\documentclass[10pt,a4paper]{article} % Packages \usepackage{fancyhdr} % For header and footer \usepackage{multicol} % Allows multicols in tables \usepackage{tabularx} % Intelligent column widths \usepackage{tabulary} % Used in header and footer \usepackage{hhline} % Border under tables \usepackage{graphicx} % For images \usepackage{xcolor} % For hex colours %\usepackage[utf8x]{inputenc} % For unicode character support \usepackage[T1]{fontenc} % Without this we get weird character replacements \usepackage{colortbl} % For coloured tables \usepackage{setspace} % For line height \usepackage{lastpage} % Needed for total page number \usepackage{seqsplit} % Splits long words. %\usepackage{opensans} % Can't make this work so far. Shame. Would be lovely. \usepackage[normalem]{ulem} % For underlining links % Most of the following are not required for the majority % of cheat sheets but are needed for some symbol support. \usepackage{amsmath} % Symbols \usepackage{MnSymbol} % Symbols \usepackage{wasysym} % Symbols %\usepackage[english,german,french,spanish,italian]{babel} % Languages % Document Info \author{spoopyy} \pdfinfo{ /Title (linear-algebra.pdf) /Creator (Cheatography) /Author (spoopyy) /Subject (Linear Algebra Cheat Sheet) } % Lengths and widths \addtolength{\textwidth}{6cm} \addtolength{\textheight}{-1cm} \addtolength{\hoffset}{-3cm} \addtolength{\voffset}{-2cm} \setlength{\tabcolsep}{0.2cm} % Space between columns \setlength{\headsep}{-12pt} % Reduce space between header and content \setlength{\headheight}{85pt} % If less, LaTeX automatically increases it \renewcommand{\footrulewidth}{0pt} % Remove footer line \renewcommand{\headrulewidth}{0pt} % Remove header line \renewcommand{\seqinsert}{\ifmmode\allowbreak\else\-\fi} % Hyphens in seqsplit % This two commands together give roughly % the right line height in the tables \renewcommand{\arraystretch}{1.3} \onehalfspacing % Commands \newcommand{\SetRowColor}[1]{\noalign{\gdef\RowColorName{#1}}\rowcolor{\RowColorName}} % Shortcut for row colour \newcommand{\mymulticolumn}[3]{\multicolumn{#1}{>{\columncolor{\RowColorName}}#2}{#3}} % For coloured multi-cols \newcolumntype{x}[1]{>{\raggedright}p{#1}} % New column types for ragged-right paragraph columns \newcommand{\tn}{\tabularnewline} % Required as custom column type in use % Font and Colours \definecolor{HeadBackground}{HTML}{333333} \definecolor{FootBackground}{HTML}{666666} \definecolor{TextColor}{HTML}{333333} \definecolor{DarkBackground}{HTML}{A3A3A3} \definecolor{LightBackground}{HTML}{F3F3F3} \renewcommand{\familydefault}{\sfdefault} \color{TextColor} % Header and Footer \pagestyle{fancy} \fancyhead{} % Set header to blank \fancyfoot{} % Set footer to blank \fancyhead[L]{ \noindent \begin{multicols}{3} \begin{tabulary}{5.8cm}{C} \SetRowColor{DarkBackground} \vspace{-7pt} {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}} } \end{tabulary} \columnbreak \begin{tabulary}{11cm}{L} \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Linear Algebra Cheat Sheet}}}} \\ \normalsize{by \textcolor{DarkBackground}{spoopyy} via \textcolor{DarkBackground}{\uline{cheatography.com/28376/cs/8341/}}} \end{tabulary} \end{multicols}} \fancyfoot[L]{ \footnotesize \noindent \begin{multicols}{3} \begin{tabulary}{5.8cm}{LL} \SetRowColor{FootBackground} \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}} \\ \vspace{-2pt}spoopyy \\ \uline{cheatography.com/spoopyy} \\ \end{tabulary} \vfill \columnbreak \begin{tabulary}{5.8cm}{L} \SetRowColor{FootBackground} \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}} \\ \vspace{-2pt}Published 4th June, 2016.\\ Updated 4th June, 2016.\\ Page {\thepage} of \pageref{LastPage}. \end{tabulary} \vfill \columnbreak \begin{tabulary}{5.8cm}{L} \SetRowColor{FootBackground} \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}} \\ \SetRowColor{white} \vspace{-5pt} %\includegraphics[width=48px,height=48px]{dave.jpeg} Measure your website readability!\\ www.readability-score.com \end{tabulary} \end{multicols}} \begin{document} \raggedright \raggedcolumns % Set font size to small. Switch to any value % from this page to resize cheat sheet text: % www.emerson.emory.edu/services/latex/latex_169.html \footnotesize % Small font. \begin{multicols*}{4} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Basis}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{A set S is a basis for V if}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{1. S spans V} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{2. S is LI.} \tn % Row Count 3 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{If S is a basis for V then every vector in V can be written in one and only one way as a linear combo of} \seqsplit{vectors in S and every set containing more than n vectors is LD.