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% Document Info
\author{afalita6}
\pdfinfo{
  /Title (linear-algrebra.pdf)
  /Creator (Cheatography)
  /Author (afalita6)
  /Subject (linear algrebra Cheat Sheet)
}

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\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}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{linear algrebra Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{afalita6} via \textcolor{DarkBackground}{\uline{cheatography.com/171839/cs/36095/}}}
\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}afalita6 \\
  \uline{cheatography.com/afalita6} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 14th December, 2022.\\
   Updated 14th December, 2022.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{3.23505 cm} x{1.74195 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Inverse of a matrix}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Triangular or diagonal matrix}} & 1/diagonal entries \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
{\bf{Permuted matrix}} & P transpose \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
{\bf{Other}} & rref ( {[}A eye(){]} ) \tn 
% Row Count 5 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.54287 cm} x{3.43413 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Multiplication of Matrix + angle}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Way 1}} & A*B full multiplication \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
{\bf{Way 2 }} & {[}row A{]}*B \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
{\bf{Way 3}} & {[}col A{]}*B \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
{\bf{Way 3}} & B11*col(A1)+B21*col(A2) \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
{\bf{Find entry 2,3}} & {[}row A2{]}*{[}columnB3{]} = 1 number \tn 
% Row Count 6 (+ 2)
% Row 5
\SetRowColor{white}
{\bf{Rank 1 matrix}} & {[}a11*rowB1; a21*rowB1;a31*rowB1{]} + ... \tn 
% Row Count 8 (+ 2)
% Row 6
\SetRowColor{LightBackground}
{\bf{Angle}} & cos(theta) = ({\bf{v}}*{\bf{w}})/(||{\bf{v}}||*||{\bf{w}}||) \tn 
% Row Count 10 (+ 2)
% Row 7
\SetRowColor{white}
{\bf{Outer Product}} & {[}column1{]}*{[}1 \# \#{]} find numbers that work \tn 
% Row Count 12 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Linear Transformation and dependency}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Linear Independent}} & Linearly independent if rref(A) -{}-{}-{}-\textgreater{} \#pivots = \#row \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
{\bf{Linear transformation (x and y given)}} & T (u + v) = T (u) + T (v), T (cu) = cT (u), where c is a number. T is one-to-one if T(u)=0⇒u=0 T is onto if Col(T) = Rm. \tn 
% Row Count 10 (+ 7)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.54287 cm} x{3.43413 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Projections or Ax=b is inconsistent}}  \tn
% Row 0
\SetRowColor{LightBackground}
formula & A'*A*xhat=A'* \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
{\bf{Step 1}} & rref ( {[}A'*A A'*b{]} ) \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
{\bf{Step 2}} & xhat = last column of rref \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
{\bf{Step 3}} & bhat = A*xhat -{}-\textgreater{} bhat is the vector spaned A closest to v and the projection of the vector onto subspace \tn 
% Row Count 7 (+ 4)
% Row 4
\SetRowColor{LightBackground}
{\bf{Step 4}} & be = b - bhat -{}-\textgreater{} be is the vector perpendicular \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
{\bf{Step 4}} & error vector/distance = norm (be) (1/sqr of components of b swuares) \tn 
% Row Count 12 (+ 3)
% Row 6
\SetRowColor{LightBackground}
{\bf{For regression}} & step 1: f(x) = {[}x{]}{[}b{]}, step 2: A = {[}x.\textasciicircum{}0 ...{]} and y = given, step 3: do LSE and find xhat which will be a,b,c \tn 
% Row Count 17 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.18988 cm} x{2.78712 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Ax = b}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Echelon form}} & Leading entries in every row are farther to the right than the row above. To do = elimination steps \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
{\bf{Reduced Echelon form (rref)}} & echelon + columns of leading entries are all 0 except the entry which must be a 1. To do = eliminations steps down to right, then left to top \tn 
% Row Count 12 (+ 7)
% Row 2
\SetRowColor{LightBackground}
{\bf{Ax=b with LU}} & L = identity but a21 = -λ1, a31 = -λ2, a32 = -λ3. U = \tn 
% Row Count 15 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.18988 cm} x{2.78712 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Ax = b (A and b specified)}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Echelon form}} & Leading entries in every row are farther to the right than the row above. To do = elimination steps \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
{\bf{Reduced Echelon form (rref)}} & echelon + columns of leading entries are all 0 except the entry which must be a 1. To do = eliminations steps down to right, then left to top \tn 
% Row Count 12 (+ 7)
% Row 2
\SetRowColor{LightBackground}
{\bf{Ax=b with LU}} & L = identity but a21 = -λ1, a31 = -λ2, a32 = -λ3. U = echelon. Then do Ly=b - given (solve for y), then Ux=y (solve for x) \tn 
% Row Count 18 (+ 6)
% Row 3
\SetRowColor{white}
