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Let $A$ be the group of invertible $2\times 2$ matrices with real entries. Let

$$B = \left\{\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}\mid ad \not=0\right\}.$$

  1. Show that $B$ is closed under multiplication.
  2. Show that $B$ is closed under inverse.
  3. Write a matrix in each left coset of $B$ in $A$. Is the set of left cosets infinte or finite?

I know that $B$ is a subgroup of $A$ if 1 and 2 are true. I also know that $1$ is asking to show that if $C$ and $D$ are $2\times 2$ matrices belonging to $B$ then I need to show $CD$ does also. I do not know how to start.

Arnaud D.
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Hopper
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1 Answers1

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Let $P = \begin{bmatrix} p&q\\&r\end{bmatrix}$ Let $S = \begin{bmatrix} s&t\\&u\end{bmatrix}$

If $B$ is closed then $PS$ is in $B.$

Is there any reason why $P,S$ would not represent all pairs of matrices in $B$?

From what you know of matrices can you find $P^{-1}$

Is $P^{-1}$ in $B$?

The left cosets of $B$ is a the trickiest part of the problem

For some $a\in A$ it has a linearly independent set of columnvectors.

The first column of $aB$ will be a scalar multiple of $a$ The second column could be just about anything.

Doug M
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  • Are the blanks in P and S zeros? I am having trouble defining two matrices that would belong to the subset B. Can I just choose any P and S with variables and then multiply them together to show that B is closed under multiplication? – Hopper Mar 27 '18 at 22:54
  • The blanks are zeros. $P, S$ are generic matrices in $B$ and the product of any $PS$ will produce a matrix in $B$ – Doug M Mar 27 '18 at 22:58