# How do row operations affect the column space?

I've been curious about this: Row operations do not affect the row space, but they affect the column space.

Is there any way to 'systematically' perform row operations to make the column space the way we want?

Let $C$ be the left matrix below, and $A$ be the right.

I'd like to find $C$ to make a column of $AC$ to be a constant multiplication of another column. Given $A$ has no pair of two column vectors that is linearly dependent. Since row operations, $C$ can change the column space of $A$, I wonder if there's any systematic way of making the dependencies of the column vectors the way I want.

$\left( \begin{array}{l} c_1 & c_2 \\ c_3 & c_4 \end{array} \right)$$\left( \begin{array}{l} 1 & 3 & 2 \\ 3 & 1 & 2\end{array} \right)$

Thanks.

• What does "make the column space the way we want" mean? – Jonas Meyer Dec 23 '14 at 10:19
• @JonasMeyer Hi, thanks for your comment. It's better to give an example, I think. I will edit the original post. – PurplePenguin Dec 23 '14 at 10:30

If $c^1,\ldots, c^n$ are the columns of a matrix and $d^1,\ldots, d^n$ the columns of a transformed matrix after some row operations, then $d^i$ is the expression of $c^i$ in a different basis. The transition matrix to this new basis is formed by the columns of $R^{-1}$ where $D=RC$ is the relation between $D=(d^1,\ldots, d^n)$ and $C=(c^1,\ldots, c^n)$.
(It follows from this that, if $c^1$ and $c^2$ are independent -- for example -- then $d^1$ and $d^2$ will be also independent, etc.)
• Thanks for your answer. Now, I see that through elementary operations, the dependency relations are not changed. My goal is actually to achieve the following. Given $H$, which is a 7-by-6 matrix, I'd like to design $D$, which is a 6-by-5 matrix in such a way that $HD = L$, where the first two rows of $L$ are dependent only on the third and fourth rows, and the rank of $L$ is 5. Now, things become different since column operations, $D$, are not elementary ones. The dependency relations between rows of $H$ can be changed. – PurplePenguin Dec 23 '14 at 11:07
• As far as I know, if $L$ is in the column space of $H$, then I can set $D = (H^T H)^{-1} H^T$ so that $HD$ becomes the projection of $L$ into the column space of $H$. Since $L$ lies in the column space of $H$, $HD$ results in $L$. The question is can I find such $D$ and $L$ in a systematic way? – PurplePenguin Dec 23 '14 at 11:12