diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml index 89c31b8..fc3402a 100755 --- a/config/en/mkdocs.yaml +++ b/config/en/mkdocs.yaml @@ -87,6 +87,7 @@ nav: - 'Elementary matrices': mathematics/linear-algebra/matrices/elementary-matrices.md - 'Determinants': mathematics/linear-algebra/determinants.md - 'Vector spaces': mathematics/linear-algebra/vector-spaces.md + - 'Linear transformations': mathematics/linear-algebra/linear-transformations.md - 'Calculus': - 'Limits': mathematics/calculus/limits.md - 'Continuity': mathematics/calculus/continuity.md diff --git a/docs/en/mathematics/linear-algebra/linear-transformations.md b/docs/en/mathematics/linear-algebra/linear-transformations.md new file mode 100644 index 0000000..546cbc6 --- /dev/null +++ b/docs/en/mathematics/linear-algebra/linear-transformations.md @@ -0,0 +1,61 @@ +# Linear transformations + +## Definition + +> *Definition*: let $V$ and $W$ be vector spaces, a mapping $L: V \to W$ is a **linear transformation** or **linear map** if +> +> $$ +> L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2), +> $$ +> +> for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in \mathbb{K}$. + +In the case that the vector spaces $V$ and $W$ are the same; $V=W$, a linear transformation $L: V \to V$ will be reffered to as a **linear operator** on $V$. + +## The image and kernel + +Let $L: V \to W$ be a linear transformation from a vector space $V$ to a vector space $W$. In this section the effect is considered that $L$ has on subspaces of $V$. Of particular importance is the set of vectors in $V$ that get mapped into the zero vector of $W$. + +> *Definition*: let $L: V \to W$ be a linear transformation. The **kernel** of $L$, denoted by $\ker(L)$, is defined by +> +> $$ +> \ker(L) = \{\mathbf{v} \in V \;|\; L(\mathbf{v}) = \mathbf{0}\}. +> $$ + +The kernel is therefore a set consisting of vectors in $V$ that get mapped into the zero vector of $W$. + +> *Definition*: let $L: V \to W$ be a linear transformation and let $S$ be a subspace of $V$. The **image** of $S$, denoted by $L(S)$, is defined by +> +> $$ +> L(S) = \{\mathbf{w} \in W \;|\; \mathbf{w} = L(\mathbf{v}) \text{ for } \mathbf{v} \in S \}. +> $$ +> +> The image of the entire vector space $L(V)$, is called the **range** of $L$. + +With these definitions the following theorem may be posed. + +> *Theorem*: if $L: V \to W$ is a linear transformation and $S$ is a subspace of $V$, then +> +> 1. $\ker(L)$ is a subspace of $V$. +> 2. $L(S)$ is a subspace of $W$. + +??? note "*Proof*:" + + Let $L: V \to W$ be a linear transformation and $S$ is a subspace of $V$. + + To prove 1, let $\mathbf{v}_{1,2} \in \ker(L)$ and let $\lambda, \mu \in \mathbb{K}$. Then + + $$ + L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = \lambda \mathbf{0} + \mu \mathbf{0} = \mathbf{0}, + $$ + + therefore $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in \ker(L)$ and hence $\ker(L)$ is a subspace of $V$. + + To prove 2, let $\mathbf{w}_{1,2} \in L(S)$ then there exist $\mathbf{v}_{1,2} \in S$ such that $\mathbf{w}_{1,2} = L(\mathbf{v}_{1,2})$ For any $\lambda, \mu \in \mathbb{K}$ we have + + $$ + \lambda \mathbf{w}_1 + \mu \mathbf{w}_2 = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2), + $$ + + since $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in S$ it follows that $\lambda \mathbf{w}_1 + \mu \mathbf{w}_2 \in L(S)$ and hence $L(S)$ is a subspace of $W$. + diff --git a/docs/en/mathematics/linear-algebra/vector-spaces.md b/docs/en/mathematics/linear-algebra/vector-spaces.md index ecd7f7b..a59d265 100644 --- a/docs/en/mathematics/linear-algebra/vector-spaces.md +++ b/docs/en/mathematics/linear-algebra/vector-spaces.md @@ -411,7 +411,8 @@ So a single nonzero vector must span one-dimension exactly. For multiple vectors \mathbf{v} = a_1 \mathbf{v}_1 + \dots a_n \mathbf{v}_n, $$ - with + with + $$ a_i = - \frac{c_i}{c_{n+1}} $$ @@ -444,17 +445,19 @@ With the definition of a row space the following theorem may be posed. Let $A$ and $B$ be two matrices, if $B$ is row equivalent to $A$ then $B$ can be formed from $A$ by a finite sequence of row operations. Thus the row vectors of $B$ must be linear combinations of the row vectors of $A$. Consequently, the