Proving that a product of certain matrices is not identity I have a very quick question regarding matrices. Consider $ x=  \left( \begin{array}{cc}
1 & 2  \\
0 & 1  \\
 \end{array} \right) $ and $ y = \left( \begin{array}{cc}
1 & 0  \\
2 & 1  \\
 \end{array} \right)$
Clearly no power except $0$ of $x$ or $y$ is equal to $I$, and clearly $x^k y^j$ is not identity for nonzero $k,j$. My question: Is any product $x^{k_1}y^{k_2}...x^{k_{n-1}}y^{k_n}$ not equal to $I$? I suspect so, but I am not sure how to prove it. It seems like induction but I don't see how it should go
 A: See Sanov subgroup in $SL(2,\mathbf Z)$ is free of rank two
A: Let $P$ be the product of $x$s and $y$s. Consider the sum $S$ of all entries in $P$. If all entries are non-negative, multiplying by either $x$ or $y$ will increase it.
Base cases:
$P = \left(\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right)$, $S=4$.
$P = \left(\begin{matrix}1 & 0 \\ 2 & 1\end{matrix}\right)$, $S=4$.
Recursive cases:
Assume that $S \ge 4$, and all entries are non-negative.
$P = \left(\begin{matrix}a & b \\ c & d \end{matrix}\right)$
$P^\prime = Px = \left(\begin{matrix}a & 2a + b \\ c & 2c + d \end{matrix}\right)$, $S^\prime = 3a+b+3c+d \ge S + 2a + 2c \ge 4$
$P^\prime = Px = \left(\begin{matrix}a + 2b & b \\ c + 2d & d \end{matrix}\right)$, $S^\prime = a+3b+c+3d \ge S + 2b + 2d \ge 4$
Thus $P^\prime$ has all non-negative entries, and a sum at least 4.
Since $I$ has $S = 2$, no $P$ can ever equal $I$.
A: Suppose we have $x^{k_1}y^{k_2}\ldots x^{k_{n-1}}y^{k_n}=I$.  It follows that $y$ is the inverse of $x^{k_1}y^{k_2}\ldots x^{k_{n-1}}y^{k_n-1}$.  If
$$x^{k_1}y^{k_2}\ldots x^{k_{n-1}}y^{k_n-1}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then we know that $a,b,c,d\ge 0$ from the forms of $x$ and $y$.  However, we have
$$y^{-1}=\begin{pmatrix}1&0\\-2&1\end{pmatrix}$$
a contradiction.
