TITLE: How to express $x_{ij}$ using $x_{k,k+1}$?
QUESTION [0 upvotes]: Let 
$$
x_{i,j}(t) = \left( \begin{matrix} 1 & 0 & \cdots & 0 & 0 \\
0 & 1 & \cdots & 0 & 0 \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & t & 0 \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & 1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
 \end{matrix} \right)_{n \times n},
$$
$t$ is at (i,j). 
I know that $x_{i,i+2}(t) = x_{i,i+1}(t)x_{i+1,i+2}(1)x_{i,i+1}(-t)x_{i+1,i+2}(-1)$. For general $i,j$, is there a general formula which express $x_{ij}(t)$ using $x_{k,k+1}(t)$? Thank you very much.

REPLY [1 votes]: Hint.  This is essentially a question about expressing row operations in terms of multiplication by elementary matrices.  Take, for example, $n=4$ and $i=2$, and write your expression for $x_{i,i+2}$ in the form
$$x_{2,4}(t)=x_{2,3}(t)x_{3,4}(1)x_{2,3}(-t)x_{3,4}(-1)I\ .$$
If we now work out the RHS one product at a time, starting at the right, we have
$$\eqalign{
  I&=\pmatrix{1&0&0&0\cr0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr}\cr
  x_{3,4}(-1)I&=\pmatrix{1&0&0&0\cr0&1&0&0\cr0&0&1&-1\cr0&0&0&1\cr}\cr
  x_{2,3}(-t)x_{3,4}(-1)I&=\pmatrix{1&0&0&0\cr0&1&-t&t\cr0&0&1&-1\cr0&0&0&1\cr}\cr
  x_{3,4}(1)x_{2,3}(-t)x_{3,4}(-1)I&=\pmatrix{1&0&0&0\cr0&1&-t&t\cr0&0&1&0\cr0&0&0&1\cr}\cr
  x_{2,3}(t)x_{3,4}(1)x_{2,3}(-t)x_{3,4}(-1)I
    &=\pmatrix{1&0&0&0\cr0&1&0&t\cr0&0&1&0\cr0&0&0&1\cr}\cr}$$
and if you look carefully at the sequence of intermediate results I think you will soon understand how to place the $t$ anywhere you want above the diagonal.  However, it is not possible to put the $t$ below the diagonal unless you also use matrices of the form $x_{k+1,k}(t)$.
Good luck!
Edit.  The expression for $x_{2,4}(t)$, and in general $x_{i,j}(t)$, is not unique.  I think that if you confirm the identity
$$x_{2,4}(t)=x_{2,3}(-t)x_{3,4}(-1)x_{2,3}(t)x_{3,4}(1)$$
by multiplying it out step by step as in the above example, you will find it even easier to understand and generalise.
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