# Basis of Column Space of matrix that is in Row Reduced Echelon form

I'm considerably new to proving things(and to Linear Algebra too).So, I hope someone would help me with this.

While proving Row Rank of Matrix = Column Rank of Matrix

Proof used the point that

For a Row Reduced Echelon Matrix, Basis of Column Space is just set of columns that contain leading non-zero entries.

Can someone provide a proof of this (Hints would more be appreciated)?

• If you mean the column space of the rref and not the original matrix, this should be pretty obvious. Each such column has a different single nonzero entry that corresponds to a nonzero row of the rref. – amd Feb 25 '18 at 7:42

## 1 Answer

For the proof you are looking for take a look to Prove that pivot columns of row reduced form of any matrix forms a basis in column space of that row reduced matrix. for the simpler case and here for the other Pivot columns of A are a basis for Col(A).

For the proof of this fundamental result and property of matrices you can take a look here Proofs that column rank = row rank.

For the intuition behind take a look here intuitive explanation why the row rank is equal to the column rank for a matrix.

• my question is not about proof of column rank=row rank.It's about basis of column space when a matrix is in row reduced echelon form. – manifold Feb 25 '18 at 7:33
• @viru Yes sorry but at first I didn't uderstand what you were looking for, I hope the given reference are useful to clarify the point. Bye – user Feb 25 '18 at 7:52
• Thanks, that pdf has answer to exactly what I was looking for.Besides it's simpler too :). – manifold Feb 25 '18 at 7:59