i\}$. You then have a map $f:\mathbb{C} \to M_2(\mathbb{R})$ (i.e., $2$-dimensional real matrices) defined by $f(z)w = zw$ (with respect to this basis). It's clear that this map preserves addition and multiplication. If you write out $f$ explicitly, it's exactly the map $a + bi \to \begin{pmatrix} a & b\\-b & a\end{pmatrix}$ you describe (at least modulo a sign flip).

Answer 5

Consider the ordered basis $\beta=\{1,i\},$ as in the answer by anomaly. Then, the range $$R(\phi) = span(\phi(\beta)) = span(\{\phi(1, 0), \phi(0, 1)\}) = span(\{ \left (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right ), \left (\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array}\right )\}),$$ is equal to the codomain $\{ \left (\begin{array}{cc} a & -b \\ b & a \end{array}\right ) : a,b \in \mathbb{R}\}.$ Hence, the linear transformation is surjective.

Now, if we apply the dimension theorem, then we observe that the null space is zero-dimensional. This implies the linear transformation is injective. Thus, we conclude it is an isomorphism.

Alternatively, if we find ordered bases $\beta$ and $\gamma$ for the domain and codomain, then the linear transformation $\phi$ is invertible if and only if the matrix representation $[\phi]_{\beta}^{\gamma}$ is invertible.

Using $\beta,$ as above, and $\gamma = \{ \left (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right ), \left (\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array}\right )\},$ then we have $$[\phi]_{\beta}^{\gamma} = \left (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right ),$$ since $\phi(1, 0) = 1 \cdot \left (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right ) + 0 \cdot \left (\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array}\right )$ and $\phi(0, 1) = 0 \cdot \left (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right ) + 1 \cdot \left (\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array}\right ).$ Clearly, $[\phi]_{\beta}^{\gamma}$ is invertible.