Let $G = \begin{Bmatrix} \begin{bmatrix}a & a \\a & a \end{bmatrix} \mid a\in \mathbb{R},a \neq 0 \end{Bmatrix}.$

Show that $G$ is a group under matrix multiplication.

Explain why each element of $G$ has an inverse even though the matrices have determinant $0$.

I've managed to show that closure, associativity and identity indeed holds for this general linear group. The problem I'm facing is showing that inverses holds and what explanation can be provided for the existence of the inverse.

Only hint to be provided

  • $\begingroup$ Note the edit to the entries of the matrix $\endgroup$ – Mathematicing Jan 20 '16 at 6:11
  • $\begingroup$ I cannot see any way how this set forms a group under standard matrix multiplication. What is the neutral element? The identity matrix is not in the set. Should it be a different group operation? Am I missing something? $\endgroup$ – Nephente Jan 20 '16 at 6:18
  • 2
    $\begingroup$ Yes, re-read my statement. $I_2$ is not in $G$. The definition of group I am familiar with demands the identity to be an element of that group. $\endgroup$ – Nephente Jan 20 '16 at 6:27
  • $\begingroup$ I was under the assumption that $I_{2}$ is in G by definition of what Group meant. But the condition set on the set G requires a=/= 0. So this is another quandary. $\endgroup$ – Mathematicing Jan 20 '16 at 6:30
  • $\begingroup$ There could exist another "identity" in the group though. One that doesn't really look like $I_2$ but still meets the definition of $Me = eM = M$. $\endgroup$ – Decaf-Math Jan 20 '16 at 6:31

You know the identity, so try solving for the inverse directly: $$\begin{bmatrix}a&a\\a&a\end{bmatrix}\begin{bmatrix}b&b\\b&b\end{bmatrix} = \begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$$ $$\begin{bmatrix}2ab&2ab\\2ab&2ab\end{bmatrix} =\begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$$

Do you see where to go from here? (remember we are solving for $b$ :))

I think that after you see the calculation of the inverse matrix, it will make more sense to you why the inverse is well-defined even though all the matrices have determinant zero.

  • $\begingroup$ We have $b=\frac{1}{4a}$ with $a=/=$ so b is well defined. $\endgroup$ – Mathematicing Jan 20 '16 at 6:39
  • $\begingroup$ Does it make sense why it doesn't matter that the determinant is 0? $\endgroup$ – TokenToucan Jan 20 '16 at 6:40
  • $\begingroup$ Yes. Any matrix, even if det=0, will have entries b=1/(4a) with a non-zero $\endgroup$ – Mathematicing Jan 20 '16 at 6:43
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    $\begingroup$ That's true, but I think what is more important to note is that the identity in your group is not the `usual' identity for $2\times 2$ matrices, and the fact about invertibility and determinant only applies if we are aiming to get $I_2$, rather than the identity for this group $G$. $\endgroup$ – TokenToucan Jan 20 '16 at 6:45
  • $\begingroup$ Maybe think about the isomorphism $G \rightarrow \mathbb Z$ by $\phi(\begin{bmatrix}a&a\\a&a\end{bmatrix}) = a$ with a group operation $a*b = 2ab$. It's easy to see why this is (commutative) group, and the question about inverses and determinant doesn't come up because it's simply irrelevant to this group. $\endgroup$ – TokenToucan Jan 20 '16 at 6:47

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