Rank-nullity is the fundamental accounting identity for a linear map. Quotient spaces explain why the kernel is exactly the information lost by the map.
Rank and nullity
Definition. For a linear map \(T:V\to W\), the rank is \(\operatorname{rank}T=\dim\operatorname{im}T\), and the nullity is \(\operatorname{nullity}T=\dim\ker T\). For a matrix \(A\), these are the dimensions of its column space and null space.
The rank-nullity theorem
Theorem. If \(V\) is finite-dimensional and \(T:V\to W\) is linear, then \[ \dim V=\dim\ker T+\dim\operatorname{im}T. \]
Proof. Choose a basis \((k_1, \ldots,k_r)\) of \(\ker T\), and extend it to a basis \((k_1, \ldots,k_r,v_1, \ldots,v_s)\) of \(V\). The vectors \(T(v_1), \ldots,T(v_s)\) span \(\operatorname{im}T\), because the \(k_i\)-components vanish under \(T\). They are independent: if \(\sum a_iT(v_i)=0\), then \(\sum a_iv_i\in\ker T\), so \(\sum a_iv_i\) is a combination of the \(k_j\)'s; basis independence forces all \(a_i=0\). Hence \(s=\dim\operatorname{im}T\).
Quotient spaces
Definition. If \(U\subset V\) is a subspace, the quotient space \(V/U\) is the set of cosets \[ v+U=\{v+u:u\in U\}, \] with operations \((v+U)+(w+U)=(v+w)+U\) and \(a(v+U)=av+U\). The canonical projection is \(\pi:V\to V/U\), \(v\mapsto v+U\).
Proposition. The quotient operations are well-defined, \(\pi\) is linear, \(\ker\pi=U\), and if \(V\) is finite-dimensional, then \[ \dim(V/U)=\dim V-\dim U. \]
Proof. If \(v-v'\in U\) and \(w-w'\in U\), then \((v+w)-(v'+w')\in U\), so addition does not depend on representatives; scalar multiplication is similar. The kernel condition is \(v+U=U\), equivalent to \(v\in U\). Extend a basis of \(U\) to a basis of \(V\); the added basis vectors give a basis of \(V/U\).
The first isomorphism theorem
Theorem. If \(T:V\to W\) is linear, then the map \[ \overline T:V/\ker T\to\operatorname{im}T, \qquad \overline T(v+ \ker T)=T(v), \] is a well-defined isomorphism.
Proof. If \(v-v'\in\ker T\), then \(T(v)=T(v')\), so \(\overline T\) is well-defined. It is surjective by definition of image. Its kernel consists of cosets \(v+ \ker T\) with \(T(v)=0\), i.e. the zero coset. Hence it is injective.
Row reduction and four fundamental subspaces
Calculation. For an \(m\times n\) matrix \(A\), row reduction finds pivot columns and free variables. Pivot columns give \(\operatorname{rank}A\), free variables give \(\dim\ker A=n-\operatorname{rank}A\), and the linear map \(x\mapsto Ax\) identifies \(F^n/\ker A\) with the column space of \(A\).
Remark. Quotients should be read as spaces where a chosen subspace has been declared invisible. Rank-nullity says: input dimension equals invisible dimension plus visible output dimension.
A quotient example from a matrix
Example. Let \(T:F^3\to F^2\) be \[ T(x,y,z)=(x+y,\ y+z). \] Solving \(x+y=0\) and \(y+z=0\) gives \(\ker T=\operatorname{span}\{(1,-1,1)\}\). The image has dimension \(2\), since \((1,0)\) and \((0,1)\) are obtained from \((1,0,0)\) and \((-1,1,0)\). Rank-nullity reads \(3=1+2\), and the quotient \(F^3/\ker T\) is two-dimensional.
Calculation. In the quotient, all vectors on the line \((x,y,z)+t(1,-1,1)\) represent the same class. The map \[ \overline T:F^3/\ker T\to F^2,\qquad \overline T([v])=T(v), \] is well-defined because adding a vector from \(\ker T\) does not change \(T(v)\). This is the first isomorphism theorem in a concrete coordinate form.