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I would like to show that $G=\left\{\begin{bmatrix} a&a\\a&a \end{bmatrix}\mid a\in\mathbb{R}\setminus\{0\}\right\}$ is a group under matrix multiplication.

I've already verified that associativity holds and that the identity element exists, which is $I_2$. However I'm having trouble understanding why this is a group, since I don't see how to get the inverse, because the determinant of the matrix is $0$.

Any help would be appreciated. Thanks.

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  • $\begingroup$ How did you show that $I_2\in G$? Or perhaps what are you calling $I_2$, since what it usually is isn’t in $G$? $\endgroup$
    – Steve Kass
    Oct 15, 2014 at 3:54
  • $\begingroup$ Ah, you're right. I was thinking $I_2$ as the $2\times 2$ identity matrix, but the identity has some entries are $0$, so its not part of the group. Thanks for that catch. I have to find the correct identity element now too. $\endgroup$ Oct 15, 2014 at 3:56
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    $\begingroup$ I think $\begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\end{bmatrix}$ works as the identity. $\endgroup$ Oct 15, 2014 at 4:00
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    $\begingroup$ @SujaanKunalan it seems you have it, then. $\endgroup$ Oct 15, 2014 at 4:03

1 Answer 1

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Hint: note that $$ \pmatrix{1/2&1/2\\1/2&1/2} \pmatrix{a&a\\a&a} = \pmatrix{a&a\\a&a} \pmatrix{1/2&1/2\\1/2&1/2} = \pmatrix{a&a\\a&a} $$ Now, $$ \pmatrix{a&a\\a&a} \pmatrix{b&b\\b&b} = 2 \pmatrix{ab&ab\\ab&ab} $$ For what $b$ does $2ab = 1/2$?

Alternatively: consider the map $$a \mapsto a \pmatrix{1/2&1/2\\1/2&1/2} = \pmatrix{a/2&a/2\\a/2&a/2}$$

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