Linear Maps
1 Linear Maps
A linear map (or linear operator, linear transformation) is a function between vector spaces that respects their structure. In finite dimensions every linear map is represented by a matrix, and the choice of basis changes the matrix but not the map.
For vector spaces \(\mathcal{V},\mathcal{W}\) over the same field \(F\), a map \(P:\mathcal{V}\to\mathcal{W}\) is linear if for all \(\vec{v}_1,\vec{v}_2\in\mathcal{V}\) and \(\alpha_1,\alpha_2\in F\) \[ P(\alpha_1\vec{v}_1+\alpha_2\vec{v}_2)=\alpha_1P(\vec{v}_1)+\alpha_2P(\vec{v}_2). \]
A linear map preserves addition and scalar multiplication: it does not matter whether we apply \(P\) before or after those operations. In particular \(P(\vec{0})=\vec{0}\). Linear maps are the morphisms between vector spaces.
2 Matrix of a Linear Map
Let \(\mathcal{B}=(\vec{b}_1,\dots,\vec{b}_m)\) be a basis of \(\mathcal{V}\) and \(\mathcal{C}=(\vec{c}_1,\dots,\vec{c}_n)\) a basis of \(\mathcal{W}\). The matrix of \(P:\mathcal{V}\to\mathcal{W}\) is the \(n\times m\) matrix \[ [P]_{\mathcal{C},\mathcal{B}}=\begin{bmatrix}[P(\vec{b}_1)]_{\mathcal{C}}&\cdots&[P(\vec{b}_m)]_{\mathcal{C}}\end{bmatrix}, \] whose \(j\)-th column is the coordinate vector of \(P(\vec{b}_j)\) in the basis \(\mathcal{C}\). When \(\mathcal{W}=\mathcal{V}\) and \(\mathcal{C}=\mathcal{B}\) we write \([P]_{\mathcal{B}}\).
By linearity \([P\vec{v}]_{\mathcal{C}}=[P]_{\mathcal{C},\mathcal{B}}[\vec{v}]_{\mathcal{B}}\), so once bases are fixed, \(n\times m\) matrices are in bijective correspondence with linear maps \(\mathcal{V}\to\mathcal{W}\).
3 Change of Basis
Let \(\mathcal{B},\mathcal{B}'\) be two bases of \(\mathcal{V}\) and let \(S\) be the invertible matrix whose columns are the coordinates of the vectors of \(\mathcal{B}'\) in the basis \(\mathcal{B}\), so that \([\vec{v}]_{\mathcal{B}}=S[\vec{v}]_{\mathcal{B}'}\).
For a linear operator \(P:\mathcal{V}\to\mathcal{V}\), \[ [P]_{\mathcal{B}'}=S^{-1}[P]_{\mathcal{B}}S. \]
Proof. For any \(\vec{v}\), \([P\vec{v}]_{\mathcal{B}'}=S^{-1}[P\vec{v}]_{\mathcal{B}}=S^{-1}[P]_{\mathcal{B}}[\vec{v}]_{\mathcal{B}}=S^{-1}[P]_{\mathcal{B}}S[\vec{v}]_{\mathcal{B}'}\).
Matrices related by \(A'=S^{-1}AS\) are called similar; they represent the same operator in different bases. Choosing a basis in which the matrix is diagonal is the goal of diagonalisation.
4 Image, Kernel and Rank
For a linear map \(P:\mathcal{V}\to\mathcal{W}\), the image and kernel are \[ \operatorname{im}(P)=\{P(\vec{x}):\vec{x}\in\mathcal{V}\}\subseteq\mathcal{W},\qquad \ker(P)=\{\vec{x}\in\mathcal{V}:P(\vec{x})=\vec{0}\}\subseteq\mathcal{V}. \] The rank is \(\operatorname{rank}(P)=\dim\operatorname{im}(P)\) and the nullity is \(\dim\ker(P)\). The image and kernel of a matrix are those of the map \(\vec{x}\mapsto A\vec{x}\).
The image and kernel are subspaces (if \(P\vec{x}=P\vec{y}=\vec{0}\) then \(P(\alpha\vec{x}+\vec{y})=\vec{0}\), and similarly for images), and \(P\) is injective if and only if \(\ker(P)=\{\vec{0}\}\), since \(P\vec{x}=P\vec{y}\iff P(\vec{x}-\vec{y})=\vec{0}\).
For a matrix \(A\in M_{n\times m}(F)\), elementary row operations do not change the kernel, and elementary column operations do not change the image.
Proof. A row operation replaces \(A\) by \(EA\) with \(E\) invertible, and \(EA\vec{x}=\vec{0}\) if and only if \(A\vec{x}=\vec{0}\). A column operation replaces \(A\) by \(AE\), and \(\operatorname{im}(AE)=\operatorname{im}(A)\) since \(E\) is invertible.
5 Rank-Nullity
For a linear map \(P:\mathcal{V}\to\mathcal{W}\) with \(\mathcal{V}\) finite-dimensional, \[ \dim\operatorname{im}(P)+\dim\ker(P)=\dim\mathcal{V}. \]
Proof. Let \(\vec{u}_1,\dots,\vec{u}_k\) be a basis of \(\ker(P)\) and extend it to a basis \(\vec{u}_1,\dots,\vec{u}_k,\vec{w}_1,\dots,\vec{w}_r\) of \(\mathcal{V}\), so \(\dim\mathcal{V}=k+r\). We claim \(P\vec{w}_1,\dots,P\vec{w}_r\) is a basis of \(\operatorname{im}(P)\).
Spanning: for \(\vec{v}=\sum a_i\vec{u}_i+\sum b_j\vec{w}_j\) we have \(P\vec{v}=\sum b_jP\vec{w}_j\), since \(P\vec{u}_i=\vec{0}\).
Independence: if \(\sum b_jP\vec{w}_j=\vec{0}\) then \(\sum b_j\vec{w}_j\in\ker(P)\), so \(\sum b_j\vec{w}_j=\sum c_i\vec{u}_i\) for some \(c_i\). Independence of the basis of \(\mathcal{V}\) forces all \(b_j=0\).
Hence \(\dim\operatorname{im}(P)=r\) and \(r+k=\dim\mathcal{V}\).