From 15bfb7730801809704c6561e20c5ca47627b2d79 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 20:53:46 +0100
Subject: Add "Gram-Schmidt method"

---
 latex/know/concept/gram-schmidt-method/source.md   | 35 ++++++++++++++++++++++
 latex/know/concept/hilbert-space/source.md         | 13 ++++----
 .../partial-fraction-decomposition/source.md       | 12 ++++----
 3 files changed, 46 insertions(+), 14 deletions(-)
 create mode 100644 latex/know/concept/gram-schmidt-method/source.md

(limited to 'latex')

diff --git a/latex/know/concept/gram-schmidt-method/source.md b/latex/know/concept/gram-schmidt-method/source.md
new file mode 100644
index 0000000..b0c7b3b
--- /dev/null
+++ b/latex/know/concept/gram-schmidt-method/source.md
@@ -0,0 +1,35 @@
+% Gram-Schmidt method
+
+
+# Gram-Schmidt method
+
+Given a set of linearly independent non-orthonormal vectors
+$\ket*{V_1}, \ket*{V_2}, ...$ from a Hilbert space, the **Gram-Schmidt method**
+turns them into an orthonormal set $\ket*{n_1}, \ket*{n_2}, ...$ as follows:
+
+1.  Take the first vector $\ket*{V_1}$ and normalize it to get $\ket*{n_1}$:
+    
+    $$\begin{aligned}
+        \ket*{n_1} = \frac{\ket*{V_1}}{\sqrt{\braket*{V_1}{V_1}}}
+    \end{aligned}$$
+    
+2.  Begin loop. Take the next non-orthonormal vector $\ket*{V_j}$, and
+    subtract from it its projection onto every already-processed vector:
+    
+    $$\begin{aligned}
+        \ket*{n_j'} = \ket*{V_j} - \ket*{n_1} \braket*{n_1}{V_j} - \ket*{n_2} \braket*{n_2}{V_j} - ... - \ket*{n_{j-1}} \braket*{n_{j-1}}{V_{j-1}}
+    \end{aligned}$$
+    
+    This leaves only the part of $\ket*{V_j}$ which is orthogonal to
+    $\ket*{n_1}$, $\ket*{n_2}$, etc. This why the input vectors must be
+    linearly independent; otherwise $\ket{n_j'}$ may become zero at some
+    point.
+    
+3.  Normalize the resulting ortho*gonal* vector $\ket*{n_j'}$ to make it
+    ortho*normal*:
+    
+    $$\begin{aligned}
+                \ket*{n_j} = \frac{\ket*{n_j'}}{\sqrt{\braket*{n_j'}{n_j'}}}
+    \end{aligned}$$
+    
+4.  Loop back to step 2, taking the next vector $\ket*{V_{j+1}}$.
diff --git a/latex/know/concept/hilbert-space/source.md b/latex/know/concept/hilbert-space/source.md
index 7d2ea05..3f6ceb5 100644
--- a/latex/know/concept/hilbert-space/source.md
+++ b/latex/know/concept/hilbert-space/source.md
@@ -41,12 +41,11 @@ other. Otherwise, they would be **linearly dependent**.
 
 A vector space $\mathbb{V}$ has **dimension** $N$ if only up to $N$ of
 its vectors can be linearly indepedent. All other vectors in
-$\mathbb{V}$ can then be written as a **linear combination** of these $N$
-so-called **basis vectors**.
+$\mathbb{V}$ can then be written as a **linear combination** of these $N$ **basis vectors**.
 
 Let $\vu{e}_1, ..., \vu{e}_N$ be the basis vectors, then any
 vector $V$ in the same space can be **expanded** in the basis according to
-the unique "weights" $v_n$, known as the **components** of the vector $V$
+the unique weights $v_n$, known as the **components** of $V$
 in that basis:
 
 $$\begin{aligned}
@@ -89,9 +88,9 @@ Two vectors $U$ and $V$ are **orthogonal** if their inner product
 $\braket{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and
 $|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors.
 
-Orthonormality is a desirable property for basis vectors, so if they are
+Orthonormality is desirable for basis vectors, so if they are
 not already orthonormal, it is common to manually derive a new
-orthonormal basis from them using e.g. the Gram-Schmidt method.
+orthonormal basis from them using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).
 
 As for the implementation of the inner product, it is given by:
 
@@ -157,7 +156,7 @@ The concept of orthonormality must be also weakened. A finite function
 $f(x)$ can be normalized as usual, but the basis vectors $x$ themselves
 cannot, since each represents an infinitesimal section of the real line.
 
-The rationale in this case is that the identity operator $\hat{I}$ must
+The rationale in this case is that action of the identity operator $\hat{I}$ must
 be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):
 
 $$\begin{aligned}
@@ -172,7 +171,7 @@ $$\begin{aligned}
     = \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi}
 \end{aligned}$$
 
-For the latter integral to turn into $f(x)$, it is clear that
+For the latter integral to turn into $f(x)$, it is plain to see that
 $\braket{x}{\xi}$ must be a [Dirac delta function](/know/concept/dirac-delta-function/),
 i.e $\braket{x}{\xi} = \delta(x - \xi)$:
 
diff --git a/latex/know/concept/partial-fraction-decomposition/source.md b/latex/know/concept/partial-fraction-decomposition/source.md
index aa03f9c..69428e7 100644
--- a/latex/know/concept/partial-fraction-decomposition/source.md
+++ b/latex/know/concept/partial-fraction-decomposition/source.md
@@ -3,7 +3,7 @@
 
 # Partial fraction decomposition
 
-*Partial fraction decomposition* or *expansion* is a method to rewrite a
+**Partial fraction decomposition** or **expansion** is a method to rewrite a
 quotient of two polynomials $g(x)$ and $h(x)$, where the numerator
 $g(x)$ is of lower order than $h(x)$, as a sum of fractions with $x$ in
 the denominator:
@@ -21,9 +21,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Then the constant coefficients $c_n$ can either be found the hard way,
+The constants $c_n$ can either be found the hard way,
 by multiplying the denominators around and solving a system of $N$
-equations, or the easy way by using the following trick:
+equations, or the easy way by using this trick:
 
 $$\begin{aligned}
     \boxed{
@@ -31,8 +31,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the
-form of:
+If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the form of:
 
 $$\begin{aligned}
     \boxed{
@@ -41,8 +40,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $c_{1,j}$ are found by putting the terms on a common denominator,
-e.g.:
+Where $c_{1,j}$ are found by putting the terms on a common denominator, e.g.
 
 $$\begin{aligned}
     \frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2}
-- 
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