Fix vol. 1 overfull hboxes

Signed-off-by: zeramorphic <[email protected]>

cmd/files.go

@@ -41,8 +41,8 @@ var BookFiles []TexFile = []TexFile{
 	{FilePath: "ia/book", Name: "IA Book"},
 	{FilePath: "ib/book", Name: "IB Book"},
 	{FilePath: "ii/book", Name: "II Book"},
-	{FilePath: "ia/book1", Name: "Book 1"},
-	{FilePath: "ia/book2", Name: "Book 2"},
+	{FilePath: "ia/vol1", Name: "Volume 1"},
+	{FilePath: "ia/vol2", Name: "Volume 2"},
 }
 
 var FilesWithBook []TexFile

ia/de/01_differentiation.tex

@@ -121,8 +121,8 @@ \subsection{Rules for differentiation}
 		f''' & = u'''v + 3u''v' + 3u'v'' + uv''' \nonumber
 	\end{align}
 	This is analogous to Pascal's triangle and the binomial expansion.
-	The coefficients are \(n!/m!(n-m)!
-	\), more often written \(n \choose m\).
+	The coefficients are
+	\[n \choose m = \frac{n!}{m!(n-m)!}\]
 \end{definition}
 
 \subsection{Order of magnitude}
@@ -255,4 +255,3 @@ \subsection{L'H\^opital's rule}
 Note that l'H\^opital's rule can be applied recursively, using higher-order derivatives.
 For example, consider \(f(x) = 3\sin x - \sin 3x\); \(g(x) = 2x - \sin 2x\).
 The limit approaches 3 as \(x \to 0\).
-

ia/de/12_multivariate_calculus.tex

@@ -187,7 +187,7 @@ \subsection{Signature of Hessian}
 If \(\abs{H} = 0\), we need higher order terms in the Taylor series.
 
 \subsection{Contours near stationary points}
-Consider a coordinate system aligned with the principal axes of the Hessian \(H\) in two-dimensional space, so
+Consider a coordinate system aligned with the principal axes of the Hessian \(H\) in two-dimen\-sional space, so
 \[
 	H = \begin{pmatrix}
 		\lambda_1 & 0 \\ 0 & \lambda_2

ia/de/13_systems_of_odes.tex

@@ -195,7 +195,7 @@ \subsection{Phase portraits}
 	\item \(\lambda_1, \lambda_2\) form a complex conjuate pair.
 	      If the real parts are negative, %TODO graph
 	      This is a stable spiral; the curves tend towards zero.
-	      If the real parts are positive we have an unstable spiral:
+	      If the real parts are positive we have an unstable spiral.
 	      %TODO graph
 	      If the real part is zero, the solution curves are circles.
 	      This is known as a centre.

ia/dr/02_dimensional_analysis.tex

@@ -19,7 +19,7 @@ \subsection{Choice of units}
 If, however, we used SI units for length, mass and time, but the imperial unit pound-force as the unit for force, the equations would be inconsistent.
 
 \subsection{Scaling and dimensional independence}
-Suppose that a dimensional quantity \(Y\) depends on a set of dimensional quantities \(X_1, X_2, \dots, X_n\), so the dimension of \(Y\) is \(L^a M^b T^c\) and the dimension of the \(X_i\) are \(L^{a_i} M^{b_i} T^{c_i}\).
+Suppose that a dimensional quantity \(Y\) depends on a set of dimensional quantities \(X_1, \dots, X_n\), so the dimension of \(Y\) is \(L^a M^b T^c\) and the dimension of the \(X_i\) are \(L^{a_i} M^{b_i} T^{c_i}\).
 
