Rank-nullity theorem ($\dim V=\dim \ker T+\dim \mathrm{im}T$)

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

For finite-dimensional linear maps $T:U\to V$ and $S:V\to W$, the restriction of $T$ to $\ker (ST)$ maps onto $\mathrm{im}T\cap \ker S$ and has kernel $\ker T$. The rank-nullity theorem on this restriction proves the title identity. Since the intersection has dimension at most $\dim \ker S$, it implies $n(ST)\le n(T)+n(S)$. Equivalently, $r(ST)\ge r(T)+r(S)-\dim V$, the rank inequality for a composition. Also $\mathrm{im}(ST)=S(\mathrm{im}T)$ directly gives the upper bounds $r(ST)\le \min (r(S),r(T))$.

For endomorphisms $S,T$ of a finite-dimensional space of dimension $d$, $r(S+T)\le r(S)+r(T)$ and $n(ST)\le n(S)+n(T)$. If both equalities hold, the first gives $r(S)+r(T)\le d$, while the second gives $2d-r(S)-r(T)\le d$. Thus the rank sum is $d$, $S+T$ is an isomorphism, and $ST=0$. Conversely, $ST=0$ implies $\mathrm{im}T\subseteq \ker S$ and hence rank sum at most $d$, whereas invertibility of $S+T$ gives rank sum at least $d$. This proves both equalities.

For a matrix whose entries depend on a parameter, a nonzero maximal minor certifies a fixed matrix rank away from its zeros. At a zero, other minors must still be checked: one vanishing minor does not itself imply a rank drop. Once the matrix rank is determined, the rank-nullity theorem gives the dimension of the kernel of a linear map. Exceptional parameters can then be solved separately to exhibit the additional kernel of a linear map directions.