Linear maps are the structure-preserving functions of linear algebra. Matrices are not the maps themselves; they are coordinate descriptions of maps after bases have been chosen.
Linear maps and their basic subspaces
Definition. A function \(T:V\to W\) between vector spaces over the same field is linear if \[ T(av+bv')=aT(v)+bT(v') \] for all \(v,v'\in V\) and \(a,b\in F\). Its kernel and image are \[ \ker T=\{v:T(v)=0\},\qquad \operatorname{im}T=\{T(v):v\in V\}. \]
Proposition. The kernel is a subspace of \(V\), the image is a subspace of \(W\), and \(T\) is injective exactly when \(\ker T=\{0\}\).
Proof. If \(T(v)=T(v')=0\), then \(T(av+bv')=0\), so \(\ker T\) is a subspace. If \(w=T(v)\) and \(w'=T(v')\), then \(aw+bw'=T(av+bv')\), so the image is a subspace. Finally, \(T(v)=T(v')\) is equivalent to \(v-v'\in\ker T\).
Matrix representation
Definition. Let \(B=(v_1, \ldots,v_n)\) be a basis of \(V\) and \(C=(w_1, \ldots,w_m)\) a basis of \(W\). The matrix of \(T\) from \(B\) to \(C\), written \([T]_{C\leftarrow B}\), is the \(m\times n\) matrix whose \(j\)-th column is \([T(v_j)]_C\). It satisfies \[ [T(v)]_C=[T]_{C\leftarrow B}[v]_B. \]
Example. The derivative map \(D:F_3[x]\to F_2[x]\), \(D(p)=p'\), has matrix \[ \begin{pmatrix} 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3 \end{pmatrix} \] relative to the monomial bases \((1,x,x^2,x^3)\) and \((1,x,x^2)\). This matrix is a coordinate expression of differentiation on a finite-dimensional polynomial space.
Composition and change of coordinates
Proposition. If \(S:U\to V\) and \(T:V\to W\) are linear, then \(T\circ S\) is linear and \[ [T\circ S]_{D\leftarrow A}=[T]_{D\leftarrow B}[S]_{B\leftarrow A}. \] If \(T:V\to V\) and \(B,C\) are bases, then \[ [T]_{C\leftarrow C}=P_{C\leftarrow B}[T]_{B\leftarrow B}P_{B\leftarrow C}. \]
Proof. Apply each matrix identity to a coordinate vector and use uniqueness of coordinates. Composition is read right-to-left: first express the input, then apply \(S\), then apply \(T\). The change-of-basis formula inserts the identity maps that convert between \(B\)-coordinates and \(C\)-coordinates.
Isomorphisms
Theorem. For finite-dimensional spaces \(V,W\) with \(\dim V=\dim W\), the following are equivalent for \(T:V\to W\): \(T\) is injective; \(T\) is surjective; \(T\) is bijective; the matrix of \(T\) in any bases is invertible.
Proof. Rank-nullity gives \(\dim V=\dim\ker T+\dim\operatorname{im}T\). If the kernel is zero, then the image has dimension \(\dim W\), hence equals \(W\). If the image is \(W\), then the kernel has dimension zero. Matrix invertibility is the same statement after identifying \(V\) and \(W\) with coordinate spaces.
Maps determined by basis values
Proposition. Let \(B=(v_1, \ldots,v_n)\) be a basis of \(V\). A linear map \(T:V\to W\) is uniquely determined by the vectors \(T(v_1), \ldots,T(v_n)\). Conversely, for any chosen vectors \(w_1, \ldots,w_n\in W\), there is a unique linear map \(T\) with \(T(v_i)=w_i\).
Proof. Every \(v\in V\) has a unique expansion \(v=\sum_i a_iv_i\). Linearity forces \(T(v)=\sum_i a_iT(v_i)\). This formula also defines a linear map for any prescribed values \(w_i\), and uniqueness of coordinates makes it well-defined.
Rank under coordinate changes
Proposition. If matrices \(A\) and \(B\) represent the same linear map in different bases, then \(B=PAQ\) for invertible matrices \(P,Q\). Hence \(A\) and \(B\) have the same rank. Rank is a property of the linear map, not of a particular coordinate description.