{\bf{Ax=b with CR}} & tMaybe not full rank. C = columns of A that have a pivot in R. R = rref form. To find x -{}-\textgreater{} using R to find FV, pivots, and special solutions (if b not 0 do rref({[}A b{]})), if one soln is given then add that in gen sol and just do rref(A) \tn 
% Row Count 29 (+ 11)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Eigenvectors and Eigenvalues}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{v}} & eigenvector \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
{\bf{λ}} & eigenvalue \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
{\bf{Finding λ}} & 1. Diag or triang = entries of diag. 2. 2x2 do λ = m +- sqrt (m\textasciicircum{}2 - p), where m = (a11+a22)/2, and p = a11*a22 - a12*a21 \tn 
% Row Count 8 (+ 6)
% Row 3
\SetRowColor{white}
{\bf{ Finding v}} & rref ( {[}A - λ*eye {]} ) and find FV, pivots, and ss \tn 
% Row Count 11 (+ 3)
% Row 4
\SetRowColor{LightBackground}
{\bf{Diagonalization }} & A = P*D*P\textasciicircum{}(-1), where P = {[}eigenvectors{]} , D = diag(λ) \tn 
% Row Count 14 (+ 3)
% Row 5
\SetRowColor{white}
{\bf{When can we diagonalize}}* & Only when: square, real λ, and if repeated λ - look rref ( {[}A - λ*eye {]} ) and only 1 pivot. \tn 
% Row Count 19 (+ 5)
% Row 6
\SetRowColor{LightBackground}
{\bf{ A = Q*D*Q'}} & Q = special solutions form rref ( {[}A - λ*eye {]} ) for every λ, and then doing norm(q1) for all of them. D = diag(λs) \tn 
% Row Count 25 (+ 6)
% Row 7
\SetRowColor{white}
{\bf{ Is λ an eigenvalue}} & Do rref ( {[}A - λ*eye {]} ) and has to be only 1 pivot (linearly dependent) \tn 
% Row Count 29 (+ 4)
% Row 8
\SetRowColor{LightBackground}
{\bf{Positive definite}} & λs all positive \tn 
% Row Count 31 (+ 2)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Eigenvectors and Eigenvalues (cont)}}  \tn
% Row 9
\SetRowColor{LightBackground}
{\bf{Semipositive definite}} & λs all positive and at least a 0 \tn 
% Row Count 2 (+ 2)
% Row 10
\SetRowColor{white}
{\bf{Indefinite}} & λ at least one is negative \tn 
% Row Count 4 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Vector Spaces and Basis}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Subspace}} & If u and v are in W , then u + v are in W , and cu is in W \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
{\bf{Basis B for V}} & A linearly independent set such that Span (B) = V To show sthg is a basis, show it is linearly independent (rref(A) has NO FV) and spans(no row of 0's). \tn 
% Row Count 10 (+ 7)
% Row 2
\SetRowColor{LightBackground}
{\bf{Row(A)}} & Space spanned by the rows of A: Row-reduce A and choose the rows that contain the pivots. Row(A) = R\textasciicircum{}n, dim = rank, Basis of Row = R in A = CR \tn 
% Row Count 17 (+ 7)
% Row 3
\SetRowColor{white}
{\bf{Col(A)}} & Space spanned by columns of A: Row-reduce A and choose the columns of A that contain the pivots. Col(A) = R\textasciicircum{}m, dim = rank, Basis of Col = C in A = CR \tn 
% Row Count 24 (+ 7)
% Row 4
\SetRowColor{LightBackground}
{\bf{Null(A) / Vector in Null}} & Solutions of Ax = 0. Row-reduce A. Null(A) = R\textasciicircum{}n, dim = n-rank, Basis of Null = rref(A), FV, pivots, special solutions \tn 
% Row Count 30 (+ 6)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Vector Spaces and Basis (cont)}}  \tn
% Row 5
\SetRowColor{LightBackground}
{\bf{LeftNull(A)}} & Solutions of A'x = 0. Row-reduce A'. LeftNull(A) = R\textasciicircum{}m, dim = m-rank, Basis of LeftNull = rref(A'), FV, pivots, special solutions \tn 
% Row Count 6 (+ 6)
% Row 6
\SetRowColor{white}
{\bf{Rank(A)}} & number of pivots \tn 
% Row Count 7 (+ 1)
% Row 7
\SetRowColor{LightBackground}
{\bf{Is v in Null}} & do A*v and it needs to equal to vector 0 \tn 
% Row Count 9 (+ 2)
% Row 8
\SetRowColor{white}
{\bf{find v in ColA}} & same vectors as in matrix \tn 
% Row Count 11 (+ 2)
% Row 9
\SetRowColor{LightBackground}
{\bf{Is v in col space of B}} & is B*x=v consistent? do rref({[}B v{]}) and see if consistent \tn 
% Row Count 14 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.16956 cm} x{1.04425 cm} p{0.54301 cm} x{1.42018 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Gram-Schmidt steps}}  \tn
% Row 0
\SetRowColor{LightBackground}
A & q1 = A(:,1) & Q = q1 & xhat =(q1'*A(:,2))/(q1'*q1) \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
ahat = Q*xhat & q2 = A(:,2) - ahat & Q(:,2) = q2 & Q(:,1) = \seqsplit{1/(q'1*q1)*q1} \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Q(:,2) = \seqsplit{1/(q'2*q2)*q2} & Q = {[} Q(:,1) Q(:,2) {]} & R = Q'*A & if 3x3 keep going \tn 
% Row Count 9 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Orthogonality}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{ v and u are othogonal}} & if {\bf{v}}*{\bf{u}} = 0 \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
{\bf{W⊥: }} & Set of v which are orthogonal to every w in W. \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
{\bf{Orthogonal projection:}} & If \{u1 · · · uk \} is a basis for W , then orthogonal projection of y on W is: \seqsplit{yˆ=(y·u1/u1*u1)+···+(y·u1/uk*uk)}, and  y − yˆ is orthogonal to yˆ, shortest distance btw y and W is ||y−yˆ|| \tn 
% Row Count 13 (+ 9)
% Row 3
\SetRowColor{white}
{\bf{Basis of W⊥: }} & basis of Null(Mw) \tn 
% Row Count 15 (+ 2)
% Row 4
\SetRowColor{LightBackground}
{\bf{Equalities between basis}} & (RowA)' = NullA and vice versa. (ColA)'=LeftNullA and vice versa \tn 
% Row Count 18 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}