Let $S$ be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form
$\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$
The matrix $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is right identity in $S$ if and only if $x+y=1$. Fine, I can see that.
But I cannot see why "If $x+y=1$ , then $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is not a left identity in $S$".
I have tried that, if $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is a left inverse then : $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\
y(a+b) & y(a+b)
\end{pmatrix}=\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$ in which case we have $x(a+b)=a$ and $y(a+b)=b$. What can i do with $x+y=1$?
Best Answer
To prove that $S$ contains no left identity, let an arbitrary element $ A = \begin{bmatrix} x & x \\ y & y \end{bmatrix} \in S $ be given. Now, either $x = 0$ or $x \ne 0$. If $x = 0$, then note that $ \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \in S $, but \begin{equation*} A \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ y & y \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ y & y \end{bmatrix} \ne \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \end{equation*} so $A$ is not a left identity in $S$. If, on the other hand, $x \ne 0$, then note that $ \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \in S $, but \begin{equation*} A \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x & x \\ y & y \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x & x \\ y & y \end{bmatrix} \ne \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \end{equation*} so $A$ is not a left identity in $S$. Therefore, no element of $S$ can be a left identity in $S$. That is, $S$ does not contain a left identity.