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

\[ \Set{x \in \R ^n}{Ax = 0}. \]

It is the set of vectors mapped to zero by $A$. Equivalently, it is the set of vectors orthogonal to the rows of $A$.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

\[ A(x + z) = Ax + Az = Ax + 0 = y \]

Conversely, suppose there were another solution $\tilde{x} \in \R ^{n}$ to the system $(A, y)$. Then $y = Ax = A\tilde{x}$, so\[ 0 = y - y = Ax - A\tilde{x} = A(x - \tilde{x}). \]

Consequently, $(x - \tilde{x}) \in \null(A)$, and so $\tilde{x}$ is the solution $x$ plus some vector in the null space of $A$. Consequently we are interested in whether $A$ has vectors in its nullspace.
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 $\R ^m$.