# Prove that G is a group under matrix multiplication.

Let G be the set of all $2x2$ invertible matrices whose columns add up to 1. So, G is the set of all matrices $A=\begin{bmatrix}a&c\\b&d\end{bmatrix}$ such that $a+b=1$ and $c+d=1$. Prove that G is a group under matrix multiplication.

So I know I have to prove closeness, associativity (which I've done), identity element, and being invertible everywhere, which I'm not sure how to use another matrix B to do so.

• Well, not just $a+b=1$ and $c+d=1$. Also $ad-bc\neq0$, since you need invertible. – Thomas Andrews Mar 2 '17 at 0:05
• @AloneAndConfused So can the second matrix B just be the identity matrix of a 2x2 matrix? And then just multiply A and B to get closure? – pmk1007 Mar 2 '17 at 0:11
• No, another matrix of the same form: $$A \, =\, \begin{pmatrix} a & 1-a \\ b & 1-b \end{pmatrix}$$ and $$B \, =\, \begin{pmatrix} a' & 1-a' \\ b' & 1-b' \end{pmatrix}.$$ Note what I wrote in my first comment was total rubbish (and hence deleted!) : if $a+b=1$ then $b=1-a$ – AloneAndConfused Mar 2 '17 at 0:18
• @AloneAndConfused $a'$? Doesn't that mean the derivative? – pmk1007 Mar 2 '17 at 0:21
• @AloneAndConfused So is $c=b$? Because if $b=1-a$ then shouldn't that equation go where $b$ is in the matrix and not in the place of $c$? – pmk1007 Mar 2 '17 at 0:23

Let $S$ be a set of $2\times 2$ matrices defined by:

$$S \, = \, \left\{\begin{pmatrix} a & c \\ b & d\end{pmatrix} \; : \; a+b=1, \, c+d=1, \, ad\neq bc\right\}.$$

Then $A\in S$ is of the form

$$A = \begin{pmatrix} a & c\\ 1-a & 1-c \end{pmatrix}$$

where $a, c\in\mathbb{R}$ with $a\neq c$. Now take two matrices in $A$:

$$A = \begin{pmatrix} a & c \\ 1-a & 1-c \end{pmatrix}, \quad B = \begin{pmatrix} a' & c' \\ 1-a' & 1-c' \end{pmatrix}.$$

with $a, c, a', c' \in\mathbb{R}$ with $a\neq c, \, a'\neq c'$. For closure, we require $AB\in S$:

$$AB = \begin{pmatrix} a & c \\ 1-a & 1-a \end{pmatrix}\begin{pmatrix} a' & c' \\ 1-a' & 1-c' \end{pmatrix} \, =\, \begin{pmatrix} aa' +(1-a')c & ac' +(1-c')c \\ (1-a)a' + (1 - c) (1 - a') & (1-a)c' + (1 - c)(1-c') \end{pmatrix}$$

which after simplifying gives

$$AB = \begin{pmatrix} aa' +(1-a')c & ac' +(1-c')c \\ 1-[aa'+(1-a')c] & 1 - [ac' +(1-c')c] \end{pmatrix}$$

which is of the same form (elements in first column add to 1, as do the elements in the second column; elements in first row can't be equal since $a\neq c$) and $AB\in S$. Hence $S$ is closed under matrix multiplication. That should get you started.

• Looks good, but again you've shown it for rows not columns or am I just completely confused? Just change it to $\left\{\begin{pmatrix} a & b \\ c & d\end{pmatrix} \; : \; a+c=1, \, b+d=1, \, ad\neq bc\right\}$ and massage a bit and it's done. – Arby Mar 2 '17 at 1:03
• But I need the columns to equal 1, not the rows. – pmk1007 Mar 2 '17 at 1:03
• But the OP question states for $a+b=1$ and $c+d=1$ i.e. the rows being equal not the columns – AloneAndConfused Mar 2 '17 at 1:05
• Oh I'm sorry I wrote the problem wrong. I corrected the matrix. – pmk1007 Mar 2 '17 at 1:08
• Ah okay. I will enter this answer accordingly. – AloneAndConfused Mar 2 '17 at 1:09

It may be more illuminating (possibly also easier) to proceed as follows.

Let $X$ be the $1\times2$ matrix $X=(1,1)$, a vector if you prefer. We observe that $$G=\{A\in M_{2\times2}(\Bbb{R})\mid \det A\neq0, XA=X\}.$$ It is then straightforward to prove the following facts, using only what you know about matrix multiplication.

1. If $XA=X$ and $XB=X$ for some $2\times2$ matrices $A$ and $B$ then also $X(AB)=X$
2. If $XA=X$ then also $X=XA^{-1}$.

These facts immediately imply that $G$ is a subgroup of the group of invertible $2\times2$-matrices.

In a later course you may learn about group actions and stabilizers. Then you can reinterpret this as stating that $G$ is the stabilizer of the vector $X$, when the group of invertible matrices acts on the (row) vectors by matrix muliplication.

The "easier" way of doing this is to note the condition on $A$ is equivalent to requiring $A$ to be invertible and satisfy:

$$\begin{pmatrix}1&1\end{pmatrix}A = \begin{pmatrix}1&1\end{pmatrix}$$

Then if $A,B$ satisfy this constraint, show $AB$ satisfies this constraint and that $A^{-1}$ satisfies this constraint.

For example, if $A$ is invertible, and $\begin{pmatrix}1&1\end{pmatrix}A = \begin{pmatrix}1&1\end{pmatrix}$ then $$\begin{pmatrix}1&1\end{pmatrix}A^{-1} = \left(\begin{pmatrix}1&1\end{pmatrix}A\right)A^{-1} = \begin{pmatrix}1&1\end{pmatrix} I = \begin{pmatrix}1&1\end{pmatrix}$$ So $A^{-1}$ is in our set.