Nullity of a linear map

{title2=$n(T)=\dim\ker T$}

= Nullity
{synonym}

The <nullity of a linear map> $T:V\to W$ is $n(T)=\dim\ker T$, the <dimension> of its <kernel>. It counts independent input directions that can change without changing the output. For a <finite-dimensional vector space> $V$, the <rank-nullity theorem> gives
$$
n(T)=\dim V-\operatorname{rank}T.
$$
For an $m\times n$ <matrix> $A$ of <matrix rank> $r$, this is $n-r$, using the number of columns, not the number of rows. This dimension identity is <Linear Algebra Done Right by Sheldon Axler (4th edition)>[Sheldon Axler, Linear Algebra Done Right, fourth edition, §3B, theorem 3.21]{ref}.

The <nullity> answers a practical question about a <linear system>: once one input is known to produce the requested output, how much freedom remains? If $T(x_0)=b$, its <fiber of a function> over $b$ is
$$
\boxed{T^{-1}(\{b\})=x_0+\ker T.}
$$
Indeed, $T(x)=T(x_0)$ exactly when $T(x-x_0)=0$. Thus every nonempty <fiber> is an <affine subspace> of <dimension> $n(T)$, with the same direction space as the <kernel>. The <kernel> itself is the <fiber> over zero. Outputs outside the <image of a linear map> have empty <fibers>; the <nullity> does not guarantee existence, and we assign no solution-space <dimension> to an empty <fiber> here.

\Image[/singularity-nullity-fibers.png]
{title=Nullity one gives parallel solution lines for $F(x,y)=(x+y,0)$}
{description=Each colored input <straight line> $x+y=c$ is the <fiber> over the corresponding output $(c,0)$. The blue line through the origin is the <kernel>. The orange inputs $(1,0)$ and $(0,1)$ give the same output; their difference belongs to the <kernel>. The output $(1,1)$ lies outside the <image of a linear map> and has no input. Exact toy geometry, not measured data.}

In the diagram, $F:\mathbb R^2\to\mathbb R^2$ has <matrix> $\begin{pmatrix}1&1\\0&0\end{pmatrix}$, <matrix rank> one and <nullity> one. Its <kernel> has <basis> $(1,-1)$, and the full solution of $F(x,y)=(c,0)$ is $(x,y)=(c,0)+s(1,-1)$. One free <scalar> moves the solution along a <straight line>; the output $(c,1)$ is impossible.

The existing three-input example $T(x,y,z)=(x+y,z)$ instead has <rank> two and <nullity> one. Its <image of a linear map> is all of $\mathbb R^2$, and
$$
T^{-1}(\{(a,b)\})=(a,0,b)+\operatorname{span}\{(1,-1,0)\}.
$$
Every output has a solution <straight line>. In contrast, $U(x,y,z)=(x+y,0)$ has <rank> one and <nullity> two, with <kernel> <basis> $(1,-1,0),(0,0,1)$. Its reachable output $(a,0)$ has the solution <affine subspace>
$$
(x,y,z)=(a,0,0)+s(1,-1,0)+u(0,0,1),\qquad s,u\in\mathbb R,
$$
a plane, while $(a,b)$ with $b\ne0$ has no solution. These two <linear maps> have the same input and output dimensions but different freedom and reachability. Computing the <kernel> separates what an output determines from what an extra constraint would have to fix. For either example, a <basis> of the <kernel> gives one independent free <scalar> per <basis vector>, which is why <nullity> measures freedom rather than the number of solutions.


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