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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 givesFor 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 ^[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 $b$ isIndeed, $T(x)=T(x_0)$ 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.

In the diagram, $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ has matrix $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, 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 ${\mathbb{R}}^2$, andEvery 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 subspacea plane, while $(a,b)$ with $b\ne 0$ 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.

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem:

The rank of a linear map and nullity of a linear map areThe rank-nullity theorem states thatTo prove it, choose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and use basis extension to obtain a basisof $V$. The vectors $T(v_1),\dots ,T(v_{n-k})$ span the image of a linear map: applying $T$ to an arbitrary linear combination of the displayed basis removes all the $u_i$ terms. They are also linearly independent. Indeed,so ${\sum }_j{a}_j{v}_j={\sum }_i{b}_i{u}_i$ for some scalars $b_i$. The linear independence of the extended basis forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty basis when one of these dimensions is zero.