}} \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Basis Test}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{1. If S is a LI set of vectors in V}, then S is a basis for V} \tn % Row Count 2 (+ 2) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{2. If S spans V, then S is a basis for V} \tn % Row Count 3 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Change of Basis}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{P{[}x{]}\_B' = {[}x{]}\_B} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{{[}x{]}\_B' = P\textasciicircum{}-1\textasciicircum{} {[}x{]}\_B} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{{[}B B'{]} -\textgreater{} {[} I P\textasciicircum{}-1\textasciicircum{} {]}} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{{[}B' B {]} -\textgreater{} {[} I P {]}} \tn % Row Count 4 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Cross Product}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{if u = u1i + u2j + u3k} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{AND} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{v = v1i + v2j + v3k} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{THEN} \tn % Row Count 4 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{u x v = (u2v3 - u3v2)i - (u1v3 -u3v1)j + (u1v2 - u2v1)k} \tn % Row Count 6 (+ 2) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Definition of a Vector Space}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{u + v is within V} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{u+v = v+u} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{u+(v+w) = (u+v)+w} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{u+0 = u} \tn % Row Count 4 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{u-u = 0} \tn % Row Count 5 (+ 1) % Row 5 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{cu is within V} \tn % Row Count 6 (+ 1) % Row 6 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{c(u+v) = cu+cv} \tn % Row Count 7 (+ 1) % Row 7 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{(c+d)u = cu+du} \tn % Row Count 8 (+ 1) % Row 8 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{c(du) = (cd)u} \tn % Row Count 9 (+ 1) % Row 9 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{1*u = u} \tn % Row Count 10 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Diagonalizable Matrices}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{A is diagonalizable when A is similar to a diagonal matrix}. \newline % Row Count 2 (+ 2) That is, A is diagonalizable when there exists an invertible matrix P such that P\textasciicircum{}-1\textasciicircum{}AP is a diagonal matrix% Row Count 5 (+ 3) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Dot Products Etc.}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{length/norm ||v|| = sqrt(v\_1\textasciicircum{}2\textasciicircum{} +...+ v\_n\textasciicircum{}2\textasciicircum{}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{||cv|| = |c| ||v||} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{v / ||v|| is the unit vector} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{distance d(u,v) = ||u-v||} \tn % Row Count 4 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{Dot product u•v = (u\_1{\emph{v\_1 +...+ u\_n}}v\_n)} \tn % Row Count 5 (+ 1) % Row 5 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{n cos(theta) = u•v / (||u|| ||v||)} \tn % Row Count 6 (+ 1) % Row 6 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{u\&v are orthagonal when dot(u},v) = 0} \tn % Row Count 7 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Eigenshit}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{The scalar lambda(Y) is called an {\bf{Eigenvalue}} of A \seqsplit{when there is a nonzero vector x such that Ax = Yx}.} \tn % Row Count 3 (+ 3) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{Vector x is an {\bf{Eigenvector}} of A corresponding to Y.} \tn % Row Count 5 (+ 2) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{The set of all eigenvectors with the zero vector is a } subspace of R\textasciicircum{}n\textasciicircum{} called the {\bf{Eigenspace}} of Y.} \tn % Row Count 8 (+ 3) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{1. Find Eigenvalues: det(YI -} A) = 0} \tn % Row Count 9 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{2. Find Eigenvectors: (YI -} A)x = 0} \tn % Row Count 10 (+ 1) % Row 5 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{If A is a triangular matrix then its eigenvalues are on } its main diagonal} \tn % Row Count 12 (+ 2) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Gram-Schmidt Orthonormalization}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{1. B = \{v1, v2, ..., vn\}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{2. B' = \{w1, w2, ..., wn\}:} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{} \tn % Row Count 2 (+ 0) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{w1 = v1} \tn % Row Count 3 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{w2 = v2 - projw1v2} \tn % Row Count 4 (+ 1) % Row 5 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{w3 = v3 - projw1v3 - projw2v3} \tn % Row Count 5 (+ 1) % Row 6 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{wn = vn - ...} \tn % Row Count 6 (+ 1) % Row 7 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{} \tn % Row Count 6 (+ 0) % Row 8 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{3. B'' = \{u1, u2, ..., un\}:} \tn % Row Count 7 (+ 1) % Row 9 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{} \tn % Row Count 7 (+ 0) % Row 10 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{ui = wi/||wi||} \tn % Row Count 8 (+ 1) % Row 11 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{} \tn % Row Count 8 (+ 0) % Row 12 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{B'' is an orthonormal basis for V}} \tn % Row Count 9 (+ 1) % Row 13 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{span(B) = span(B'')} \tn % Row Count 10 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Important Vector Spaces}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{R\textasciicircum{}n\textasciicircum{}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{C(-inf, +inf)} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{C{[}a, b{]}} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{P} \tn % Row Count 4 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{P\_n} \tn % Row Count 5 (+ 1) % Row 5 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{M\_m,n} \tn % Row Count 6 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Inner Products}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{||u|| = sqrt\textless{}u,u\textgreater{}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{d(u,v) = ||u-v||} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{cos(theta) = \textless{}u,v\textgreater{} / (||u|| ||v||)} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{u\&v are orthagonal when \textless{}u,v\textgreater{} = 0} \tn % Row Count 4 (+ 1) % Row 4 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{proj\_v u = \textless{}u,v\textgreater{}/\textless{}v,v\textgreater{} * v} \tn % Row Count 5 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Kernal}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{For T:V-\textgreater{}W \seqsplit{The set of all vectors v in V that satisfies} \seqsplit{T(v)=0 is the kernal of T. ker(T) is a subspace of v.} \newline % Row Count 3 (+ 3) For T:R\textasciicircum{}n\textasciicircum{} -\textgreater{}R\textasciicircum{}m\textasciicircum{} by T(x)=Ax \seqsplit{ker(T) = solution space of Ax=0 \&} Cspace(A) = range(T)% Row Count 5 (+ 2) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Linear Combo}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{v is a linear combo of u\_1 ... u\_n} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{.} \tn % Row Count 2 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Linear Independence}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{a set of vectors S is LI if c1v1 +...+ ckvk = 0 has only the trivial solution. \newline % Row Count 2 (+ 2) If there are other solutions S is LD. A set \seqsplit{S is LI iff one of its vectors can } \newline % Row Count 4 (+ 2) \seqsplit{be written as a combo of other S vectors}.% Row Count 5 (+ 1) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Linear Transformation}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{V \& W are Vspaces. T:V-\textgreater{}W is a linear transformation of V into W if:} \tn % Row Count 2 (+ 2) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{1. T(u+v) = T(u) = T(v)} \tn % Row Count 3 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{2. T(cu) = cT(u)} \tn % Row Count 4 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Non-Homogeny}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{If xp is a solution to Ax = b then every solution to} \seqsplit{the system can be written as} x = xp% Row Count 2 (+ 2) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Nullity}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{Nullspace(A) = \{x ε R\textasciicircum{}n\textasciicircum{} : Ax = 0 \newline % Row Count 1 (+ 1) \seqsplit{Nullity(A) = dim(Nullspace(A)) } \newline % Row Count 2 (+ 1) = n - rank(A)% Row Count 3 (+ 1) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Orthogonal Sets}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{Set S in V is orthogonal when every pair of vectors} in \seqsplit{S is orthogonal. If each } vector is a unit vector, then S is orthonormal% Row Count 3 (+ 3) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{One-to-One and Onto}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{T is one-to-one {\bf{iff}} ker(T) = \{0\}} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{T is onto {\bf{iff}} rank(T) = dim(W)} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{If dim(T) = dim(W) then T is one-to-one {\bf{iff}} it is onto} \tn % Row Count 4 (+ 2) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Rank and Nullity of T}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{nullity(T) = dim(kernal)} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{rank(T) = dim(range)} \tn % Row Count 2 (+ 1) % Row 2 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{range(T) + nullity(T) = n (in m\_x} n)} \tn % Row Count 3 (+ 1) % Row 3 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{dim(domain) = dim(range) + dim(kernal)}} \tn % Row Count 4 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Rank of a Matrix}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{Rank(A) = dim(Rspace) = dim(Cspace)}% Row Count 1 (+ 1) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Similar Matrices}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{\seqsplit{For square matrices A and A' of order n}, A' is similar \seqsplit{to A when there exits an invertible matrix P such that A' =} P\textasciicircum{}-1\textasciicircum{} AP% Row Count 3 (+ 3) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Spanning Sets}} \tn \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{S = \{v1...vk\} is a subset of vector space V. S \seqsplit{spans V if every vector in v can be written as a linear} combo of vectors in S.% Row Count 3 (+ 3) } \tn \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} \begin{tabularx}{3.833cm}{X} \SetRowColor{DarkBackground} \mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Test for Subspace}} \tn % Row 0 \SetRowColor{LightBackground} \mymulticolumn{1}{x{3.833cm}}{1. u+v are in W} \tn % Row Count 1 (+ 1) % Row 1 \SetRowColor{white} \mymulticolumn{1}{x{3.833cm}}{2. cu is in w} \tn % Row Count 2 (+ 1) \hhline{>{\arrayrulecolor{DarkBackground}}-} \end{tabularx} \par\addvspace{1.3em} % That's all folks \end{multicols*} \end{document}