Is it possible to express $\left(\begin{array}{cc}a & -a \\ a-1 & 1-a \\ \end{array} \right)$ as a certain product of two matrices? Is it possible to express
$$\left(
          \begin{array}{cc}
            a & -a \\
            a-1 & 1-a \\
          \end{array}
        \right), \ \ \ \ a\in\mathbb R$$
as a certain product of two matrices? Namely,
$$\left(
          \begin{array}{cc}
            a & -a \\
            a-1 & 1-a \\
          \end{array}
        \right)=PQ$$
where $P$, $Q$ are two matrices with entries not all $1$.
I tried to do some product but without success. Any suggestions please?
 A: How about this?$$\left(
          \begin{array}{cc}
            1 & 0 \\
            1 & -1 \\
          \end{array}
        \right)\left(
          \begin{array}{cc}
            a & -a \\
            1 & -1 \\
          \end{array}
        \right)$$
A: Yes, it is.  Set
$P = \begin{bmatrix} a & a \\ a - 1 & a - 1 \end{bmatrix} \tag{1}$
and
$Q = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \tag{2}$
neither $P$ nor $Q$ have entries all equal to $1$, no matter what the value of $a$. And the product $PQ$ is . . . ?
A: Here's a start...
$$\left(
          \begin{array}{cc}
            a & -a \\
            a-1 & 1-a \\
          \end{array}
        \right)$$
This matrix is $2 \times 2$, so we will need $P$ to be $2 \times m$ and $Q$ to be $m \times 2$. This is probably simplest if we let $m=1$, so let's try that.
$$\underbrace{\left(
          \begin{array}{cc}
            w \\
            x \\
          \end{array}
        \right)}_{P} 
\underbrace{\left(
          \begin{array}{cc}
            y & z
          \end{array}
        \right)}_{Q}
= 
\left(
          \begin{array}{cc}
            wy & wz \\
            xy & xz \\
          \end{array}
        \right)$$
We want $$wy=a$$ $$wz=-a$$ $$xy=a-1$$ $$xz=1-a$$
There are four equations, and we need to solve for four variables ($w,x,y,z$).
Do you know how to proceed? (The answer is not unique.)
A: Consider the product of two matrices $A$ and $B$ in the following way.
\begin{align}
A \cdot B = \left(\begin{array}{cc} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array} \right)\left(\begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right) = \left(\begin{array}{cc} a & -a \\ a-1 & 1-a \end{array} \right).
\end{align}
Now, it can be seen that
\begin{align}
a_{1} a_{2} + b_{1} c_{2} &= a = -(a_{1} b_{2} + b_{1} d_{2}) \\
a_{2} c_{1} + d_{1} c_{2} &= a-1 = -(c_{1} b_{2} + d_{1} d_{2})
\end{align}
for which $a_{1} a_{2} + b_{1} c_{2} = -(a_{1} b_{2} + b_{1} d_{2})$ leads to
\begin{align}
a_{1} = - b_{1} \, \left( \frac{c_{2} + d_{2} }{ a_{2} + b_{2} } \right)
\end{align}
and from the second set 
\begin{align}
c_{1} = - d_{1} \, \left( \frac{c_{2} + d_{2} }{ a_{2} + b_{2} } \right).
\end{align}
It should be mentioned that $a_{2} \neq - b_{2}$. By using $a_{1} a_{2} + b_{1} c_{2} = a$ then
\begin{align}
b_{1} = \frac{a \, (a_{2} + b_{2})}{b_{2} c_{2} - a_{2} d_{2}}
\end{align}
and similarly
\begin{align}
d_{1} = \frac{(a-1) \, (a_{2} + b_{2})}{b_{2} c_{2} - a_{2} d_{2}}
\end{align}
where $b_{2} c_{2} \neq a_{2} d_{2}$. Putting this all together leads to
\begin{align}\tag{1}
\frac{1}{\left| \begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right|} \, \left(\begin{array}{cc} a \, (c_{2} + d_{2}) & -a \, ( a_{2} + b_{2}) \\ a \, ( c_{2} + d_{2}) & -(a-1) \, (a_{2} + d_{2}) \end{array} \right)\left(\begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right) = \left(\begin{array}{cc} a & -a \\ a-1 & 1-a \end{array} \right).
\end{align}
where $a_{2} d_{2} - b_{2} c_{2} \neq 0$ and $a_{2} + b_{2} \neq 0$. From here many choices can be made for the values of $\{a_{2}, b_{2}, c_{2}, d_{2}\}$. One example is $(a_{2}, b_{2}, c_{2}, d_{2}) = (1,0,0,-1)$.
