# 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

## 1 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.

• and also nowhere in our course it was pointed out that elementary row operations can change column space, really good to know!But when the matrix is in rref and we have its Col(A) and Nul(A), there isn't much to be done or modified in terms of elementary row operations then how is it that the pivot columns of rref(A) do not form a basis for Col(A)? – Jen Mar 27 '17 at 11:54
• 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