Updated vector spaces and added first section of linear transformations.

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

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 $W$. 
+

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 $n = m$.
 
+<br>
+
+> *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*:"