Final commit for archival

2 files changed, 8 insertions, 5 deletions

diff --git a/latex/know/concept/gram-schmidt-method/source.md b/latex/know/concept/gram-schmidt-method/source.md
index b0c7b3b..7920a30 100644
--- a/latex/know/concept/gram-schmidt-method/source.md
+++ b/latex/know/concept/gram-schmidt-method/source.md
@@ -4,7 +4,8 @@
 # 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**
+$\ket*{V_1}, \ket*{V_2}, ...$ from a [Hilbert space](/know/concept/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 $\ket*{n_1}$:
@@ -33,3 +34,5 @@ turns them into an orthonormal set $\ket*{n_1}, \ket*{n_2}, ...$ as follows:
     \end{aligned}$$
     
 4.  Loop back to step 2, taking the next vector $\ket*{V_{j+1}}$.
+
+If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/).
diff --git a/latex/know/concept/hilbert-space/source.md b/latex/know/concept/hilbert-space/source.md
index 3f6ceb5..780fc0a 100644
--- a/latex/know/concept/hilbert-space/source.md
+++ b/latex/know/concept/hilbert-space/source.md
@@ -89,8 +89,8 @@ $\braket{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and
 $|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors.
 
 Orthonormality is 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](/know/concept/gram-schmidt-method).
+not already like that, it is common to manually turn them into a new
+orthonormal basis using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).
 
 As for the implementation of the inner product, it is given by:
 
@@ -171,8 +171,8 @@ $$\begin{aligned}
     = \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi}
 \end{aligned}$$
 
-For the latter integral to turn into $f(x)$, it is plain to see that
-$\braket{x}{\xi}$ must be a [Dirac delta function](/know/concept/dirac-delta-function/),
+Since we want the latter integral to reduce to $f(x)$, it is plain to see that
+$\braket{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
 i.e $\braket{x}{\xi} = \delta(x - \xi)$:
 
 $$\begin{aligned}