Date modified: Wed 2024-03-06 . 11:58 PM
Date created: Wed 2024-03-06 . 03:36 PM
# 18b Commentary on the columnspace of a matrix. Recall the columnspace of a matrix $A$ is simply the span of the columns, that is, if $A=[\ a_{1}\ | \ a_{2}\ |\ \cdots\ |\ a_{k}\ ]$ with $a_{1},a_{2},\ldots,a_{k}$ the columns of $A$, then $$ \text{columnspace}(A) = CS(A)=\operatorname{span}(a_{1},a_{2},\ldots,a_{k}). $$ And combing with what we learned in the previous notes on removing redundant column vectors from a spanning set, if we row-reduce $A$ to an echelon form, where we have $r$ pivots occurring at columns $i_{1},i_{2},\ldots,i_{r}$, then $$ CS(A) = \operatorname{span} (a_{i_{1}},a_{i_{2}},\ldots ,a_{i_{r}}). $$ And we defined that this number of pivots $r$ of an echelon form of $A$ to be its **rank**, and we denote it $\text{rank}(A)=\text{rk}(A)$. **Example.** The following matrix $A$ happens to have a reduced row echelon form given by: $$ A = \begin{bmatrix}3 & 6 & 5 & 1 & 2 \\ 5 & 10 & 1 & 9 & -4 \\ 4 & 8 & 1 & 7 & -3\end{bmatrix} \stackrel{\text{row}}\sim \begin{bmatrix}\colorbox{lightblue}1 & 2 & 0 & 2 & -1 \\ 0 & 0 & \colorbox{lightblue}1 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0\end{bmatrix}. $$ So its columnspace is $$ CS(A) = \operatorname{span} (\begin{bmatrix}3\\5\\4\end{bmatrix},\begin{bmatrix}5\\1\\1\end{bmatrix}) $$with redundant vectors removed with our algorithm. Also, the rank of this matrix $A$ is $\text{rank}(A) = 2$ .$\blacklozenge$ Doing so we can be "as economical as we can be" when writing out the columnspace of a matrix as a span. Note, this is not the only possible way of writing out the span, as spanning sets are not unique, however later we will show and prove that one cannot use less than $r=\text{rank(A)}$ many vectors to span the columnspace of $A$ -- that this is as good as it gets, in terms of numbers of vectors used. --- ## An important observation about columns. Now we are going to do something very interesting regarding the span of a list of vectors. Let us make the following three observations about the spans of a list of vectors, $v_{1},v_{2},\ldots,v_{k}$ (1) Swapping a pair in that list does not change their span. That is $$ \operatorname{span} (v_{1},\ldots,\colorbox{lightblue}{\(v_{i}\)},\ldots,\colorbox{lightgreen}{\(v_{j}\)},\ldots,v_{k})=\operatorname{span} (v_{1},\ldots,\colorbox{lightgreen}{\(v_{j}\)},\ldots,\colorbox{lightblue}{\(v_{i}\)},\ldots,v_{k}) $$If these are column vectors in a matrix, then this shows $$ \text{columnspace}\left(\begin{bmatrix}\\v_{1} \cdots \colorbox{lightblue} {\(v_{i}\)} \cdots \colorbox{lightgreen}{\(v_{j}\)}\cdots v_{k}\\\ \end{bmatrix}\right) = \text{columnspace}\left(\begin{bmatrix}\\v_{1}\cdots \colorbox{lightgreen}{\(v_{j}\)}\cdots \colorbox{lightblue} {\(v_{i}\)}\cdots v_{k}\\\ \end{bmatrix}\right). $$ In other words, **swapping a pair of columns in a matrix does not change its columnspace.** (2) Let us take a nonzero number $c\neq 0$. Then $$ \operatorname{span} (v_{1},\ldots,{\colorbox{lightblue}{\(v_{i}\)}},\ldots,v_{k})=\operatorname{span} (v_{1},\ldots,{\colorbox{lightblue}{\(cv_{i}\)}},\ldots,v_{k}) $$If these are column vectors in a matrix, then this shows $$ \text{columnspace}\left(\begin{bmatrix}\\v_{1} \cdots \colorbox{lightblue} {\(v_{i}\)} \cdots v_{k}\\\ \end{bmatrix}\right) = \text{columnspace}\left(\begin{bmatrix}\\v_{1}\cdots \colorbox{lightblue} {\(cv_{i}\)}\cdots v_{k}\\\ \end{bmatrix}\right). $$In other words, **scaling a column by a nonzero scalar in a matrix does not change its columnspace.** (3) And let us consider two vectors with different indices in that list, say $v_{i}$ and $v_{j}$ with $i\neq j$, if we replace $v_{i}$ with $v_i+cv_{j}$ then the span is also not changed, for any scalar $c$. That is $$ \operatorname{span} (v_{1},\ldots,v_{i},\ldots,v_{j},\ldots,v_{k})= \operatorname{span} (v_{1},\ldots,v_{i}+cv_{j},\ldots,v_{j},\ldots,v_{k}). $$If these are column vectors in a matrix, then this shows $$ \text{columnspace}\left(\begin{bmatrix}\\v_{1} \cdots \colorbox{lightblue} {\(v_{i}\)} \cdots {v_{j}}\cdots v_{k}\\\ \end{bmatrix}\right) = \text{columnspace}\left(\begin{bmatrix}\\v_{1}\cdots \colorbox{lightblue}{\(v_{i}+cv_{j}\)}\cdots {v_{j}}\cdots v_{k}\\\ \end{bmatrix}\right). $$In other words, **replacing a column by adding to a scalar multiple of another column in a matrix does not change its columnspace.** Each of these three results can be proved using our proposition on when are two spans the same. As an exercise, I want you to do it. Let us call these three operations **elementary column operations**. What this shows is **elementary column operations preserve the columnspace of a matrix $A$**. If we denote $\stackrel{\text{col}}\sim$ for performing a sequence of elementary column operations, then >**Elementary column operations preserve columnspace.**$$ A \stackrel{\text{col}}\sim A' \ \ \ \text{implies}\ \ \ CS(A)=CS(A') \ \ \ ! $$ **Isn't this cool!** A word of **caution**, however. **Although elementary column operations preserves the columnspace of a matrix, elementary row operations do not preserve the columnspace of a matrix!** What is preserved through elementary row operations are the solution sets to the associated linear system of equation, in particular the solution set to the homogeneous equation $Ax = \vec 0$. Namely, > **Elementary row operation preserve nullspace.**$$A \stackrel{\text{row}}\sim \tilde A \ \ \ \text{implies}\ \ NS(A)= NS(\tilde A)\ \ \ ! $$ While column operations would **not** preserve the solution set to the associated linear system of equations! **Remark.** But nevertheless, since the columnspaces are the same after performing elementary column operations on a matrix, this shows the rank of these matrices are also the same. That is, if $A \stackrel{\text{col}}\sim A'$, then $\text{rank}(A)=\text{rank}(A')$.