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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 $x+y=1$?

user26857
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pigeon
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    You can't do anything with $x+y=1$. They should better have written that the ring (rng) $S$ has no left identity at all. They meant "$\begin{pmatrix} x & x\y & y\end{pmatrix}$ is not a left identity even if $x+y = 1$ (and it is hence a right identity)". – Daniel Fischer May 21 '15 at 19:22
  • @KittyL the question ,in the exersice was : "If $x+y=1$ ,show that $\begin{pmatrix} x & x \ y & y \end{pmatrix}$ is not a left identity in $S$" " – pigeon May 21 '15 at 19:30
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    Ignore my question. I agree with @DanielFischer. For the right hand side to be the same, you need $x=\frac{a}{a+b}, y=\frac{b}{a+b}$, which depends on $a,b$. So there is not a left identity. – KittyL May 21 '15 at 19:36

2 Answers2

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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.

Tyrrell McAllister
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Your result:

$\begin{pmatrix} x & x \\ y & y \end{pmatrix}\begin{pmatrix} a & a \\ b & b \end{pmatrix}=\begin{pmatrix} x(a+b) & x(a+b) \\ y(a+b) & y(a+b) \end{pmatrix}$

shows that in $S$ does not exists a left identity.

Emilio Novati
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  • Well passing from this result of mine , i could not understand the passage from here to the end result. Only when KittyL mentioned that i could see that $x$ and $y$ became dependents on $a$ and $b$, hence this is not a left identity. IMHO it is not always obvious to see such small details. – pigeon May 21 '15 at 21:15