Updated orthogonality.

docs/en/mathematics/linear-algebra/inner-product-spaces.md

@@ -88,7 +88,7 @@ Which is consistent with Euclidean geometry. According to definition 1 the dista
 
 A pair of orthogonal vectors will satisfy the theorem of Pythagoras.
 
-> *Theorem 1*: let $V$ be an inner product space and $\langle \mathbf{u}, \mathbf{v} \rangle = 0$ are orthogonal then
+> *Theorem 1*: let $V$ be an inner product space and $\mathbf{u}$ and $\mathbf{v}$ are orthogonal then
 >
 > $$
 >   \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2,
@@ -132,7 +132,7 @@ $$
     \langle \mathbf{u} - \mathbf{p}, \mathbf{p} \rangle = \langle \mathbf{u}, \mathbf{p} \rangle - \langle \mathbf{p}, \mathbf{p} \rangle = a^2 - a^2 = 0.
 $$
 
-Additionaly, it may be observed that $\mathbf{u} = \mathbf{p}$ if and only if $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$; $\mathbf{u} = b \mathbf{v}$ for some $b \in \mathbb{K}$. Since 
+Additionally, it may be observed that $\mathbf{u} = \mathbf{p}$ if and only if $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$; $\mathbf{u} = b \mathbf{v}$ for some $b \in \mathbb{K}$. Since 
 
 $$
     \mathbf{p} = \frac{\langle b \mathbf{v}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} = b \mathbf{v} = \mathbf{u}.

docs/en/mathematics/linear-algebra/orthogonality.md

@@ -1 +1,368 @@
 # Orthogonality
+
+## Orthogonal subspaces
+
+> *Definition 1*: two subspaces $S$ and $T$ of an inner product space $V$ are **orthogonal** if 
+>
+> $$
+>   \langle \mathbf{u}, \mathbf{v} \rangle = 0,
+> $$
+>
+> for all $\mathbf{u} \in S$ and $\mathbf{v} \in T$. Orthogonality of $S$ and $T$ may be denoted by $S \perp T$. 
+
+The notion of orthogonality is only valid in vector spaces with a defined inner product.
+
+> *Definition 2*: let $S$ be a subspace of an inner product space $V$. The set of all vectors in $V$ that are orthogonal to every vector in $S$ will be denoted by $S^\perp$. Which implies
+>
+> $$
+>   S^\perp = \{\mathbf{v} \in V \;|\; \langle \mathbf{v}, \mathbf{u} \rangle = 0 \; \forall \mathbf{u} \in S \}.
+> $$
+>
+> The set $S^\perp$ is called the **orthogonal complement** of $S$. 
+
+For example the subspaces $X = \mathrm{span}(\mathbf{e}_1)$  and $Y = \mathrm{span}(\mathbf{e}_2)$ of $\mathbb{R}^3$ are orthogonal, but they are not orthogonal complements. Indeed,
+
+$$
+    X^\perp = \mathrm{span}(\mathbf{e}_2, \mathbf{e}_3) \quad \text{and} \quad Y^\perp = \mathrm{span}(\mathbf{e}_1, \mathbf{e}_3). 
+$$
+
+We may observe that if $S$ and $T$ are orthogonal subspaces of an inner product space $V$, then $S \cap T = \{\mathbf{0}\}$. Since for $\mathbf{v} \in S \cap T$ and $S \perp T$ then $\langle \mathbf{v}, \mathbf{v} \rangle = 0$ and hence $\mathbf{v} = \mathbf{0}$. 
+
+Additionally, we may also observe that if $S$ is a subspace of an inner product space $V$, then $S^\perp$ is also a subspace of $V$. Since for $\mathbf{u} \in S^\perp$ and $a \in \mathbb{K}$ then 
+
+$$
+    \langle a \mathbf{u}, \mathbf{v} \rangle = a \cdot 0 = 0
+$$
+
+for all $\mathbf{v} \in S$, therefore $a \mathbf{u} \in S^\perp$. 
