Example of matrix group related to complex numbers [duplicate]

Solution 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.

Solution 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)

Solution 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.

Solution 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).

Solution 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.