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

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?

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1 Answer 1

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 $A$.

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