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According to Nering's Linear Algebra and Matrix Theory, there are three types of elementary row operations on matrices:

  • I. Multiply a row by a non-zero scalar
  • II. Add one row to another
  • III. Interchange two rows

There are three analogous elementary column operation.

The associated elementary matrices are, for example example:

$$I.\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & a & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)$$

$$II.\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$

$$III.\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$

Nering gives:

Theorem II-6.1 Any non-singular matrix can written as a product of elementary matrices.

As mention in Are the statement and proof of this theorem regarding the linear transformation of a contented set satisfactory? CH Edwards, Theorem IV 5.1, CH Edwards asserts that a standard theorem of linear algebra says a non-singular matrix can be expressed as a product of two types of elementary matrices. These elementary matrices are of type I and II given by Nering.

In his book General Theory of Relativity, Dirac says that the determinant $g$ of the flat space-time metric is negative for oblique axes because it is $g=-1$ for the Minkowski metric.

With oblique axes $g$ must still be negative, because the oblique axes can be obtained from the orthogonal ones by a continuous process, resulting in $g$ varying continuously, and $g$ cannot pass through the value zero.

I take that as intuitively obvious. But, if a set of axes (basis vectors) can be continuously transformed to and from an orthogonal set without changing the sign of the matrix whose columns are the basis vectors, can't the same thing be accomplished using elementary operations?

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You can swap rows and columns using only the first two types of operations. For simplicity let us consider the vector $(a,b)$. Here are the steps:

  • Add the second component to the first: $(a+b,b)$
  • Multiply the first component by $(-1)$: $(-a-b,b)$
  • Add the first component to the second: $(-a-b,-a)$
  • Multiply the second component by $(-1)$: $(-a-b,a)$
  • Add the second component to the first: $(-b, a)$
  • Multiply the first component by $(-1)$: $(b, a)$
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  • $\begingroup$ $D=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right];C=\left[\begin{array}{cc} 1 & 0\\ 1 & 1 \end{array}\right];B=\left[\begin{array}{cc} -1 & 0\\ 0 & 1 \end{array}\right];A=\left[\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right]; E=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right];$ $BADCBA=E$ $\endgroup$ – Steven Thomas Hatton Mar 14 at 12:14
  • $\begingroup$ That is a very good answer. Concise and insightful. $\endgroup$ – Steven Thomas Hatton Mar 14 at 13:16

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