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The rank-nullity theorem states that for a linear map with finite-dimensional domain,Here the domain is ${\mathbb{R}}^3$, and the preceding parts givewhich verifies the theorem.

The $j$th column of $B$ isThus the image is contained in $\mathrm{span}\{v,1\}$. Since $n>1$, $v$ is not constant; since $m>1$ and $a\ne 0$, two columns have a nonzero constant difference. The two spanning vectors therefore both lie in the image, so $\mathrm{rank}B=2$. By rank-nullity theorem,

For a linear map $T:V\to W$, its rank is $\dim \mathrm{im}T$ and its nullity is $\dim \ker T$. The rank-nullity theorem states that, when $V$ is finite-dimensional,To prove it, take a basis $v_1,\dots ,v_k$ of $\ker T$ and extend it to a basisof $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span $\mathrm{im}T$. They are also linearly independent: if ${\sum }_{j=k+1}^n{a}_{j}T{v}_j=0$, then ${\sum }_{j=k+1}^n{a}_j{v}_j\in \ker T$, and independence of the chosen basis forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.

For the given subspace, row reduction of the coefficient matrix givesTaking $x_3=s$ and $x_5=t$ therefore givesHenceis a basis of $W$.

Rank-nullity theorem gives $\mathrm{rank}T+\dim \ker T=n$. Equal nullities therefore give equal ranks.

Evaluation of bounded-degree multivariate polynomials on a finite set defines a linear map $E_P:f\mapsto (f(p){)}_{p\in P}$. Its rank is at most $|P|$ and its kernel is the space of polynomials vanishing on $P$. The rank-nullity theorem gives $\dim \ker E_P\ge \left(\genfrac{}{}{0ex}{}{d+n}{n}\right)-|P|$ for degree at most $d$ in $n$ variables. Dependent point constraints only enlarge the kernel.

Let a line collection have $M$ joints of a line collection and put $D=\lceil n{M}^{1/n}\rceil$. If $M>LD$, deleting each line with at most $D$ current joints of a line collection, together with those joints of a line collection, leaves a nonempty configuration in which every line has more than $D$ joints of a line collection. A nonzero polynomial of degree at most $D$ vanishes on the retained joints of a line collection by rank-nullity theorem. Choose one of minimum degree. Its restriction vanishes identically on every retained line, so its gradient is orthogonal to a spanning set of directions at every joint of a line collection and hence vanishes there. Each nonzero partial derivative would be a lower-degree polynomial with the same zeros, contradicting minimality. In characteristic zero, all derivatives vanishing forces a constant polynomial, another contradiction. Therefore $M\le LD$, giving the joints theorem. The argument's characteristic-zero assumption is essential to its derivative step.

A linear subspace $W\subseteq X^{\ast }$ is separating if every nonzero $x\in X$ has $T(x)\ne 0$ for some $T\in W$. For finite-dimensional $X$, an evaluation map $X\to F^{\dim W}$ is injective, so the rank-nullity theorem forces $\dim W\ge \dim X=\dim X^{\ast }$ and $W=X^{\ast }$. For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.