 If \(n \leq 3\), then \(Y = C \cdot X_1^{p_1}X_2^{p_2}X_3^{p_3}\), and \(p_1, p_2, p_3\) can be found by balancing the dimensions.
 Hence \(a = a_1p_1 + a_2p_2 + a_3p_3\) and so forth for \(b\) and \(c\).

ia/dr/05_friction.tex

@@ -1,4 +1,3 @@
-\subsection{Definition}
 Friction is a contact force, unlike the forces we have discussed previously.
 It is a convenient encapsulation of many complicated molecular phenomena; it is not a fundamental force.
 

ia/dr/15_geometry_of_spacetime.tex

@@ -114,18 +114,20 @@ \subsection{Rapidity}
 \]
 This represents a boost in the \(x\) direction.
 Combining two boosts, we get
-\[
-	\Lambda[\beta_1]\Lambda[\beta_2] = \begin{pmatrix}
+\begin{align*}
+	\Lambda[\beta_1]\Lambda[\beta_2] &= \begin{pmatrix}
 		\gamma_1         & -\gamma_1\beta_1 \\
 		-\gamma_1\beta_1 & \gamma_1
 	\end{pmatrix}\begin{pmatrix}
 		\gamma_2         & -\gamma_2\beta_2 \\
 		-\gamma_2\beta_2 & \gamma_2
-	\end{pmatrix} = \begin{pmatrix}
+	\end{pmatrix} \\
+	&= \begin{pmatrix}
 		\gamma_1\gamma_2(1 + \beta_1 \beta_2) & -\gamma_1\gamma_2(\beta_1 + \beta_2)  \\
 		-\gamma_1\gamma_2(\beta_1 + \beta_2)  & \gamma_1\gamma_2(1 + \beta_1 \beta_2)
-	\end{pmatrix} = \Lambda\qty[\frac{\beta_1 + \beta_2}{1 + \beta_1\beta_2}]
-\]
+	\end{pmatrix} \\
+	&= \Lambda\qty[\frac{\beta_1 + \beta_2}{1 + \beta_1\beta_2}]
+\end{align*}
 Note the relation to the velocity transformation law.
 Recall that with spatial rotations, we can characterise a rotation \(R\) by some parameter \(\theta\), where \(R(\theta_1) R(\theta_2) = R(\theta_1 + \theta_2)\).
 This is the same kind of composition law.

ia/dr/16_relativistic_physics.tex

@@ -271,7 +271,7 @@ \subsection{Particle decay}
 Hence the rest mass of the initial particle must be \textit{at least} the sum of the rest masses of the particles that result from the decay.
 
 \subsection{Higgs to photon decay}
-Consider the decay of the Higgs particle \(h\) into two photons \(\gamma\).
+Consider the decay of the Higgs particle \(h\) into two photons \(\gamma_1, \gamma_2\).
 By conservation of 4-momentum,
 \[
 	P_h = P_{\gamma_1} + P_{\gamma_2}

ia/groups/01_axiomatic_definition.tex

@@ -251,7 +251,9 @@ \subsection{Subgroups generated by a subset}
 
 \begin{proposition}
 	Let \(X \subseteq G, X \neq \varnothing\).
-	Then \(\genset{X}\) is the set of elements of \(G\) of the form \(x_1^{\alpha_1} x_2^{\alpha_2} x_3^{\alpha_3} \cdots x_k^{\alpha_k}\) where \(x_i \in X\) (not necessarily distinct), \(\alpha_i = \pm 1\), and \(k \geq 0\).
+	Then \(\genset{X}\) is the set of elements of \(G\) of the form
+	\[x_1^{\alpha_1} x_2^{\alpha_2} x_3^{\alpha_3} \cdots x_k^{\alpha_k}\]
+	where \(x_i \in X\) (not necessarily distinct), \(\alpha_i = \pm 1\), and \(k \geq 0\).
 	By convention, the empty product \(k=0\) is defined to be \(e\).
 \end{proposition}
 \begin{proof}