+
+If $\mathbf{u}_1, \mathbf{u}_2 \in S^\perp$ then 
+
+$$
+    \langle \mathbf{u}_1 + \mathbf{u}_2, \mathbf{v} \rangle = \langle \mathbf{u}_1, \mathbf{v} \rangle + \langle \mathbf{u}_2, \mathbf{v} \rangle = 0 + 0 = 0,
+$$
+
+for all $\mathbf{v} \in S$, and hence $\mathbf{u}_1 + \mathbf{u}_2 \in S^\perp$. Therefore $S^\perp$ is a subspace of $V$.
+
+### Fundamental subspaces
+
+Let $V$ be an Euclidean inner product space $V = \mathbb{R}^n$ with its inner product defined by the [scalar product](inner-product-spaces/#euclidean-inner-product-spaces). With this definition of the inner product on $V$ the following theorem may be posed.
+
+> *Theorem 1*: let $A$ be an $m \times n$ matrix, then
+>
+> $$
+>   N(A) = R(A^T)^\perp,
+> $$
+>
+> and 
+>
+> $$
+>   N(A^T) = R(A)^\perp,
+> $$
+>
+> for all $A \in \mathbb{R}^{m \times n}$ with $R(A)$ denoting the column space of $A$ and $R(A^T)$ denoting the row space of $A$. 
+
+??? note "*Proof*:"
+
+    Let $A \in \mathbb{R}^{m \times n}$ with $R(A) = \mathrm{span}(\mathbf{\vec{a}}_i^T)$ for $i \in \mathbb{N}[i \leq n]$ denoting the column space of $A$ and $R(A^T) = \mathrm{span}(\mathbf{a}_i)$ for $i \in \mathbb{N}[i \leq m]$ denoting the row space of $A$. 
+
+    For the first equation, let $\mathbf{v} \in R(A^T)^\perp$ then $\mathbf{v}^T \mathbf{\vec{a}}_i^T = \mathbf{0}$ which obtains
+
+    $$
+        \mathbf{0} = \mathbf{v}^T \mathbf{\vec{a}}_i^T = \big(\mathbf{v}^T \mathbf{\vec{a}}_i^T \big)^T = \mathbf{\vec{a}}_i \mathbf{v},
+    $$
+
+    so $A \mathbf{v} = \mathbf{0}$ and hence $\mathbf{v} \in N(A)$. Which implies that $R(A^T)^\perp \subseteq N(A)$. Similarly, let $\mathbf{w} \in N(A)$ then $A \mathbf{w} = \mathbf{0}$ which obtains
+
+    $$
+        \mathbf{0} = \mathbf{\vec{a}}_i \mathbf{v} = \big(\mathbf{v}^T \mathbf{\vec{a}}_i^T \big)^T = \mathbf{v}^T \mathbf{\vec{a}}_i^T,
+    $$
+
+    and hence $\mathbf{w} \in R(A^T)^\perp$ which implies that $N(A) \subseteq R(A^T)^\perp$. Therefore $N(A) = R(A^T)^\perp$.
+
+    For the second equation, let $\mathbf{v} \in R(A)^\perp$ then $\mathbf{v}^T \mathbf{a}_i = \mathbf{0}$ which obtains
+
+    $$
+        \mathbf{0} = \mathbf{v}^T \mathbf{a}_i = \big(\mathbf{v}^T \mathbf{a}_i \big)^T = \mathbf{a}_i^T \mathbf{v},
+    $$
+
+    so $A^T \mathbf{v} = \mathbf{0}$ and hence $\mathbf{v} \in N(A^T)$. Which implies that $R(A)^\perp \subseteq N(A^T)$. Similarly, let $\mathbf{w} \in N(A^T)$ then $A^T \mathbf{w} = \mathbf{0}$ which obtains
+
+    $$
+        \mathbf{0} = \mathbf{a}_i^T \mathbf{w} = \big(\mathbf{a}_i^T \mathbf{w} \big)^T = \mathbf{w}^T \mathbf{a}_i, 
+    $$
+
+    and hence $\mathbf{w} \in R(A)^\perp$ which implies that $N(A^T) \subseteq R(A)^\perp$. Therefore $N(A^T) = R(A)^\perp$. 
