Prove that if $A$ row reduces to $B$, removing a column from $A$ and row reducing will result in $B$ with the same column removed.

Question

How do we know that removing a column from a matrix is equivalent to removing the same column from the matrix's row reduced form?

Answer 1

For notation, let's fix any $k$, and for any matrix $M$ write $f(M)$ for the matrix $M$ with its $k$th column removed. There are three small things to check here:

1. Examine each one of the elementary row operations individually.
   For example, let $P$ be a row permutation. It's straightforward to see that $Pf(M) = f(PM)$. You can check this analogously for the other two elementary row operations, multiplying by a constant or adding a multiple of a row to another. So you can conclude that the operation of column deletion commutes with the elementary row operations. So, inductively, applying any sequence of row operations $E_i$ to a matrix commutes with column deletion.

2. Column deletion preserves the property of being row-reduced: if $B$ is in row-reduced form, then so is $f(B)$ (check this!).

3. Finally note that row-reduction always ends in the same place. If $B$ is row-reduced and there is a sequence of elementary row-operations bringing $A$ to $B$, then row-reducing $A$ always yields $B$.

Putting all of this together... let $E$ be the product of all elementary row operations bringing $A$ to $B$, so $B = EA$

$(1)$ says that $Ef(A) = f(EA) = f(B)$.

$(2)$ says that $f(B)$ is in row-reduced form. Hence $E$ is a sequence of elementary row operations bringing $f(A)$ to a row-reduced matrix $f(B)$.

$(3)$ says that by row-reducing $f(A)$ we must get to $f(B)$. $f(B)$ is the row-reduction of $f(A)$.