ia/groups/03_types_of_groups.tex

@@ -11,7 +11,8 @@ \subsection{Direct products of groups}
 	\begin{itemize}
 		\item (closure) For a pair of elements \((g_1, h_1)\) and \((g_2, h_2)\) in \(G \times H\), the product \((g_1 \ast_G g_2, h_1 \ast_H h_2)\) is clearly in \(G \times H\), because the first entry is in \(G\) and the second entry is in \(H\), which is the requirement for being a member of \(G \times H\).
 		\item (identity) The element \((e_G, e_H)\) is an identity.
-		\item (inverses) Given an element \((g, h) \in G \times H\), the element \((g^{-1}, h^{-1})\) satisfies \((g^{-1}, h^{-1})(g, h) = (e_G, e_H) = e_{G \times H}\).
+		\item (inverses) Given an element \((g, h) \in G \times H\), the element \((g^{-1}, h^{-1})\) satisfies
+		\[(g^{-1}, h^{-1})(g, h) = (e_G, e_H) = e_{G \times H}\]
 		\item (associativity) Given three elements \((g_i, h_i)\), \(i \in \{1, 2, 3\}\), we have
 		      \begin{align*}
 			      ((g_1, h_1) \ast (g_2, h_2)) \ast (g_3, h_3) & = (g_1 \ast g_2, h_1 \ast h_2) \ast (g_3, h_3)       \\

ia/groups/04_permutation_groups.tex

@@ -141,7 +141,9 @@ \subsection{Products of transpositions}
 \end{proposition}
 \begin{proof}
 	It is enough to prove this for just a cycle, then we can use the disjoint cycle decomposition to create a transposition product for the whole \(\sigma\).
-	\((a_1\ a_2\cdots a_n) = (a_1\ a_2)(a_2\ a_3)\cdots(a_{n-1}\ a_n)\), so the result is immediate.
+	We have
+	\[(a_1\ a_2\cdots a_n) = (a_1\ a_2)(a_2\ a_3)\cdots(a_{n-1}\ a_n)\]
+	so the result is immediate.
 \end{proof}
 Note that this decomposition is not unique in general.
 

ia/groups/07_normal_subgroups_and_quotients.tex

@@ -199,7 +199,7 @@ \subsection{Examples and properties}
 	      We have \(\faktor{G}{H} \cong K\) and \(\faktor{G}{K} \cong H\) % (TODO: proof as exercise).
 	\item Consider \(N := \genset{r^2} \trianglelefteq D_8\).
 	      We can check that it is normal by trying \(r^{-1}r^2r^{-1} \in N\), and also \(s^{-1}r^2s = r^{-2} = r^2 \in N\).
-	      Since \(\genset{r, s} = D_8\), and the generators obey this normal subgroup relation, it follows that \(g^{-1}ng\) for all \(g \in D_8\) (TODO proof as exercise).
+	      Since \(\genset{r, s} = D_8\), and the generators obey this normal subgroup relation, it follows that \(g^{-1}ng\) for all \(g \in D_8\).
 	      We know \(\abs{N} = 2\), so \(\abs{\faktor{D_8}{N}} = \abs{D_8 : N} = \frac{\abs{D_8}}{\abs{N}}\) by Lagrange's Theorem.
 	      So \(\abs{\faktor{D_8}{N}} = 4\).
 	      We know that any group of order 4 is isomorphic either to \(C_4\) or \(C_2 \times C_2\).

ia/groups/08_isomorphism_theorems.tex

@@ -57,7 +57,7 @@ \subsection{Correspondence theorem}
 	So the correspondence is bijective (this satisfies the property that \(ff^{-1}\) and \(f^{-1}f\) are the identity maps on the relevant sets).
 \end{proof}
 This correspondence preserves lots of structure: for example, indices, normality, containment.
-One example is the group \(C_4 \times C_2\), where \(C_4 = \genset a\) and \(C_2 = \genset b\).
+% One example is the group \(C_4 \times C_2\), where \(C_4 = \genset a\) and \(C_2 = \genset b\).
 % The subgroups of this are (TODO draw subgroup lattice)
 % TODO draw subgroup lattice
 Now, let \(N := \genset{(a^2, b)}\).
@@ -107,7 +107,7 @@ \subsection{Third isomorphism theorem}
 			  \[
 				  \faktor{H}{H \cap N} \cong \faktor{HN}{N} \implies \faktor{3\mathbb Z}{15\mathbb Z} \cong \faktor{\mathbb Z}{5\mathbb Z} \cong \mathbb Z_5
 			  \]
-		\item (TODO see \(C_4 \times C_2\) example from last time) Let \(C_4 = \genset a\), \(C_2 = \genset b\), \(G = C_4 \times C_2\), \(N = \genset{(a^2, b)}\), \(M = \genset{(e, b), (a^2, e)}\).
+		\item Let \(C_4 = \genset a\), \(C_2 = \genset b\), \(G = C_4 \times C_2\), \(N = \genset{(a^2, b)}\), \(M = \genset{(e, b), (a^2, e)}\).
 			  Then \(N \leq M \leq G\).
 			  By the Third Isomorphism Theorem,
 			  \[

