Let's say I have the following matrix $A$:
$$A = \begin{bmatrix}1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{bmatrix}$$
and I row reduce it to $A'$:
$$A' = \begin{bmatrix}1 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
Correct me if I'm wrong, but elementary row operations show us if the rows are linearly dependant/independant, not if the columns are linearly dependant/independant. They don't affect the row space $R(A)$. but they do change the column space $C(A)$.
$C(A) = \left\{\lambda \begin{bmatrix}1\\1\\1\end{bmatrix} \ni \lambda \in \mathbb{R}\right\}$ whereas $C(A') = \left\{\lambda \begin{bmatrix}1\\0\\0\end{bmatrix} \ni \lambda \in \mathbb{R}\right\}$
Now since elementary row operations change the columns space $C(A)$, how can we find the linearly independent columns of $A$ and express $C(A)$ as linear combinations of those linearly independent columns?
Would we have to revert to elementary column operations (if there is such a thing I've never seen it before), to column reduce $A$ and find it's linearly independent columns?
Furthermore, what is the best way/method to find the columns space $C(A)$? Is it better to find the linearly independent columns and express $C(A)$ in terms of those, or do you just take the linear combinations of all the columns, and simplify from there?
Best Answer
You can obtain a basis for the column space by performing row operations.
After you obtain the row-echelon form, note down those columns that are pivotal columns.
Look at the original matrix, those columns should be linearly independent.