$Let \ A = \left[\begin{array}{cccc}1&2&-1&1\\2&4&-3&0\\1&2&1&5\end{array}\right]$

Using Gauss elimination, lead matrix A to row reduced echelon form:


And now using that form I am supposed to say what are column and row space basis. I think that Column space basis is: $\left[\begin{array}{cc}1&-1\\2&-3\\1&1\end{array}\right]$ But what is rowspace basis???

I dont know why I am asking all these question, when wikipedia exist but whatever http://en.wikipedia.org/wiki/Row_space says that row space are all non zero row vectors in (reduced row) echelon form. So i Guess its: $\left[\begin{array}{cccc}1&2&0&3\\0&0&1&2\end{array}\right]$ Good?


2 Answers 2


Those two rows do provide a basis of the row space, yes - it's not the only one, but it answers the question so it will do.

The point is that the rows have the same span after Gaussian elimination as they do before, but in row reduced echelon form the non-zero rows are linearly independent, so they are a basis, rather than just a spanning set.

If you know the sifting algorithm you could also just apply that to the rows of the original matrix, to get a basis for the row space whose members are actually rows of $A$.


Please note that elementary row operations as well as column operations don't alter the row space, column space and rank of a matrix.

The objective of performing echelon transformation on a matrix, row wise (column wise), is to identify the linearly dependent rows (columns). The rows (columns) which are linearly dependent on other rows (columns) ultimately become null under such operations.

Again the definition of rank of a matrix is the maximum number of linearly independent rows or columns i.e. the non-null rows or column in the echelon matrix. That's why row rank of a matrix is the number of non-null rows in row echelon form or row reduced echelon form.

Same concept goes for column rank, but here you need to perform elementary column operations.

  • $\begingroup$ You wrote this Answer two years after the Question was posted, so there should have been no urgency to justify a short sloppy explanation. Voting to delete. $\endgroup$
    – hardmath
    Jan 12 at 2:09

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