ia/groups/09_group_actions.tex

@@ -32,8 +32,10 @@ \subsection{Definition}
 	\(\forall g \in G, \alpha_g: X \to X, x \mapsto \alpha_g(x)\) is a bijection.
 \end{lemma}
 \begin{proof}
-	We have that \(\alpha_g(\alpha_{g^{-1}}(x)) = \alpha_{g g^{-1}}(x) = \alpha_e(x) = x\).
-	Similarly, \(\alpha_{g^{-1}}(\alpha_g(x)) = \alpha_{g^{-1} g}(x) = \alpha_e(x) = x\).
+	We have that
+	\[\alpha_g(\alpha_{g^{-1}}(x)) = \alpha_{g g^{-1}}(x) = \alpha_e(x) = x\]
+	Similarly,
+	\[\alpha_{g^{-1}}(\alpha_g(x)) = \alpha_{g^{-1} g}(x) = \alpha_e(x) = x\]
 	So the composition \(\alpha_g \circ \alpha_{g^{-1}}\) is the identity on \(X\), and \(\alpha_{g^{-1}} \circ \alpha_g\) is also the identity on \(X\), so \(\alpha_g\) is a bijection.
 \end{proof}
 We can also define actions by linking \(G\) to \(\Sym(X)\).

ia/groups/11_action_of_the_m_obius_group.tex

@@ -181,7 +181,7 @@ \subsection{Geometric properties of M\"obius maps}
 		\item \(z \mapsto \frac{1}{z}\)
 	\end{itemize}
 	So it is enough to check that each of these generating maps preserves circles.
-	Writing \(S(A, B, C)\) for the circle satisfying
+	We will write \(S(A, B, C)\) for the circle satisfying
 	\[
 		Az\overline z + \overline B z + B \overline z + C = 0 \tag{\(\clubsuit\)}
 	\]

ia/groups/12_matrix_groups.tex

@@ -109,7 +109,7 @@ \subsection{Matrix encoding of M\"obius maps}
 
 \subsection{Actions of matrices on vector spaces}
 All of the groups defined above act on the corresponding vector spaces.
-For example, \(GL_n(\mathbb F) \acts \mathbb F^n\).
+For example, we have \(GL_n(\mathbb F) \acts \mathbb F^n\).
 As an example, let \(G \leq GL_2(\mathbb R) \acts \mathbb R^2\).
 What are the orbits of this action?
 Clearly, \(\{ \vb 0 \}\) is a singleton orbit since we are acting by linear maps.

ia/ns/03_modular_arithmetic.tex

@@ -63,13 +63,13 @@ \subsection{Euler's totient function}
 	      Note that \(\varphi(ab) = \varphi(a)\varphi(b)\).
 \end{itemize}
 
-\subsection{Fermat's little theorem and Fermat-Euler theorem}
+\subsection{Fermat's little theorem and Fermat--Euler theorem}
 \begin{theorem}
 	Let \(p\) be a prime.
 	Then in \(\mathbb Z_p\), \(a \neq 0 \implies a^{p-1} = 1\).
 \end{theorem}
 \noindent This is actually a special case of the following theorem:
-\begin{theorem}[Fermat-Euler Theorem]
+\begin{theorem}[Fermat--Euler Theorem]
 	Let \(n \geq 2\).
 	Then in \(\mathbb Z_n\), any unit \(a\) satisfies \(a^{\varphi(n)} = 1\).
 \end{theorem}
@@ -227,7 +227,7 @@ \subsection{RSA encryption}
 For example, we can find \(x, x^2, x^4, x^8, x^{16}\) through repeated squaring, and then for example we can calculate \(x^{19} = x^{16} x^{2} x^{1}\).
 