+
+Known as the fundamental theorem of linear algebra. Which can be used to prove the following theorem.
+
+> *Theorem 2*: if $S$ is a subspace of the inner product space $V = \mathbb{R}^n$, then 
+>
+> $$
+>   \dim S + \dim S^\perp = n.
+> $$
+>
+> Furthermore, if $\{\mathbf{v}_i\}_{i=1}^r$ is a basis of $S$ and $\{\mathbf{v}_i\}_{i=r+1}^n$ is a basis of $S^\perp$ then $\{\mathbf{v}_i\}_{i=1}^n$ is a basis of $V$. 
+
+??? note "*Proof*:"
+
+    If $S = \{\mathbf{0}\}$, then $S^\perp = V$ and 
+
+    $$
+        \dim S + \dim S^\perp = 0 + n = n.
+    $$
+
+    If $S \neq \{\mathbf{0}\}$, then let $\{\mathbf{x}_i\}_{i=1}^r$ be a basis of $S$ and define $X \in \mathbb{R}^{r \times m}$ whose $i$th row is $\mathbf{x}_i^T$ for each $i$. Matrix $X$ has rank $r$ and $R(X^T) = S$. Then by theorem 2
+
+    $$
+        S^\perp = R(X^T)^\perp = N(X),
+    $$
+
+    from the [rank nullity theorem](vector-spaces/#rank-and-nullity) it follows that
+
+    $$
+        \dim S^\perp = \dim N(X) = n - r.
+    $$
+
+    and therefore 
+
+    $$
+        \dim S + \dim S^\perp = r + n - r = n.
+    $$
+
+    Let $\{\mathbf{v}_i\}_{i=1}^r$ be a basis of $S$ and $\{\mathbf{v}_i\}_{i=r+1}^n$ be a basis of $S^\perp$. Suppose that 
+
+    $$
+        c_1 \mathbf{v}_1 + \dots + c_r \mathbf{v}_r + c_{r+1} \mathbf{v}_{r+1} + \dots + c_n \mathbf{v}_n = \mathbf{0}.
+    $$
+
+    Let $\mathbf{u} = c_1 \mathbf{v}_1 + \dots + c_r \mathbf{v}_r$ and let $\mathbf{w} = c_{r+1} \mathbf{v}_{r+1} + \dots + c_n \mathbf{v}_n$. Then we have
+
+    $$
+        \mathbf{u} + \mathbf{w} = \mathbf{0},
+    $$
+
+    implies $\mathbf{u} = - \mathbf{w}$ and thus both elements must be in $S \cap S^\perp$. However, $S \cap S^\perp = \{\mathbf{0}\}$, therefore
+
+    $$
+    \begin{align*}
+        c_1 \mathbf{v}_1 + \dots + c_r \mathbf{v}_r &= \mathbf{0}, \\
+        c_{r+1} \mathbf{v}_{r+1} + \dots + c_n \mathbf{v}_n &= \mathbf{0},
+    \end{align*}
+    $$
+
+    since $\{\mathbf{v}_i\}_{i=1}^r$ and $\{\mathbf{v}_i\}_{i=r+1}^n$ are linearly independent, we must also have that $\{\mathbf{v}_i\}_{i=1}^n$ are linearly independent and therefore form a basis of $V$.
+
+> *Proposition 1*: let $S$ be a subspace of $V$, then $(S^\perp)^\perp = S$. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Proposition 2*: let $A \in \mathbb{R}^{m \times n}$ and $\mathbf{b} \in \mathbb{R}^m$, then either there is a vector $\mathbf{x} \in \mathbb{R}^n$ such that 
+>
+> $$
+>   A \mathbf{x} = \mathbf{b}, 
+> $$
+>
+> or there is a vector $\mathbf{y} \in \mathbb{R}^m$ such that
+>
+> $$
+>   A^T \mathbf{y} = \mathbf{0} \;\land\; \mathbf{y}^T \mathbf{b} \neq 0 .
+> $$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Orthonormal sets
+
+In working with an inner product space $V$, it is generally desirable to have a basis of mutually orthogonal unit vectors.
