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I know that for each matrix there is only one reduced row echelon form. However, there can be multiple row echelon forms (not reduced) for a single matrix.

Can one matrix have one row echelon form with an arbitrary number of nonzero rows (for ex. 2), and another row echelon form with a different, arbitrary number of nonzero rows (for ex. 1)?

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  • $\begingroup$ The number of nonzero rows in row echelon form must be the same for all different row echelon forms of the same matrix. The number of nonzero rows in a matrix in row echelon form is, in fact, the rank of the matrix. But rank is not changed by row operations. $\endgroup$ – Will Orrick Sep 3 '14 at 3:43
  • $\begingroup$ Another way to look at it: suppose one row echelon form of a matrix had one nonzero row and another row echelon form of the same matrix had two nonzero rows. The row operations that lead to row echelon form are reversible. So there would be a sequence of row operations that turned the matrix with one nonzero row into the row echelon matrix with two nonzero rows. But the only thing row operations can produce in a matrix with one nonzero row are scalar multiples of the nonzero row. You could therefore never end up with a row echelon matrix with two nonzero rows. $\endgroup$ – Will Orrick Sep 3 '14 at 3:45
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So I can find two ways to reduce the matrix $A$ into echelon (i.e. upper triangular) form:

1) I can factorize the matrix as $A=LU$, where $L$ is lower triangular and $U$ is upper triangular. This is probably the first factorization that gets taught in a linear algebra class (it is deeply connected to the the echelon form of the matrix).

2) Alternately, I can factorize the matrix as $A=QR$, where $Q$ is orthonormal or unitary (depends if your matrices are over the real or the complex numbers) and $R$ is right (i.e. upper) triangular.

In general, these are completely different factorizations. To convince yourself for the 2 x 2 case, I'd suggest something like this:

1) Start with the matrix $$A=\pmatrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}\pmatrix{a&0\\0&b}$$

2) Multiply $A$ out and figure out its LU factorization

3) Compare the LU factorization to the original expression for $A$

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  • $\begingroup$ I'm not seeing how this addresses the original question, which was concerned with the number of nonzero rows in a row echelon matrix. In the context of the OP's question, you need to insist that $L$ be nonsingular. If you do that, then $U$ and $R$ will have the same number of nonzero rows. In other words, the answer to the OP's question (Can different row echelon forms of the same matrix have different numbers of nonzero rows?) is no. $\endgroup$ – Will Orrick Sep 3 '14 at 4:16
  • $\begingroup$ Fair point. Skipped the piece about the factorizations being rank revealing. $\endgroup$ – Brendan Drew Sep 3 '14 at 6:47

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