 To decode \(x^e\), we ideally want some number \(d\) such that \((x^e)^d = x\).
-By the Fermat-Euler Theorem, we have \(x^{\varphi(n)} = 1\), so clearly \(x^{k\varphi(n) + 1} = x\).
+By the Fermat--Euler Theorem, we have \(x^{\varphi(n)} = 1\), so clearly \(x^{k\varphi(n) + 1} = x\).
 In other words, we want \(ed \equiv 1 \mod \varphi(n)\).
 By running Euclid's algorithm on \(e\) and \(\varphi(n)\), we can find such a \(d\).
 Note that this requires \(e\) and \(\varphi(n)\) to be coprime; in practice we would choose \(e\) after we have chosen \(n\) such that this is the case.

ia/ns/04_the_reals.tex

@@ -181,7 +181,8 @@ \subsection{Sequences and limits}
 	      Then for all \(n \geq N\), either \(x_n = \frac{1}{n}\) or 0.
 	      In either case, \(\abs{x_n - 0} \leq \frac{1}{n} \leq \frac{1}{N} < \varepsilon\).
 \end{enumerate}
-We can denote the entirety of a sequence \(x_1, x_2, \dots\) as \((x_n)\) or \((x_n)_{n=1}^\infty\).
+We can denote the entirety of a sequence \(x_1, x_2, \dots\) as
+\[(x_n) \quad \text{or} \quad (x_n)_{n=1}^\infty\]
 For example, \(\left( (-1)^n \right)_{n=1}^{\infty}\) is divergent.
 This isn't saying that it goes to infinity, just that it doesn't converge.
 Note also that if \(x_n \to c\) and \(x_n \to d\), then \(c=d\).
@@ -223,7 +224,8 @@ \subsection{Testing convergence of a sequence}
 This is a very important theorem that we will refer back to time and time again.
 \begin{note}
 	If we were in \(\mathbb Q\), this would not necessarily hold.
-	For example, \(1, 1.4, 1.41, 1.414, 1.4142, \dots\) (the decimal expansion of \(\sqrt{2}\)).
+	For example, consider the decimal expansion of \(\sqrt{2}\).
+	\[1, 1.4, 1.41, 1.414, 1.4142, \dots\]
 	They don't converge to a limit in \(\mathbb Q\).
 	So our proof will have to be more rigorous than just `they have to tend to somewhere below the upper bound'; we must use a property that \(\mathbb R\) has that \(\mathbb Q\) does not have, i.e.\ the least upper bound axiom.
 \end{note}

ia/ns/07_countability.tex

@@ -240,15 +240,17 @@ \subsection{Comparing sizes of sets}
 	      Let \(f(a) = a'\) for each \(a\in A\), and \(f\) is injective.
 \end{itemize}
 
-\subsection{Schr\"oder-Bernstein theorem}
+\subsection{Schr\"oder--Bernstein theorem}
 Further, we must also have that if `\(A\) is at most as large as \(B\)' and `\(B\) is at most as large as \(A\)', then they must be the same size.
 Otherwise this size intuition would not make sense.
-\begin{theorem}[Schr\"oder-Bernstein Theorem]
+\begin{theorem}[Schr\"oder--Bernstein Theorem]
 	If \(f\colon A\to B\) and \(g\colon B\to A\) are injections, then there exists a bijection \(h\colon A\to B\).
 \end{theorem}
 \begin{proof}
 	For \(a\in A\), we will write \(g^{-1}(a)\) to denote the unique \(b \in B\) such that \(g(b) = a\), if such a \(b\) exists (and similarly for \(f^{-1}(b)\)).
-	The `ancestor sequence' of \(a \in A\) is \(g^{-1}(a), f^{-1}g^{-1}(a), g^{-1}f^{-1}g^{-1}(a), \dots\) which may terminate.
+	The `ancestor sequence' of \(a \in A\) is
+	\[g^{-1}(a), f^{-1}g^{-1}(a), g^{-1}f^{-1}g^{-1}(a), \dots\]
+	which may terminate.
 	So for any ancestor, after undergoing the relevant function \(f\) or \(g\) repeatedly, we will end up at \(a\).
 	There are three possible behaviours:
 	\begin{itemize}

ia/vm/04_index_notation_and_the_summation_convention.tex

@@ -46,8 +46,8 @@ \subsection{Kronecker \texorpdfstring{\( \delta \)}{𝛿} and Levi-Civita \texor
 
