Recently, I learned that for each matrix, there are several row echelon forms, and a singular reduced row echelon form that each matrix can be reduced down to.

However, the conditions under which a matrix can be deemed reduced to RREF or REF seem really convoluted and complicated.

I'll focus on RREF for this question --

Pivots are $1$, are the leftmost nonzero in their row, are the only nonzero in their column? And then all rows full of zeros are on the bottom? And the matrix should be upper triangular? Am I missing anything, and is their some sort of pneumonic, trick, or picture in mind I should have when reducing?

  • 1
    $\begingroup$ If you solve $Ax=b$ using Gaussian elimination, and you try to be kind of systematic about it, you will probably find yourself putting $A$ in reduced row echelon form even if you don't know what RREF means. You just systematically eliminate variables one by one. Understanding the Gaussian elimination strategy allows you to remember what RREF should look like. $\endgroup$ – littleO Jun 12 '17 at 0:58
  • $\begingroup$ Before I even knew what these matrices were, the way I did systems of equations seemed similar to RREF. I had each leading variable to be one, and I would end up with simple quick substitutions @littleO $\endgroup$ – Saketh Malyala Jun 12 '17 at 1:02
  • $\begingroup$ As for a picture you should have in your mind, for each pivot, (which you say is the furthest left nonzero entry of a row, technically a little more complicated than that, but its close enough not to correct here) you need everything to the left, below, and below and to the left to be zero. The matrix $\begin{bmatrix}1&0&0&0\\0&0&0&\color{blue}{1}\\0&0&\color{red}{1}&0\\0&0&0&0\end{bmatrix}$ is not in RREF despite being upper triangular since rows two and three are in the wrong order, the red $1$ is below and to the left of the blue $1$. $\endgroup$ – JMoravitz Jun 12 '17 at 1:56

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