+
+> *Definition 3*: the set of vectors $\{\mathbf{v}_i\}_{i=1}^n$ in an inner product space $V$ is **orthogonal** if 
+>
+> $$
+>   \langle \mathbf{v}_i, \mathbf{v}_j \rangle = 0,
+> $$
+>
+> whenever $i \neq j$. Then $\{\mathbf{v}_i\}_{i=1}^n$ is said to be an **orthogonal set** of vectors.
+
+For example the trivial set $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ is an orthogonal set in $\mathbb{R}^3$. 
+
+> *Theorem 3*: if $\{\mathbf{v}_i\}_{i=1}^n$ is an orthogonal set of nonzero vectors in an inner product space $V$, then $\{\mathbf{v}_i\}_{i=1}^n$ are linearly independent.
+
+??? note "*Proof*:"
+
+    Suppose that $\{\mathbf{v}_i\}_{i=1}^n$ is an orthogonal set of nonzero vectors in an inner product space $V$ and 
+
+    $$
+        c_1 \mathbf{v}_1 + \dots + c_n \mathbf{v}_n = \mathbf{0},
+    $$
+
+    then
+
+    $$
+        c_1 \langle \mathbf{v}_j, \mathbf{v}_1 \rangle + \dots + c_n \langle \mathbf{v}_j, \mathbf{v}_n \rangle = 0,
+    $$
+
+    for $j \in \mathbb{N}[j \leq n]$ obtains $c_j \|\mathbf{v}_j\| = 0$ and hence $c_j = 0$ for all $j \in \mathbb{N}[j \leq n]$. 
+
+We may even go further and define a set of vectors that are orthogonal and have a length of $1$, a unit vector by definition. 
+
+> *Definition 4*: an **orthonormal** set of vectors is an orthogonal set of unit vectors.
+
+For example the set $\{\mathbf{u}_i\}_{i=1}^n$ will be orthonormal if and only if
+
+$$
+    \langle \mathbf{u}_i, \mathbf{u}_j \rangle = \delta_{ij},
+$$
+
+where 
+
+$$
+    \delta_{ij} = \begin{cases} 1 &\text{ for } i = j, \\ 0 &\text{ for } i \neq j.\end{cases}
+$$
+
+> *Theorem 4*: let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$. If
+>
+> $$
+>   \mathbf{v} = \sum_{i=1}^n c_i \mathbf{u}_i,
+> $$
+>
+> then $c_i = \langle \mathbf{v}, \mathbf{u}_i \rangle$ for all $i \in \mathbb{N}[i \leq n]$. 
+
+??? note "*Proof*:"
+
+    Let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$ and let 
+
+    $$
+        \mathbf{v} = \sum_{i=1}^n c_i \mathbf{u}_i,
+    $$
+
+    we have
+
+    $$
+        \langle \mathbf{v}, \mathbf{u}_i \rangle = \Big\langle \sum_{j=1}^n c_j \mathbf{u}_j, \mathbf{u}_i \Big\rangle = \sum_{j=1}^n c_j \langle \mathbf{u}_j, \mathbf{u}_i \rangle = \sum_{j=1}^n c_j \delta_{ij} = c_i.
+    $$
+
+Implying that it is much easier to calculate the coordinates of a given vector with respect to an orthonormal basis.
+
+> *Corollary 1*: let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$. If 
+>
+> $$
+>   \mathbf{v} = \sum_{i=1}^n a_i \mathbf{u}_i,
+> $$ 
+>
+> and 
+>
+> $$
+>   \mathbf{w} = \sum_{i=1}^n b_i \mathbf{u}_i,
+> $$
+>
+> then $\langle \mathbf{v}, \mathbf{w} \rangle = \sum_{i=1}^n a_i b_i$. 
+
+??? note "*Proof*:"
+
+    Let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$ and let 
+
+    $$
+        \mathbf{v} = \sum_{i=1}^n a_i \mathbf{u}_i,
+    $$
+
+    and 
+
+    $$
+        \mathbf{w} = \sum_{i=1}^n b_i \mathbf{u}_i,
+    $$
+
+    by theorem 4 we have 
+
+    $$
+        \langle \mathbf{v}, \mathbf{w} \rangle = \Big\langle \sum_{i=1}^n a_i \mathbf{u}_i, \mathbf{w} \Big\rangle = \sum_{i=1}^n a_i \langle \mathbf{w}, \mathbf{u}_i \rangle = \sum_{i=1}^n a_i b_i. 
