The set of vector which map to 0 under a given linear transformation is a subspace.
The nullspace (or
kernel) of a matrix $A
\in \R ^{m \times n}$ is the set
\[
\Set{x \in \R ^n}{Ax = 0}.
\]
We denote the nullspace of $A \in \R ^{m \times n}$ by $\null(A) \subset \R ^{n}$. Some authors denote the nullspace of $A$ by $\mathcal{N} (A)$.
Of course, the nullspace of a matrix is a subspace There are a few routes to see this.
The first is direct. If $w, z \in \null(A)$, then $Aw = 0$ and $Az = 0$. So then $A(w + z) = Aw + Az = 0$. So $\null(A)$ is closed under vector addition. Also $A(\alpha w) = \alpha (Aw) = 0$ for all $\alpha \in \R $. [In particular $A0 = 0$, so $0 \in \null(A)$; i.e., $\null(A)$ contains the origin.] So $\null(A)$ is closed under scalar multiplication.
The second is by thinking about orthogonal complements. Second, we can view the $\null(A)$ as the set of vectors orthgonal to all the rows of $A$. In other words, $\null(A) = \set{\tilde{a}_1, \dots , \tilde{a}_m}^\perp $. The orthogonal complement of any set is a subspace (see Orthogonal Real Subspaces).
The dimension of the nullspace is called the nullity.
Suppose we have a solution to the system of
linear equation with data $(A, y)$.
In other words, we have a vector $x \in
\R ^n$ so that $y = Ax$.
If we have a vector $z \in \null(A)$, then $x
+ z$ is also a solution to the system $(A,
y)$, since
\[
A(x + z) = Ax + Az = Ax + 0 = y
\] \[
0 = y - y = Ax - A\tilde{x} = A(x - \tilde{x}).
\]
The origin $0$ is always in the nullspace of $A$. However, this vector does not mean that we can find different solutions, since $x + 0 = x$ for all $x \in \R ^n$. If, on the other hand, there is a nonzero vector $z \in \null(A)$, then $x + z \neq x$, and $x+z$ is a solution for $(A, y)$. We think about $A$ as a function from $\R ^n$ to $\R ^m$. In the case that there is a nonzero element in the nullspace, $A$ maps different vectors to the same vector. Here, $x$ and $x + z$ both map to $y$. In this case, the function is not invertible, because it is not one-to-one. If, however, zero is the only element of the null space, the function is one-to-one. So call $A$ one-to-one if $\null(A) = 0$.
A matrix $A \in \R ^{m \times n}$ is one-to-one if the linear function $f: \R ^n \to \R ^m$ defined by $f(x) = Ax$ is one-to-one. In this case, if there exists $x \in \R ^n$ so that $y = Ax$, then there is only one such $x$. Different elements in $\R ^n$ map to different elements in $\R ^m$.