# [Reduced] Column Echelon Form

• Sep 23rd 2008, 01:22 PM
Chris L T521
[Reduced] Column Echelon Form
I'm trying to understand this a little better.

If we have a $n\times n$ matrix, $\left[\begin{array}{cccc}a_{11}&a_{12}&\cdots&a_{1n}\\a_ {21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots& a_{nn}\end{array}\right]$, would the CEF have the form $\left[\begin{array}{ccccc}1&0&0&\cdots&0\\a_{21}&1&0&\cd ots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\a_{(n-1)1}&a_{(n-1)2}&\cdots&1&0\\a_{n1}&a_{n2}&a_{n3}&\cdots&1\end {array}\right]$??

And would the RCEF of this matrix be $I_{n\times n}$??\

I'd appreciate any clarification!

--Chris
• Sep 23rd 2008, 07:45 PM
NonCommAlg
Quote:

Originally Posted by Chris L T521
I'm trying to understand this a little better.

If we have a $n\times n$ matrix, $\left[\begin{array}{cccc}a_{11}&a_{12}&\cdots&a_{1n}\\a_ {21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots& a_{nn}\end{array}\right]$, would the CEF have the form $\left[\begin{array}{ccccc}1&0&0&\cdots&0\\a_{21}&1&0&\cd ots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\a_{(n-1)1}&a_{(n-1)2}&\cdots&1&0\\a_{n1}&a_{n2}&a_{n3}&\cdots&1\end {array}\right]$??

And would the RCEF of this matrix be $I_{n\times n}$??\

I'd appreciate any clarification!

--Chris

i guess by "column echelon form" of a matrix $A$ you mean the row echelon form of $A^T.$ i'm saying this because row echelon form is what we normally use. well, the answer to your question

is "yes" only if your (square) matrix is invertible. for example $\begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 0 \end{bmatrix}$ is also in column echelon form. besides your matrix can be any $m \times n$ matrix.