# Pivot columns and basis for column space

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!

• It is certainly true that the column spaces of $A$ and $A^t$ need not be the same. If $A$ isn't square, they can't be the same, and even if $A$ is square, consider the column spaces of $$\pmatrix{1&1\cr0&0\cr}$$ and its transpose. – Gerry Myerson Mar 10 '17 at 8:40
• Are you still there, Jen? Anything to say? – Gerry Myerson Mar 12 '17 at 7:54
• @GerryMyerson yes, sorry had a chaotic semester but I did read your answer that night and it definitely contributed tremendously to my understanding of null spaces!Especially so that we've moved to eigenvectors after the spring break! – Jen Mar 27 '17 at 11:50

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.
• If the matrix $A$ is in RREF, then its pivot columns (which are all of the form, $(1,0,\dots,0)$, $(0,1,0,\dots,0)$, and so on) do form a basis for its column space. – Gerry Myerson Mar 27 '17 at 22:31