Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The book asks you to prove that $SL_n(\mathbb{R})$ is generated by elementary (row operation) matrices in which one nonzero off-diagonal entry is added to the identity matrix. For example,

$$ \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} $$

acts by left multiplication on $2\times2$ matrices by adding $a$ times (row 2) to (row 1). Considering a simple example:

$$ M= \begin{bmatrix} a & 0 \\ 0 & 1/a \end{bmatrix}, a \neq 0$$

you can see that the matrix $M$ does belong to $SL_n(\mathbb{R})$. However, the elementary matrices composing $M$ are of the below type and not of the first type (nonzero off-diagonal entry).

$$ \begin{bmatrix} c & 0 \\ 0 & 1 \end{bmatrix} $$ i.e. one nonzero diagonal entry added to the identity matrix. So $$M = \begin{bmatrix} a & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1/a \end{bmatrix} $$

So $M$ clearly isn't generated by the first type. What is going wrong?

share|cite|improve this question
up vote 4 down vote accepted

The point is, you can do without using elementary row operations of the form $R_i\leftarrow \lambda R_i$. Given a matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ such that its first column is nonzero, we can always reduce its first column to $(1,0)^T$ using only elementary row operations of the form $R_i\leftarrow R_i+kR_j$. In fact, let $k$ be a scalar such that $ka+c\neq0$. Then $$ \begin{pmatrix}1&0\\-(ka+c)&1\end{pmatrix} \begin{pmatrix}1&\frac{1-a}{ka+c}\\0&1\end{pmatrix} \begin{pmatrix}1&0\\k&1\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} =\begin{pmatrix}1&\ast\\0&\ast\end{pmatrix}. $$ The first elementary row operation (i.e. the rightmost one) modifies the $(2,1)$-th entry to nonzero; the second one makes the $(1,1)$-th entry becomes $1$ and the third (the leftmost one) kills off the $(2,1)$-th entry.

For your particular example, we may put $k=1$ and get $$ \begin{pmatrix}1&0\\-a&1\end{pmatrix} \begin{pmatrix}1&\frac{1-a}{a}\\0&1\end{pmatrix} \begin{pmatrix}1&0\\1&1\end{pmatrix} \begin{pmatrix}a&0\\0&\frac1a\end{pmatrix} =\begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}. $$ Hence $$ \begin{align*} \begin{pmatrix}a&0\\0&\frac1a\end{pmatrix} &= \begin{pmatrix}1&0\\1&1\end{pmatrix}^{-1} \begin{pmatrix}1&\frac{1-a}{a}\\0&1\end{pmatrix}^{-1} \begin{pmatrix}1&0\\-a&1\end{pmatrix}^{-1} \begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}\\ &= \begin{pmatrix}1&0\\-1&1\end{pmatrix} \begin{pmatrix}1&\frac{a-1}{a}\\0&1\end{pmatrix} \begin{pmatrix}1&0\\a&1\end{pmatrix} \begin{pmatrix}1&\frac{1-a}{a^2}\\0&1\end{pmatrix}. \end{align*} $$

share|cite|improve this answer
Is there a proof for the case of general $n$? – AlphaGo May 14 at 6:55
@Mathaholic This is already a proof for the general case. If the first column is nonzero, you only need to kill the off-diagonal entries one by one. – user1551 May 14 at 14:17

In the case of $SL_2(\Bbb R)$, let $L(a) = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ and $R(a) = \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$.

Since $L(a)L(b) = L(a+b)$ and $R(a)R(b) = R(a+b)$, you need to find a way to write $\begin{pmatrix} a & 0 \\ 0 & 1/a \end{pmatrix}$ as some finite composition $L(a_1)R(a_2)L(a_3)\ldots$, with $a_i \neq 0$.

It is clearly not of the form $L(a_1)$.
Neither of the form $L(a_1)R(a_2) = \begin{pmatrix} 1 + a_1a_2 & a_1 \\ a_2 & 1 \end{pmatrix}$.
Neither of the form $L(a_1)R(a_2)L(a_3) = \begin{pmatrix} 1 + a_1a_2 & a_1+a_3+a_1a_2a_3 \\ a_2 & 1+a_2a_3 \end{pmatrix}$.
$L(a_1)R(a_2)L(a_3)R(a_4) = \begin{pmatrix} 1 + a_1a_2+a_1a_4+a_3a_4+a_1a_2a_3a_4 & a_1+a_3+a_1a_2a_3 \\ a_2 +a_4+a_2a_3a_4 & 1+a_2a_3 \end{pmatrix}$.

If this is a diagonal matrix, then we need $a_2+a_4+a_2a_3a_4 = a_1+a_3+a_1a_2a_3 = 0$, and then $L(a_1)R(a_2)L(a_3)R(a_4) = \begin{pmatrix} 1 + a_3a_4 & 0 \\ 0 & 1+a_2a_3 \end{pmatrix}$.

So, we have to pick $a_1 = -a_3/(1+a_2a_3) = -a_3a$, $a_2 = -a_4/(1+a_3a_4) = -a_4/a$. Then we are left to pick $a_3$ and $a_4$ such that $a_3a_4 = a-1$.

Pick $a_4=1$. Then $a_3 = a-1, a_2 = -1/a, a_1 = (1-a)/(1+(-1/a)(a-1)) = a(1-a)$. And you can check that $L(a-a^2)R(-1/a)L(a-1)R(1) = \begin{pmatrix} a & 0 \\ 0 & 1/a \end{pmatrix}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.