I learned from my lectures that it is not true that the pivot columns of $rref(A)$ form a basis for $Col(A)$. Now I am trying to fully understand why this is not true and my questions are:
- Is it because if the columns are linearly independent, it does not prove they are a basis in $\mathbb{R}^n$ ?
- Is it because there are some cases when $Col(A^T) \neq Col(A)$ Col(A) i.e. the column space does not equal the row space
- Or does the explanation lie in a scenario described in this question Could non pivot columns form the basis for the column space of a matrix?
I would sincerely appreciate any clarification because I am so confused at this point (after reading too many SE questions with different explanation on this subject) and any examples, thank you!
Best Answer
You get to reduced row echelon form by doing elementary row operations. Elementary row operations don't change the row space or the nullspace of a matrix, but they sure can change the column space. Think, for example, of using an elementary row operation to go from $$\pmatrix{1&0\cr1&0\cr}{\rm\quad to\quad}\pmatrix{1&0\cr0&0\cr}$$ and look at what happens to the column space.