Prove that matrices of the form $\begin{pmatrix} x & x \\ x & x \end{pmatrix}$ are a group under matrix multiplication.

Question

G is the set of matrices of the form $G=$$\begin{pmatrix} x & x \\ x & x \end{pmatrix}$. So for this set to be a group I know it needs to be:

• Closed under matrix multiplication
• The Associative Property holds
• Contains an Identity Element
• Every element needs to have an inverse

So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?

Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Isn't the subgroup supposed to have the same identity element as its parent group?

Answer 1

Hint

1) What is $\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}$?

2) The multiplication of matrices is associative.

3) When you are looking for the identity you want

$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} e & e \\ e & e \end{bmatrix}=\begin{bmatrix} x & x \\ x & x \end{bmatrix}$$

Now, do the multiplication on the left, what do you get?

4) With the $e$ from $3)$ solve

$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}=\begin{bmatrix} e & e \\ e & e \end{bmatrix}$$

for $y$. Again, all you need to do is doing the multiplication...

P.S. In order for this to be a group, you need $x \neq 0$.

P.P.S Since $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=2\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, you can prove that

$$F: \mathbb R \backslash\{0 \} \to G$$ $$F(x) =\frac{x}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

is a bijection and it preserves multiplications. Since $\mathbb R \backslash\{0 \}$ is a group it follows that G must also be a group and $F$ is an isomorphism... But this is probably beyond what you covered so far...