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/hilbert-space/source.md | 13 ++++++-------
 1 file changed, 6 insertions(+), 7 deletions(-)

(limited to 'latex/know/concept/hilbert-space/source.md')

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 $\braket{x}{\xi} = \delta(x - \xi)$:
 
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