\[ \newcommand{R}{\mathbb{R}} \newcommand{N}{\mathbb{N}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]
Let \(X\) and \(Y\) be vector spaces (both real, or both complex), and \(T \colon X \to Y\) a mapping between them.
We say that \(T\) is a linear transformation (or just linear) if it preserves the linear structure of a vector space: \[ T \text{ linear } \defarrow T(\lambda x+ \mu y) = \lambda Tx + \mu Ty, \qquad x,y \in X, \: \mu, \lambda \in \mathbb R \: (\text{or } \mathbb C). \]
The set of linear transformations \(X \to Y\) is denoted by \(L(X,Y)\): \[ L(X,Y) \stackrel{\text{def.}}{=} \{ T \colon X \to Y \text{ linear}\} \]
IF \(X = Y\), we may abbreviate \(L(X,X)\) by \(L(X)\).
If, for all \(S,T \in L(X,Y)\), we define \[ (T+S)(x) := Tx + Sx \quad\text{ and }\quad (\lambda T)x := \lambda (Tx),\] for all \(x \in X\) and \(\lambda \in \R\) (or \(\mathbb C\)), it is easily checked that \(L(X,Y)\) becomes a vector space. In particular, \(\mu T + \lambda S \in L(X,Y)\) for any \(S,T \in L(X,Y)\).
Let \(X\) be a finite-dimensional 2) vector space with basis \(\{e_1, \ldots, e_n\}\). For any values \(y_1, \ldots, y_n \in Y\) there exists exactly one linear transformation \(T \in L(X,Y)\) such that \[ Te_j = y_j, \qquad j =1,\ldots,n.\]
Let \(X,Y\) be real vector spaces of dimension \(n\) and \(m\), respectively. Then \(L(X,Y) \cong M_{m\times n}(\R)\).
N.b. The corresponding statement holds for complex vector spaces \(X,Y\), with \(M_{m\times n}(\mathbb C)\) also complex-valued.
Let \(e = \{e_1, \ldots, e_n\}\) (standard basis) and \(f = \{f_1, \ldots, f_n\}\) (new basis) be two bases for \(\R^n\), and \([f]\) the matrix with \([f_j]\), \( j=1,\ldots, n\), as column vectors (expressed in the standard basis \(e\)). Then \[ x_e = [f] x_f \quad\text{ and }\quad x_f = [f]^{-1}x_e. \] Hence, if \[ T \in L(\R^n) \colon\quad T \text{ is realised by } A_e \in M_{n \times n}(\R) \text{ in the basis } e,\] what is its realisation \(A_f\) in the basis \(f\)? We have \[ y_e = A_e x_e \quad\Longleftrightarrow\quad y_f = [f]^{-1} y_e = [f]^{-1} A_e x_e = [f]^{-1} A_e [f] x_f. \] Thus \[ A_f = [f]^{-1} A_e [f] \] is the realisation of \(T\) in the basis \(f\).
Let \(T \in L(X,Y)\). The set of vectors for which \(T\) vanishes is called the kernel of T. \[ \mathrm{ker}(T) \stackrel{\text{def.}}{=} \{ x \in X \colon Tx = {\bf 0} \text{ in } Y\}. \]
Let \(T \in L(X,Y)\). Then \(\mathrm{ker}(T) \subset X\) is a linear subspace of \(X\), and \(\mathrm{ran}(T) \subset Y\) is a linear subspace of \(Y\).
N.b. The dimension of \(\mathrm{ran}(T)\) is called the rank of \(T\), \(\mathrm{rank}(T)\).
Let \(T \in L(X,Y)\). Then \[ T \text{ injective } \quad \Longleftrightarrow\quad \mathrm{ker}(T) = \{\bf{0}\}.\]
Let \(A = (a_{ij})_{ij} \in M_{m\times n}(\R)\) be the matrix realisation of a linear map \(\R^n \to \R^m\).7)
In this case the kernel of \(A\) is also called the null space of \(A\): \[ \begin{align*} x \in \mathrm{ker}(A) \quad&\Longleftrightarrow\quad Ax = 0 \quad\Longleftrightarrow\quad \sum_{j=1}^n a_{ij} x_j = 0 \quad \forall i = 1, \ldots, m\\ \quad&\Longleftrightarrow\quad (x_1, \ldots, x_n) \perp (a_{i1}, \ldots, a_{in}) \quad \text{ for all } i = 1, \ldots, n. \end{align*} \] Thus, the kernel is the space of vectors \(x \in \R^n\) which are orthogonal to the row vectors of \(A\).
