Linear Algebra: Preserving the null space 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.
 A: Say we have $V, W$ vector spaces and a linear transformation $T : V \to W$. The null space of $T$ is defined $N(T) = \{x \in V : T(x) = 0\}$.
When talking about row operations, we are talking about the matrix representation of the linear transformation. So if $\alpha = \{ x_1, ..., x_n \}$ is a basis of $V$ and $\beta = \{ y_1, ..., y_n \}$ is a basis of $W$ we have a matrix, say $A$, such that $A = [T]_\alpha^\beta$.
Say $T(x_i) = a_{i1}y_1 + ... + a_{in}y_n$. Then the matrix A will be the following
$$ A = \left[\begin{array}{ccc}
a_{11} & ... & a_{1n} \\
... & ... & ... \\
a_{n1} & ... & a_{nn}
\end{array}\right]. $$
Row operations are as follows
(i) Swap two rows.
(ii) Multiply row by non zero constant.
(iii) Add multiple of one row to a different row.
Say we swap the $i$th and $j$th row and call the new matrix $A'$. This new matrix defines a new linear transformation $T':V \to W$ such that $T'(x) = A'x$. We want to show that $N(T) = N(T')$. For any $x \in V$, say $$T(x) = b_1y_1 + ... + b_iy_i + ... + b_jy_j + ... + b_ny_n.$$ Then since $T'$ is just rows swapped $$T'(x) = b_1y_1 + ... +b_jy_i + ... + b_iy_j + ... b_ny_n.$$
Thus if $T(x) = 0$, $T'(x) = 0$, as $b_i = 0$ for all $i$. So $N(T) = N(T')$.
For (ii), think about if we multiply the $i$th row by $\beta$, then if the $i$th scalar for $T(x)$ is $c_i$, then the $i$th scalar for $T'(i)$ is $\beta c_i$. So if $T(x) = 0$, then $c_i = 0$, so $\beta c_i = 0$. All other scalars stay the same, thus $T'(x) = 0$. And if $T'(x) = 0$, $\beta^{-1}c_i = 0$, so $T(x) = 0$. Once again $N(T) = N(T')$.
I leave (iii) to you.
A: I think the most natural way to think about this is to note that the kernel of a matrix is also the orthogonal complement to the column-space of it's transpose. (this is the fundamental theorem of linear algebra)
So when you do row operations on a matrix, this is the same as doing linear combinations of the columns in it's transpose. This doesn't change the overall space spanned by those columns in the transpose, so it doesn't change the kernel of the original matrix.
A: It means that performing an elementary row operation on a matrix does not change the null space of the matrix. That is, if $A$ is a matrix, and $E$ is an elementary matrix of the appropriate size, then the matrix $EA$ has the same null space as $A$.
To see why this is true, suppose first that $x$ is in the null space of $A$. This means that $Ax=\vec 0$. Multiplying both sides of this equation by $E$, we see that $(EA)x=E\vec 0 =\vec 0$, meaning that $x$ is also in the null space of $EA$. Now suppose that $x$ is in the null space of $EA$, so that $(EA)x=\vec 0$. As you mentioned, $E$ is invertible, so we can multiply this equation by $E^{-1}$:
$$Ax=IAx=(E^{-1}E)Ax=E^{-1}(EA)x=E^{-1}\vec 0=\vec 0\;,$$
showing that $x$ is in the null space of $A$. In other words, a vector is in the null space of $EA$ if and only if it is in the null space of $A$, and $EA$ and $A$ have the same null space.
