# Show that the set of all matrices of the form $\bigl(\begin{smallmatrix} a & 0 \\ 0 & 0 \end{smallmatrix} \bigr)$, $a \in R$ − {0} forms an abelian group under matrix multiplication.

## 1 Answer

Toolbox:
• A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
• (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
• (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
• (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a^ {-1} G.$
• (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
• e is called the identity element of $G$ and $a^{?1}$ is called the inverse of a in $G.$
Step 1:
Let $G=\{\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),a\in R-\{0\}\}$
To show $(G,o)$ is an abelian group where $'o'$ stands for matrix multiplication.
Step 2:
Closure :
Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)\in G$
$\therefore a,b\neq 0$
$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ab &0\\0&0\end{smallmatrix} \bigr)\in G$
Since $a\neq 0,b\neq 0\Rightarrow ab\neq 0$
Closure is satisfied.
Step 3:
Associativity :
Matrix multiplication is associative.
Step 4:
Existence of identity :
Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)(a\neq 0)$
Such that $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow \bigl(\begin{smallmatrix}ae &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow \bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow \bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$
It can be seen that $\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$
$\therefore \bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)\in G$ is the identity element.
Step 5:
Existence of inverse :
Let $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)\in G(a\neq 0)$
Consider $\bigl(\begin{smallmatrix}a' &0\\0&0\end{smallmatrix} \bigr)$ such that
$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}a' &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow \bigl(\begin{smallmatrix}aa' &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow aa'=1\Rightarrow a'=\large\frac{1}{a}$(which exists since $a\neq 0)$
Now $\large\frac{1}{a}\neq 0$
$\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)\in G$
$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$
$\therefore$ the inverse of $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$ is $\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)$ in $G$
Thus every element in G has an inverse.
Step 6:
Let $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)\in G$
Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ab &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ba &0\\0&0\end{smallmatrix} \bigr)$
$\Rightarrow \bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$
Since multiplication of real numbers is commutative.
$\therefore$ commutativity is satisfied.
The four group axioms being satisfied and the commutative property being satisfied $(G,0)$ is an abelian group.
answered Sep 16, 2013

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