luc/notes
1
0
Fork 0
Code Issues Pull requests Activity

port from mathematics-physics notes

Browse source
This commit is contained in:
Luc Bijl 2025-08-26 15:48:53 +02:00
parent a4e106ce02
commit c009ea53f0
124 changed files with 13224 additions and 0 deletions
Download patch file Download diff file Expand all files Collapse all files

467
docs/mathematics/linear-algebra/orthogonality.md Normal file
View file

@ -0,0 +1,467 @@
# 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 $V$.
We may further extend this with the notion of a direct sum.
> *Definition 3*: if $U$ and $V$ are subspaces of a vector space $W$ and each $\mathbf{w} \in W$ can be written uniquely as
>
> $$
> \mathbf{w} = \mathbf{u} + \mathbf{v},
> $$
>
> with $\mathbf{u} \in U$ and $\mathbf{v} \in V$ then $W$ is a **direct sum** of U and $V$ denoted by $W = U \oplus V$.
In the following theorem it will be posed that the direct sum of a subspace and its orthogonal complement make up the whole vector space, which extends the notion of theorem 2.
> *Theorem 3*: if $S$ is a subspace of the inner product space $V = \mathbb{R}^n$, then
>
> $$
> V = S \oplus S^\perp.
> $$
??? note "*Proof*:"
Will be added later.
The following results emerge from these posed theorems.
> *Proposition 1*: let $S$ be a subspace of $V$, then $(S^\perp)^\perp = S$.
??? note "*Proof*:"
Will be added later.
Recall that the system $A \mathbf{x} = \mathbf{b}$ is consistent if and only if $\mathbf{b} \in R(A)$ since $R(A) = N(A^T)^\perp$ we have the following result.
> *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 4*: 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 4*: 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 5*: 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 5*: 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 5 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 6*: 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 6*: 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
A standard technique in mathematical and statistical modeling is to find a least squares fit to a set of data points. This implies that the sum of squares fo errors between the model and the data points are minimized. A least squares problem can generally be formulated as an overdetermined linear system of equations.
For a system of equations $A \mathbf{x} = \mathbf{b}$ with $A \in \mathbb{R}^{m \times n}$ with $m, n \in \mathbb{N}[m>n]$ and $\mathbf{b} \in \mathbb{R}^m$ then for each $\mathbf{x} \in \mathbb{R}^n$ a *residual* $\mathbf{r}: \mathbb{R}^n \to \mathbb{R}^m$ can be formed
$$
\mathbf{r}(\mathbf{x}) = \mathbf{b} - A \mathbf{x}.
$$
The distance between $\mathbf{b}$ and $A \mathbf{x}$ is given by
$$
\| \mathbf{b} - A \mathbf{x} \| = \|\mathbf{r}(\mathbf{x})\|,
$$
We wish to find a vector $\mathbf{x} \in \mathbb{R}^n$ for which $\|\mathbf{r}(\mathbf{x})\|$ will be a minimum. A solution $\mathbf{\hat x}$ that minimizes $\|\mathbf{r}(\mathbf{x})\|$ is a *least squares solution* of the system $A \mathbf{x} = \mathbf{b}$. Do note that minimizing $\|\mathbf{r}(\mathbf{x})\|$ is equivalent to minimizing $\|\mathbf{r}(\mathbf{x})\|^2$.
> *Theorem 7*: let $S$ be a subspace of $\mathbb{R}^m$. For each $b \in \mathbb{R}^m$, there exists a unique $\mathbf{p} \in S$ that suffices
>
> $$
> \|\mathbf{b} - \mathbf{s}\| > \|\mathbf{b} - \mathbf{p}\|,
> $$
>
> for all $\mathbf{s} \in S\backslash\{\mathbf{p}\}$ and $\mathbf{b} - \mathbf{p} \in S^\perp$.
??? note "*Proof*:"
Will be added later.
If $\mathbf{p} = A \mathbf{\hat x}$ in $R(A)$ that is closest to $\mathbf{b}$ then it follows that
$$
\mathbf{b} - \mathbf{p} = \mathbf{b} - A \mathbf{x} = \mathbf{r}(\mathbf{\hat x}),
$$
must be an element of $R(A)^\perp$. Thus, $\mathbf{\hat x}$ is a solution to the least squares problem if and only if
$$
\mathbf{r}(\mathbf{\hat x}) \in R(A)^\perp = N(A^T).
$$
Thus to solve for $\mathbf{\hat x}$ we have the *normal equations* given by
$$
A^T A \mathbf{x} = A^T \mathbf{b}.
$$
Uniqueness of $\mathbf{\hat x}$ can be obtained if $A^T A$ is nonsingular which will be posed in the following theorem.
> *Theorem 8*: let $A \in \mathbb{R}^{m \times n}$ be an $m \times n$ matrix with rank $n$, then $A^T A$ is nonsingular.
??? note "*Proof*:"
Let $A \in \mathbb{R}^{m \times n}$ be an $m \times n$ matrix with rank $n$. Let $\mathbf{v}$ be a solution of
$$
A^T A \mathbf{x} = \mathbf{0},
$$
then $A \mathbf{v} \in N(A^T)$, but we also have that $A \mathbf{v} \in R(A) = N(A^T)^\perp$. Since $N(A^T) \cap N(A^T)^\perp = \{\mathbf{0}\}$ it follows that
$$
A\mathbf{v} = \mathbf{0},
$$
so $\mathbf{v} = \mathbf{0}$ by the nonsingularity of $A$.
It follows that
$$
\mathbf{\hat x} = (A^T A)^{-1} A^T \mathbf{b},
$$
is the unique solution of the normal equations for $A$ nonsingular and consequently, the unique least squares solution of the system $A \mathbf{x} = \mathbf{b}$.