Proving that a product of certain matrices is not identity

Question

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

I suppose you are only allowing integer exponents, and I think you are only allowing positive ones, but please be specific about this. — Marc van Leeuwen, Apr 13 '13 at 17:13
Sorry. I am allowing integers, and I am allowing negative ones. — ziangzao, Apr 13 '13 at 17:14
You meant nonzero integers? (We have $xy^0x^{-1}y^0=I$, for instance.) — user1551, Apr 13 '13 at 17:22
@ziango: Then you must be forbidding $0$, as putting just one such exponent in an intermediate position clearly allows getting back to the identity. — Marc van Leeuwen, Apr 13 '13 at 17:24

score 2 · Answer 1 · answered Apr 13 '13 at 17:39

2

See Sanov subgroup in $SL(2,\mathbf Z)$ is free of rank two

answered Apr 13 '13 at 17:39

Marc van Leeuwen

115,048

score 1 · Answer 2 · answered Apr 13 '13 at 17:02

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 $I$.

score 0 · Answer 3 · answered Apr 13 '13 at 16:59

0

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.

answered Apr 13 '13 at 16:59

Jared

31,451

1

What if we allow for negative k though? – ziangzao Apr 13 '13 at 17:07
If you allow inverses, then we have $xx^{-1}=I$. I guess I wouldn't consider something like $x^{-k}y^{-j}$ to be a "product" of the matrices $x$ and $y$. – Jared Apr 13 '13 at 17:11

Proving that a product of certain matrices is not identity

3 Answers3