 We use the `summation convention' to abbreviate the many summation symbols used throughout linear algebra.
 \begin{enumerate}
-	\item An index which occurs exactly once in some term, denoted a `free index', must appear once in every term in that equation.
-	\item An index which occurs exactly twice in a given term, denoted a `repeated/contracted/dummy index', is implicitly summed over.
+	\item An index which occurs exactly once in some term, called a `free' index, must appear once in every term in that equation.
+	\item An index which occurs exactly twice in a given term, called a `repeated', `contracted', or `dummy' index, is implicitly summed over.
 	\item No index can occur more than twice in a given term.
 \end{enumerate}
 

ia/vm/10_adjugates_and_alternating_forms.tex

@@ -97,9 +97,10 @@ \subsection{Levi-Civita \texorpdfstring{\( \varepsilon \)}{𝜀} in higher dimen
 \subsection{Properties}
 \begin{enumerate}
 	\item The alternating form is multilinear.
-	      \[
-		      [ \vb v_1, \cdots, \vb v_{p-1}, \alpha \vb u + \beta \vb w, \vb v_{p+1} \cdots, \vb v_n ] = \alpha [ \vb v_1, \cdots, \vb v_{p-1}, \vb u, \vb v_{p+1} \cdots, \vb v_n ] + \beta [ \vb v_1, \cdots, \vb v_{p-1}, \vb w, \vb v_{p+1} \cdots, \vb v_n ]
-	      \]
+	      \begin{align*}
+			[ \vb v_1, \cdots, \vb v_{p-1}, \alpha \vb u + \beta \vb w, \vb v_{p+1} \cdots, \vb v_n ] &= \alpha [ \vb v_1, \cdots, \vb v_{p-1}, \vb u, \vb v_{p+1} \cdots, \vb v_n ] \\
+			&+ \beta [ \vb v_1, \cdots, \vb v_{p-1}, \vb w, \vb v_{p+1} \cdots, \vb v_n ]
+		  \end{align*}
 	\item It is totally antisymmetric.
 	      \([ \vb v_{\sigma(1)}, \vb v_{\sigma(2)}, \cdots, \vb v_{\sigma(n)} ] = \varepsilon(\sigma) [ \vb v_1, \cdots, \vb v_n ]\)
 	\item Standard basis vectors give a positive result: \([\vb e_i, \cdots, \vb e_n] = 1\).
@@ -111,7 +112,7 @@ \subsection{Properties}
 	      \[
 		      [\vb v_1, \cdots, \vb v_p, \cdots, \vb v_q, \cdots, \vb v_n ] = 0
 	      \]
-	\item If \(v_p\) can be written as a non-trivial linear combination of the other vectors, then
+	\item If \(\vb v_p\) can be written as a non-trivial linear combination of the other vectors, then
 	      \[
 		      [\vb v_1, \cdots, \vb v_p, \cdots, \vb v_n ] = 0
 	      \]
@@ -121,10 +122,13 @@ \subsection{Properties}
 To justify (ii) above, it suffices to check a transposition \(\tau = (p\ q)\) where (without loss of generality) \(p < q\), then since transpositions generate all permutations the result follows.
 \begin{align*}
