# Why do $\mathbb{C}$ and $\mathbb{H}$ generate all of $M_2(\mathbb{C})$?

For this question, I'm identifying the quaternions $\mathbb{H}$ as a subring of $M_2(\mathbb{C})$, so I view them as the set of matrices of form $$\begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix}.$$

I'm also viewing $\mathbb{C}$ as the subfield of scalar matrices in $M_2(\mathbb{C})$, identifying $z\in\mathbb{C}$ with the diagonal matrix with $z$ along the main diagonal.

Since $\mathbb{H}$ contains $j=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$ and $k=\begin{pmatrix} 0 & i \\ i & 0\end{pmatrix}$, I know that $$ij+k=\begin{pmatrix} 0 & 2i\\ 0 & 0 \end{pmatrix}$$ and $$-ij+k= \begin{pmatrix} 0 & 0\\ 2i & 0 \end{pmatrix}$$ are in the generated subring. I'm just trying to find matrices of form $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0\\ 0 & d \end{pmatrix}$ for $a,d\neq 0$ to conclude the generated subring is the whole ring. How can I get these remaining two pieces? Thanks.

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Hint: Use linear combinations of $$jk=\pmatrix{i&0\cr0&-i\cr}\qquad\text{and the scalar}\qquad \pmatrix{i&0\cr 0&i\cr}.$$
Note that $jk$ is the quaternionic $i$ that is distinct from the scalar $i$. –  Jyrki Lahtonen Jun 13 '12 at 6:40
Ah, I forgot about $jk=i$. That makes it easy, thanks Jyrki! –  Noomi Holloway Jun 13 '12 at 6:45