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.

Answer 6

Since there are already many correct and rigorous answers, I'll just mention a simple way of thinking about this.

Notice that the multiplication of two complex numbers is essentially about adding the angles of complex numbers and multiplying their magnitudes. In order to find the same structure among matrices, what's the most natural choice?

When multiplying two $2\times2$ rotation matrices, you get another rotation matrix which is merely the rotation matrix corresponding to the sum of angles. That is $R(\theta) R(\phi) = R(\theta+\phi)$ . Note that it only works with $2$ dimensional rotation matrices. In $3$ dimensions, the rotation matrices do not commute. We also wish to encode the information about length of the complex number in these matrices. How do we do that? Just multiply it by a scalar of the length of complex number.

$r R(\theta) = \begin{bmatrix} r\cos \theta & -r\sin \theta \\ r\sin \theta & r\cos \theta \end{bmatrix} $

So a rotation matrix (having the same angle as complex number) multiplied by a scalar equal to the length of the complex numbers is precisely that complex number (in terms of matrices). It is precisely the matrix in your problem statement (just identify $a$ and $b$ appropriately). You just need to check whether the addition also works fine with it. But that's trivial.