TITLE: Linear Algebra: Preserving the null space
QUESTION [13 upvotes]: What does it mean when a book says that row operations preserve the null space? And why should that be true? I have read that row operations are equivalent to multiplying a vector on the left by an invertible elementary matrix. And I think I understand that the nullspace is the set of all vectors from $u \in U$ which get mapped to the zero vector in $V$ if $T:U\rightarrow V$ is linear. But I'm still not sure what this means.

REPLY [9 votes]: For what it's worth, I think the following way of thinking about why row operations do not change the null space is worthwhile:
An important observation to make: 

One may observe that multiplication of a matrix $A$ by a column vector
  $\bf x$ amounts to taking a linear combination of the columns of $A$. 
  The coefficients of the particular linear combination are the
  coordinates of $\bf x$.

For example:
$$\tag{1}
\underbrace{\left[
\matrix{a_{11}&a_{12}&a_{13} \cr 
        a_{21}&a_{22}&a_{23}\cr 
        a_{31}&a_{32}&a_{33} }\right]}_{A}
\left[\matrix{x\cr y\cr z}\right]
=\left[\matrix{x a_{11} +ya_{12}+ z a_{13}    \cr 
 x a_{21} +ya_{22}+ z a_{23}   \cr
 x a_{31} +ya_{32}+ z a_{33}}
\right]
= x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1}
 +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2}
 +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3}  
 
$$ 
A second important observation to make:

$\bf x$ is in the null space of $A$ if and only if $A{\bf x}=\bf0$.
  This means that the linear combination of the columns of $A$ whose
  coefficients are the coordinates of  ${\bf x}=\Bigl[{\textstyle{x\atop
 y}\atop\scriptstyle z }\Bigr]$ is the zero vector:
$$\tag{2} 
  x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1}
 +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2}
 +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3}  =\bf0 $$

Now on to row operations:
Here, we make our third and final observation:

If one performs  a row operation on $A$, then the corresponding right
  hand side of $(1)$ is obtained by performing the same row operation to
  each of ${\bf a_1}$,  ${\bf a_2}$, and ${\bf a_3}$.


It follows from all of this that multiplying a row of $A$ by a non-zero number does not affect the null space.
For example if row 1 of $A$ were multiplied by 2 then the right hand side of (1) would be:
$$ x\underbrace{\left[\matrix{\color{maroon}{2a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1}
 +y\underbrace{\left[\matrix{\color{maroon}{2a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2}
 +z\underbrace{\left[\matrix{\color{maroon}{2a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3}  =\bf z
$$
If (2) holds, it is easily seen that ${\bf z}=\bf0$.
It should be fairly obvious that interchanging two rows of $A$ does not affect the null space. For example if rows 1 and 3 of $A$ were interchanged, the right hand side of (1) would become
$$
x\left[\matrix{\color{darkblue}{ a_{31}}\cr  \color{darkgreen}{ a_{21}}\cr \color{maroon}{ a_{11}}}\right]
+y\left[\matrix{\color{darkblue}{ a_{32}}\cr \color{darkgreen}{ a_{22}}\cr \color{maroon}{ a_{12}}}\right]
+z\left[\matrix{\color{darkblue}{ a_{33}}\cr\color{darkgreen}{  a_{23}}\cr\color{maroon}{  a_{13}}}\right] =\bf z
$$
If (2) holds, then $\bf z$ would be $\bf 0$.
Finally, if a multiple of a row of $A$ were added to another row of $A$, the null space would be unchanged.  For example if twice row 1 of $A$ were added to to row 2, then the right hand side of $(1)$ would become:
$$\tag{3}
x\left[\matrix{\color{maroon}{ a_{11}}\cr \color{darkgreen}{ a_{21}}+2\color{maroon}{ a_{11}}\cr\color{darkblue} a_{31}}\right]
+y\left[\matrix{\color{maroon}{  a_{12}}\cr\color{darkgreen}{  a_{22}}+2 \color{maroon}{a_{12}}\cr\color{darkblue} a_{32}}\right]
+z\left[\matrix{\color{maroon}{  a_{13}}\cr\color{darkgreen}{  a_{23}}+2 \color{maroon}{a_{13}}\cr\color{darkblue} a_{33}}\right] =\bf z
$$
If $(2)$ holds, then $\bf z=0$. In particular, looking at the second component of the left hand side of $(3)$:
$$\eqalign{
x(\color{darkgreen}{a_{21}}+2\color{maroon}{a_{11}})+
y(\color{darkgreen}{a_{22}}+2\color{maroon}{a_{12}})+
z(\color{darkgreen}{a_{23}}+2\color{maroon}{a_{13}})
&=
(x \color{darkgreen}{a_{21}}+
y \color{darkgreen}{a_{22}}+ 
z \color{darkgreen}{a_{23}} )\cr &\ \ \ \ \ \ \ 
+2(  x\color{maroon}{a_{11}} + y\color{maroon}{a_{12}}+ z\color{maroon}{a_{13}} )\cr&=0+2\cdot0\cr&=0}.
$$