Problem source

\begin{questionparts}  \item Let $S$ be the set of matrices of the form
\[
\begin{pmatrix}a & a\\
a & a
\end{pmatrix},
\]
where $a$ is any real non-zero number. Show that $S$ is closed under
matrix multiplication and, further, that $S$ is a group under matrix
multiplication. 

\item Let $G$ be a set of $n\times n$ matrices which is a group
under matrix multiplication, with identity element $\mathbf{E}.$
By considering equations of the form $\mathbf{BC=D}$ for suitable
elements $\mathbf{B},$ $\mathbf{C}$ and $\mathbf{D}$ of $G$, show
that if a given element $\mathbf{A}$ of $G$ is a singular matrix
(i.e. $\det\mathbf{A}=0$), then all elements of $G$ are singular.
Give, with justification, an example of such a group of singular matrices
in the case $n=3.$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item Let $\mathbf{A} = \begin{pmatrix}1 & 1\\
1 & 1
\end{pmatrix}$, then we need to show that $(a\mathbf{A})(b\mathbf{A})$ is of the form $cA$ where $a, b, c \neq 0$.

Since $\mathbf{A}^2 = \begin{pmatrix}2 & 2\\
2 & 2
\end{pmatrix} = 2\mathbf{A}$ this is certainly the case, since $(a\mathbf{A})(b\mathbf{A}) = 2ab\mathbf{A}$.

To check that we have a group be need to check:

\begin{itemize}
\item Closure (done)
\item Associativity (inherited from matrix multiplication)
\item Identity ($\frac12 \mathbf{A}$)
\item Inverses the inverse of $a\mathbf{A}$ is $\frac{1}{4a}\mathbf{A}$
\end{itemize}

\item Suppose $\mathbf{A}$ is singular (ie $\det\mathbf{A}=0$), then $\mathbf{AA^{-1}B=B}$ (where inverse is the group inverse rather than the matrix inverse) for any matrix $\mathbf{B}$. Taking determinants we have:

$\det(\mathbf{AA^{-1}B}) =  \det(B) \Rightarrow \det(A) \det(A^{-1}B) = \det(B) \Rightarrow 0 = \det(B)$, ie all matrices are singular.

Consider the set of non-zero multiples of $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$, then the same logic as part (i) will suffice

\end{questionparts}