+    $$
+
+> *Corollary 2*: let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$ and
+>
+> $$
+>   \mathbf{v} = \sum_{i=1}^n c_i \mathbf{u}_i,
+> $$
+>
+> then
+>
+> $$
+>   \|\mathbf{v}\|^2 = \sum_{i=1}^n c_i^2.
+> $$
+
+??? note "*Proof*:"
+
+    Let $\{\mathbf{u}_i\}_{i=1}^n$ be an orthonormal basis of an inner product space $V$ and let 
+
+    $$
+        \mathbf{v} = \sum_{i=1}^n c_i \mathbf{u}_i,
+    $$
+
+    then by corollary 1 we have
+
+    $$
+        \|\mathbf{v}\|^2 = \langle \mathbf{v}, \mathbf{v} \rangle = \sum_{i=1}^n c_i \mathbf{u}_i.
+    $$
+
+### Orthogonal matrices
+
+> *Definition 5*: an $n \times n$ matrix $Q$ is an **orthogonal matrix** if 
+>
+> $$
+>   Q^T Q = I.
+> $$
+
+Orthogonal matrices have column vectors that form an orthonormal set in $V$, as may be posed in the following theorem.
+
+> *Theorem 5*: let $Q = (\mathbf{q}_1, \dots, \mathbf{q}_n)$ be an orthogonal matrix, then $\{\mathbf{q}_i\}_{i=1}^n$ is an orthonormal set.
+
+??? note "*Proof*:"
+
+    Let $Q = (\mathbf{q}_1, \dots, \mathbf{q}_n)$ be an orthogonal matrix. Then
+
+    $$
+        Q^T Q = I,
+    $$
+
+    and hence $\mathbf{q}_i^T \mathbf{q}_j = \delta_{ij}$ such that for an inner product space with a scalar product we have
+
+    $$
+        \langle \mathbf{q}_i, \mathbf{q}_j \rangle = 0,
+    $$
+
+    for $i \neq j$. 
+
+It follows then that if $Q$ is an orthogonal matrix, then $Q$ is nonsingular and $Q^{-1} = Q^T$. 
+
+In general scalar products are preserved under multiplication by an orthogonal matrix since 
+
+$$
+    \langle Q \mathbf{u}, Q \mathbf{v} \rangle = (Q \mathbf{v})^T Q \mathbf{u} = \mathbf{v}^T Q^T Q \mathbf{u} = \langle \mathbf{u}, \mathbf{v} \rangle. 
+$$
+
+In particular, if $\mathbf{u} = \mathbf{v}$ then $\|Q \mathbf{u}\|^2 = \|\mathbf{u}\|^2$ and hence $\|Q \mathbf{u}\| = \|\mathbf{u}\|$. Multiplication by an orthogonal matrix preserves the lengths of vectors.
+
+## Orthogonalization process
+
+Let $\{\mathbf{a}_i\}_{i=1}^n$ be a basis of an inner product space $V$. We may use the modified method of Gram-Schmidt to determine the orthonormal basis $\{\mathbf{q}_i\}_{i=1}^n$ of $V$.
+
+Let $\mathbf{q}_1 = \frac{1}{\|\mathbf{a}_1\|} \mathbf{a}_1$ be the first step. 
+
+Then we may induce the following step for $i \in \mathrm{range}(2,n)$:
+
+$$
+\begin{align*}
+    \mathbf{w} &= \mathbf{a}_i - \langle \mathbf{a}_i, \mathbf{q}_1 \rangle \mathbf{q}_1 - \dots - \langle \mathbf{a}_i, \mathbf{q}_{i-1} \rangle \mathbf{q}_{i-1}, \\
+    \mathbf{q}_i &= \frac{1}{\|\mathbf{w}\|} \mathbf{w}.
+\end{align*}
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Least squares solutions of overdetermined systems
+