row space of $B$ must be a subspace of the row space of $A$. Since $A$ is row equivalent to $B$, by the same reasoning, the row space of $A$ is a subspace of the row space of $B$. -> *Definition*: the **rank** of a matrix $A$, denoted as $\text{rank}(A)$, is the dimension of the row space of $A$. - -The rank of a matrix may be determined by reducing the matrix to row echelon form. The nonzero rows of the row echelon matrix will form a basis for the row space. The rank may be interpreted as a measure for singularity of the matrix. +With the definition of a column space a theorem posed in [systems of linear equations](systems-of-linear-equations.md) may be restatated as. > *Theorem*: a linear system $A \mathbf{x} = \mathbf{b}$ is consistent if and only if $\mathbf{b}$ is in the column space of $A$. ??? note "*Proof*:" - A restatement of a previous theorem in [systems of linear equations](docs/en/mathematics/linear-algebra/systems-of-linear-equations.md). + For the proof, see the initial proof in [systems of linear equations](systems-of-linear-equations.md). -> *Theorem*: let $A$ be an $m \times n$ matrix. The linear system $A \mathbf{x} = \mathbf{b}$ is consistent for every $\mathbf{b} \in \mathbb{R}^m$ if and only if the column vectors of $A$ span $\mathbb{R}^m$. The system $A \mathbf{x} = \mathbf{b}$ has at most one solution for every $\mathbf{b}$ if and only if the column vectors of $A$ are linearly independent. +With this restatement the following statements may be proposed. + +> *Proposition*: let $A$ be an $m \times n$ matrix. The linear system $A \mathbf{x} = \mathbf{b}$ is consistent for every $\mathbf{b} \in \mathbb{R}^m$ if and only if the column vectors of $A$ span $\mathbb{R}^m$. +> +> The system $A \mathbf{x} = \mathbf{b}$ has at most one solution for every $\mathbf{b}$ if and only if the column vectors of $A$ are linearly independent. ??? note "*Proof*:" @@ -466,17 +469,25 @@ The rank of a matrix may be determined by reducing the matrix to row echelon for It follows that $\mathbf{x}_1 - \mathbf{x}_2 = \mathbf{0}$ and hence $\mathbf{x}_1 = \mathbf{x}_2$. +From these propositions the following corollary emerges. + > *Corollary*: an $n \times n$ matrix $A$ is nonsingular if and only if the column vectors of $A$ form a basis for $\mathbb{R}^n$. ??? note "*Proof*:" Let $A$ be an $m \times n$ matrix. If the column vectors of $A$ span $\mathbb{R}^m$, then $n$ must be greater or equal to $m$, since no set of fewer than $m$ vectors could span $\mathbb{R}^m$. If the columns of $A$ are linearly independent, then $n$ must be less than or equal to $m$, since every set of more than $m$ vectors in $\mathbb{R}^m$ is linearly dependent. Thus, if the column vectors of $A$ form a basis for $\mathbb{R}^m$, then $n = m$. +
+ +> *Definition*: the **rank** of a matrix $A$, denoted as $\text{rank}(A)$, is the dimension of the row space of $A$. + +The rank of a matrix may be determined by reducing the matrix to row echelon form. The nonzero rows of the row echelon matrix will form a basis for the row space. The rank may be interpreted as a measure for singularity of the matrix. + > *Definition*: the **nullity** of a matrix $A$, denoted as $\text{nullity}(A)$, is the dimension of the null space of $A$. The nullity of $A$ is the number of columns without a pivot in the reduced echelon form. -> *Theorem*: if $A$ is an $m times n$ matrix, then +> *Theorem*: if $A$ is an $m \times n$ matrix, then > > $$ > \text{rank}(A) + \text{nullity}(A) = n. @@ -486,6 +497,8 @@ The nullity of $A$ is the number of columns without a pivot in the reduced echel Let $U$ be the reduced echelon form of $A$. The system $A \mathbf{x} = \mathbf{0}$ is equivalent to the system $U \mathbf{x} = \mathbf{0}$. If $A$ has rank $r$, then $U$ will have $r$ nonzero rows and consequently the system $U \mathbf{x} = \mathbf{0}$ will involve $r$ pivots and $n - r$ free variables. The dimension of the null space will equal the number of free variables. +The section of vector spaces may be finished off, with this reasonably important theorem. + > *Theorem*: if $A$ is an $m \times n$ matrix, the dimension of the row space of $A$ equals the dimension of the column space of $A$. ??? note "*Proof*:"