 	 & [\vb v_1, \cdots, \vb v_{p-1}, \vb v_q, \vb v_{p+1}, \cdots, \vb v_{q-1}, \vb v_p, \vb v_{q+1}, \cdots, \vb v_n]                                                                                                                          \\
-	 & = \sum_\sigma \varepsilon(\sigma) (\vb v_1)_{\sigma(1)} \cdots (\vb v_{p-1})_{\sigma(p-1)}(\vb v_q)_{\sigma(p)}(\vb v_{p+1})_{\sigma(p+1)} \cdots (\vb v_{q-1})_{\sigma(q-1)}(\vb v_p)_{\sigma(q)}(\vb v_{q+1})_{\sigma(q+1)}             \\
-	 & = \sum_\sigma \varepsilon(\sigma) (\vb v_1)_{\sigma'(1)} \cdots (\vb v_{p-1})_{\sigma'(p-1)}(\vb v_q)_{\sigma'(q)}(\vb v_{p+1})_{\sigma'(p+1)} \cdots (\vb v_{q-1})_{\sigma'(q-1)}(\vb v_p)_{\sigma'(p)}(\vb v_{q+1})_{\sigma'(q+1)}      \\
+	 & = \sum_\sigma \varepsilon(\sigma) (\vb v_1)_{\sigma(1)} \cdots (\vb v_{p-1})_{\sigma(p-1)}(\vb v_q)_{\sigma(p)}(\vb v_{p+1})_{\sigma(p+1)} \\
+	 &\quad\quad\quad\cdots (\vb v_{q-1})_{\sigma(q-1)}(\vb v_p)_{\sigma(q)}(\vb v_{q+1})_{\sigma(q+1)}             \\
+	 & = \sum_\sigma \varepsilon(\sigma) (\vb v_1)_{\sigma'(1)} \cdots (\vb v_{p-1})_{\sigma'(p-1)}(\vb v_q)_{\sigma'(q)}(\vb v_{p+1})_{\sigma'(p+1)} \\
+	 &\quad\quad\quad\cdots (\vb v_{q-1})_{\sigma'(q-1)}(\vb v_p)_{\sigma'(p)}(\vb v_{q+1})_{\sigma'(q+1)}      \\
 	\intertext{where \(\sigma' = \sigma\tau\)}
-	 & = -\sum_{\sigma'} \varepsilon(\sigma') (\vb v_1)_{\sigma'(1)} \cdots (\vb v_{p-1})_{\sigma'(p-1)}(\vb v_p)_{\sigma'(p)}(\vb v_{p+1})_{\sigma'(p+1)} \cdots (\vb v_{q-1})_{\sigma'(q-1)}(\vb v_q)_{\sigma'(q)}(\vb v_{q+1})_{\sigma'(q+1)} \\
+	 & = -\sum_{\sigma'} \varepsilon(\sigma') (\vb v_1)_{\sigma'(1)} \cdots (\vb v_{p-1})_{\sigma'(p-1)}(\vb v_p)_{\sigma'(p)}(\vb v_{p+1})_{\sigma'(p+1)} \\
+	 &\quad\quad\quad\cdots (\vb v_{q-1})_{\sigma'(q-1)}(\vb v_q)_{\sigma'(q)}(\vb v_{q+1})_{\sigma'(q+1)} \\
 	 & = -[\vb v_1, \cdots, \vb v_{p-1}, \vb v_p, \vb v_{p+1}, \cdots, \vb v_{q-1}, \vb v_q, \vb v_{q+1}, \cdots, \vb v_n]
 \end{align*}
 as required.

ia/vm/13_eigenvalues_and_eigenvectors.tex

@@ -225,7 +225,7 @@ \subsection{Eigenspaces and multiplicities}
 
 \subsection{Linear independence of eigenvectors}
 \begin{proposition}
-	Let \(\vb v_1, \vb v_2, \dots, \vb v_r\) be eigenvectors of an \(n\times n\) matrix \(A\) with eigenvalues \(\lambda_1, \lambda_2,\dots,\lambda_r\).
+	Let \(\vb v_1, \dots, \vb v_r\) be eigenvectors of an \(n\times n\) matrix \(A\) with eigenvalues \(\lambda_1,\dots,\lambda_r\).
 	If the eigenvalues are distinct, then the eigenvectors are linearly independent.
 \end{proposition}
 \begin{proof}