Say I have a ring $M_2(\mathbb{R})$, which is the ring of $2×2$ matrices with real entries.
Say I also know that $$S = \left\{\begin{pmatrix}
a & b\\
c & d
\end{pmatrix} \mid a,b,c,d \in \mathbb{R}, a + b = c+ d\right\}.$$
I'm trying to prove that $S$ is a subring of $M_2(\mathbb{R}).$
I assume that this is a matter of testing for $\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}$ and $\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$ in $S$ and then looking at addition and multiplication of matrices.
Am I on the right track? Is there a different way I should be looking at the proof?
Best Answer
If $I$ is the identity, $U=\begin{pmatrix}1&0\\1&0\end{pmatrix}$ and $V=\begin{pmatrix}0&1\\0&1\end{pmatrix},$ then if $a+b=c+d,$ $$\begin{pmatrix}a&b\\c&d\end{pmatrix}=(a+b-c)I+bV+cU$$
Show $S$ is exactly all $\alpha I+\beta U+\gamma V,$ with $\alpha,\beta,\gamma\in \mathbb R.$
Then use $UV=V^2=V, VU=U^2=U$ to prove $S$ is closed under multiplication.
Also show $I,\mathbf 0\in S.$
Alternatively, let $v_0=\begin{pmatrix}1\\1\end{pmatrix}.$ Then $A\in S$ if an only if, for some $\lambda\in \mathbb R,$ $Av_0=\lambda v_0.$
Then if $A_1, A_2$ have eigenvector $v_0$ with eigenvalues $\lambda_1,\lambda_2$ then show $$(A_1+A_2)v_0=(\lambda_1+\lambda_2)v_0\\ (A_1A_2)v_0=(\lambda_1\lambda_2)v_0.$$
Also, show $I$ and $\mathbf 0$ are in $S.$
This latter approach works in general. If $v_0\in\mathbb R^n$ is a fixed vector, define:
$$S_{v_0}=\left\{A\in M_n(\mathbb R)\mid \exists\lambda\in\mathbb R(Av_0=\lambda v_0)\right\}$$
$S_{v_0}$ is always a subring.
If $v_0\neq 0,$ the $\lambda$ associated for any $A\in S_{v_0}$ is unique, and the map $A\to\lambda_A$ is a homomorphism $S_{v_0}\to \mathbb R.$
The first technique is harder to generalize.