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Merge pull request #18 from pabloocal/table-of-contents-hyperref
Table of contents and hyperref
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Chapters/chapter4.tex

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@@ -2698,7 +2698,7 @@ \subsection{Adding a multiple of one row to another doesn't change
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Here we used linearity in a row and the fact that the determinant of a
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matrix with two rows the same is zero.
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\subsection{The determinant of $QA$}
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\subsection{The determinant of \texorpdfstring{$QA$}{QA}}
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\label{sec:dettheory4}
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To begin, we compute the determinants of the elementary
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\det(Q_1Q_2Q_3\cdots Q_kA)=\det(Q_1)\det(Q_2)\cdots\det(Q_k)\det(A).
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\]
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\subsection{The determinant of $A$ is zero exactly when $A$ is
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not invertible}
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\subsection{The determinant of \texorpdfstring{$A$}{A} is zero exactly when
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\texorpdfstring{$A$}{A} is not invertible}
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\label{sec:dettheory5}
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Recall that if $R$ denotes the reduced form of $A$, obtained by
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other words, we can take $R=I$. Then $\det(R)=1$. Each
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$\det(Q_i^{-1})$ is non-zero too, so $\det(A)\ne 0$.
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\subsection{The product formula: $\det(AB)=\det(A)\det(B)$}
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\subsection{The product formula:
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\texorpdfstring{$\det(AB)=\det(A)\det(B)$}{determinant of product equals
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product of determinants}}
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\label{sec:dettheory6}
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If either $A$ or $B$ is non-invertible, then $AB$ is non-invertible

notes.tex

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\usepackage{enumerate}
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\usepackage{amssymb}
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% \usepackage{amsmath}
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\usepackage{hyperref}
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% headers and footers
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\usepackage{fancyhdr}

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