The column space of \(A\) is the range of \(A\): since \[ Ax = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = x_1 \begin{bmatrix} a_{11} \\ \vdots \\ a_{m1} \end{bmatrix} + x_2 \begin{bmatrix} a_{12} \\ \vdots \\ a_{m2} \end{bmatrix} + \ldots + x_n \begin{bmatrix} a_{1n} \\ \vdots \\ a_{mn} \end{bmatrix}, \] we have that \[ \mathrm{ran}(A) = \{Ax \colon x \in \R^n\} = \Big\{ \sum_{j=1}^n x_j A_j \colon (x_1, \ldots, x_n) \in \R^n \Big\} = \mathrm{span}\{A_1, \ldots, A_n\} \] is the subspace of \(\R^m\) spanned by the column vectors \(A_j\), \(j = 1, \ldots, n\), of \(A\).
Similarly, define the row space of \(A\) to be the space spanned by the row vectors of \(A\). Then \[ \text{ row space of } A = \text{ column space of } A^t, \] where \(A^t = (a_{ji})\) is the transpose of \(A = (a_{ij})\).
Let \(A \in M_{m\times n}(\R)\). Then \[ \mathrm{ker}(A) \perp \mathrm{ran}(A^t), \] meaning that if \(x \in \mathrm{ker}(A)\) and \(y \in \mathrm{ran}(A^t)\), then \(x \cdot y = \sum_{j=1}^n x_j y_j = 0.\)
Let \(T \in L(\R^n,\R^m)\). Then \[ \mathrm{dim}\, \mathrm{ker}(T) + \mathrm{dim}\, \mathrm{ran}(T) = n. \]
N.b. The name comes from that \(\mathrm{dim}\, \mathrm{ker}(T)\) is the nullity of \(T\). Thus, the sum of the rank and the nullity of \(T\) equals the dimension of its ground space (domain).
Let \(T \in L(\R^n)\) be a linear transformation \(\R^n \to \R^n\). Then the following are equivalent:
Define the direct sum \(X \oplus Y\) of two vector spaces (both real, or both complex) as the space of pairs \((x,y)\) with the naturally induced vector addition and scalar multiplication: \[ X \oplus Y \stackrel{\text{def.}}{=} \{(x,y) \in X \times Y\}, \] where \[ (x_1,y_1) + (x_2,y_2) \stackrel{\text{def.}}{=} (x_1+x_2, y_1 + y_2) \quad\text{ and }\quad \lambda(x,y) \stackrel{\text{def.}}{=} (\lambda x, \lambda y). \] If \(X,Y \subset V\) are subspaces of a vector space \(V\), then \[ X \oplus Y = V \quad\Longleftrightarrow\quad X \cap Y = \{0\} \quad\text{ and }\quad X + Y \stackrel{\text{def.}}{=} \{ x + y \colon x \in X, y \in Y\} = V, \] where the equality \(X \oplus Y = V\) should be interpreted in terms of isomorphisms (\(V\) can be represented as \(X \oplus Y\)). Note that \[ \dim(X \oplus Y) = \dim(X) + \dim(Y). \]
With these definitions, the rank–nullity theorem can be expressed as a geometric description of the underlying space (\(\R^n\)) in terms of the matrix \(A\).
Let \(A \in M_{m\times n}(\R)\). Then \[ \R^n = \mathrm{ker}(A) \oplus \mathrm{ran}(A^t). \]
N.b. A consequence of this is that \(\mathrm{rank}(A) = \mathrm{rank}(A^t)\); another is that \(\R^m = \mathrm{ker}(A^t) \oplus \mathrm{ran}(A)\).
Let \(A \in M_{m\times n}(\R)\) be the realisation of a linear transformation \(\R^n \to \R^m\), and consider the linear equation \[Ax = b.\]