Commit 8018b48a by Jim Hefferon

### did the usage pass over gr fiels

parent 783c7ed1
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 \chapter{Chapter One: Linear Systems} \subsection{One.I.1: Linear Systems} \begin{ans}{One.I.1.17} Gauss' method can be performed in different ways, so these simply We can perform Gauss' method can in different ways, so these simply exhibit one possible way to get the answer. \begin{exparts} \partsitem Gauss' method ... ... @@ -214,7 +214,7 @@ \end{ans} \begin{ans}{One.I.1.24} Yes. For example, the fact that the same reaction can be performed For example, the fact that we can have the same reaction in two different flasks shows that twice any solution is another, different, solution (if a physical reaction occurs then there must be at least one nonzero solution). ... ... @@ -282,7 +282,7 @@ \partsitem Yes, by inspection the given equation results from $$-\rho_1+\rho_2$$. \partsitem No. The given equation is satisfied by the pair $$(1,1)$$. The pair $$(1,1)$$ satisfies the given equation. However, that pair does not satisfy the first equation in the system. \partsitem Yes. ... ... @@ -468,7 +468,7 @@ again by the definition of satisfies'. Subtract $$k$$ times the $$i$$-th equation from the $$j$$-th equation (remark:~here is where $$i\neq j$$ is needed; if $$i=j$$ then the two (remark:~here is where we need $$i\neq j$$; if $$i=j$$ then the two $$d_i$$'s above are not equal) to get that the previous compound statement holds if and only if \begin{align*} ... ... @@ -522,7 +522,7 @@ \end{ans} \begin{ans}{One.I.1.34} Swapping rows is reversed by swapping back. Reverse a row swap by swapping back. \begin{eqnarray*} \begin{linsys}{3} a_{1,1}x_1 &+ &\cdots &+ &a_{1,n}x_n &= &d_1 \\ ... ... @@ -1704,8 +1704,8 @@ \end{ans} \begin{ans}{One.I.3.23} In this case the solution set is all of $$\Re^n$$ and can be expressed in the required form In this case the solution set is all of $$\Re^n$$ and we can express it in the required form \begin{equation*} \set{c_1\colvec[r]{1 \\ 0 \\ \vdotswithin{1} \\ 0} +c_2\colvec[r]{0 \\ 1 \\ \vdotswithin{0} \\ 0} ... ... @@ -1759,7 +1759,7 @@ Gauss' method will use only rationals (e.g., $$-(m/n)\rho_i+\rho_j$$). Thus the solution set can be expressed using only rational numbers as Thus we can express the solution set using only rational numbers as the components of each vector. Now the particular solution is all rational. ... ... @@ -1858,7 +1858,7 @@ -\colvec[r]{-1 \\ 0 \\ -4} =\colvec[r]{3 \\ 0 \\ 7} \end{equation*} that plane can be described in this way. we can describe that plane in this way. \begin{equation*} \set{\colvec[r]{-1 \\ 0 \\ -4} +m\colvec[r]{1 \\ 1 \\ 2} ... ... @@ -2086,11 +2086,11 @@ \end{ans} \begin{ans}{One.II.2.14} The set We could describe the set \begin{equation*} \set{\colvec{x \\ y \\ z}\suchthat 1x+3y-1z=0} \end{equation*} can also be described with parameters in this way. with parameters in this way. \begin{equation*} \set{\colvec[r]{-3 \\ 1 \\ 0}y+\colvec[r]{1 \\ 0 \\ 1}z \suchthat y,z\in\Re} ... ... @@ -2374,7 +2374,7 @@ \end{ans} \begin{ans}{One.II.2.30} The angle between $$(a)$$ and $$(b)$$ is found We can find the angle between $$(a)$$ and $$(b)$$ (for $$a,b\neq 0$$) with \begin{equation*} \arccos(\frac{ab}{\sqrt{a^2}\sqrt{b^2}}). ... ... @@ -2468,7 +2468,7 @@ The $$\vec{z}_1+\vec{z}_2=\zero$$ case is easy. For the rest, by the definition of angle, we will be done if we show this. we will be finished if we show this. \begin{equation*} \frac{\vec{z}_1\dotprod(\vec{z}_1+\vec{z}_2)}{ \norm{\vec{z}_1}\,\norm{\vec{z}_1+\vec{z}_2} } ... ... @@ -3015,7 +3015,7 @@ +\colvec[r]{-1 \\ -1 \\ 0 \\ 1}w \suchthat z,w\in\Re} \end{equation*} (of course, the zero vector could be omitted from the description). (of course, we could omit the zero vector from the description). \partsitem The Jordan'' half \begin{equation*} \grstep{-(1/7)\rho_2} ... ... @@ -3210,7 +3210,7 @@ For symmetric, we assume $A$ has the same sum of entries as~$B$ and obviously then $B$ has the same sum of entries as~$A$. Transitivity is no harder\Dash if $A$ has the same sum of entries as $B$ and $B$ has the same sum of entries as $C$ then clearly as $B$ and $B$ has the same sum of entries as $C$ then $A$ has the same as $C$. \end{ans} ... ... @@ -29481,6 +29481,32 @@ ans = 0.017398 \end{ans} \begin{ans}{2} This \textit{Sage} session gives equal values. \begin{lstlisting} sage: H=matrix(QQ,[[0,0,0,1], [1,0,0,0], [0,1,0,0], [0,0,1,0]]) sage: S=matrix(QQ,[[1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4]]) sage: alpha=0.85 sage: G=alpha*H+(1-alpha)*S sage: I=matrix(QQ,[[1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1]]) sage: N=G-I sage: 1200*N [-1155.00000000000 45.0000000000000 45.0000000000000 1065.00000000000] [ 1065.00000000000 -1155.00000000000 45.0000000000000 45.0000000000000] [ 45.0000000000000 1065.00000000000 -1155.00000000000 45.0000000000000] [ 45.0000000000000 45.0000000000000 1065.00000000000 -1155.00000000000] sage: M=matrix(QQ,[[-1155,45,45,1065], [1065,-1155,45,45], [45,1065,-1155,45], [45,45,1065,-1155]]) sage: M.echelon_form() [ 1 0 0 -1] [ 0 1 0 -1] [ 0 0 1 -1] [ 0 0 0 0] sage: v=vector([1,1,1,1]) sage: (v/v.norm()).n() (0.500000000000000, 0.500000000000000, 0.500000000000000, 0.500000000000000) \end{lstlisting} \end{ans} \begin{ans}{3} We have this. \begin{equation*} H=\begin{mat}
 ... ... @@ -671,6 +671,11 @@ beginfig(19) % 2-flat in R3; 2x+y+z=4, paramatized %drawarrow z8--z5 withcolor shading_color; drawarrow z5--z6 withcolor shading_color; drawarrow z5--z7 withcolor shading_color; label.rt(btex {\hspace*{0.05in} \scriptsize $P=\set{\colvec{x \\ y \\ z}=\colvec[r]{2 \\ 0 \\ 0} +y\cdot\colvec[r]{-1/2 \\ 1 \\ 0} +z\cdot\colvec[r]{-1/2 \\ 0 \\ 1}\suchthat y,z\in\Re}$} etex,.618[z3,z4]); endfig; ... ...
 ... ... @@ -470,6 +470,52 @@ beginfig(10); % m drawarrow subpath(xpart(times433)+.05,xpart(times434)-.05) of p43; endfig; % Search: every site points in a circle % beginfig(11); % %numeric u; %scaling factor %numeric v; %vertical scaling factor %numeric w; %horizontal scaling factor numeric circlescale; circlescale=19pt; path node; node=fullcircle scaled circlescale; path n[]; % node paths pickup pencircle scaled line_width_light; z1=(0w,6v); n1=node shifted z1; draw n1; label(btex \small $p_1$ etex,z1); z2=(9w,y1); n2=node shifted z2; draw n2; label(btex \small $p_2$ etex,z2); z3=(x2,0v); n3=node shifted z3; draw n3; label(btex \small $p_3$ etex,z3); z4=(x1,y3); n4=node shifted z4; draw n4; label(btex \small $p_4$ etex,z4); path p[], q[]; pair times[]; % intersection times % arrow from p1 to p2 p12=z1--z2; times121=p12 intersectiontimes n1; times122=p12 intersectiontimes n2; drawarrow subpath(xpart(times121)+.05,xpart(times122)-.05) of p12; % arrow from p2 to p3 p23=z2--z3; times232=p23 intersectiontimes n2; times233=p23 intersectiontimes n3; drawarrow subpath(xpart(times232)+.05,xpart(times233)-.05) of p23; % arrow from p3 to p4 p34=z3--z4; times343=p34 intersectiontimes n3; times344=p34 intersectiontimes n4; drawarrow subpath(xpart(times343)+.05,xpart(times344)-.05) of p34; % arrow from p4 to p1 p41=z4--z1; times414=p41 intersectiontimes n4; times411=p41 intersectiontimes n1; drawarrow subpath(xpart(times414)+.05,xpart(times411)-.05) of p41; endfig; ... ...
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 ... ... @@ -56,10 +56,10 @@ be parallel, or be the same line. \end{center} \end{minipage} \end{center} These pictures aren't a short way to prove the These pictures aren't a short way to prove the results from the prior section, because those apply to any number of linear equations and any number of unknowns. But they pictures do help us to understand those results. But they do help us understand those results. This section develops the ideas that we need to express our results geometrically. In particular, while ... ... @@ -129,7 +129,7 @@ despite that those displacements start in different places. \includegraphics{ch1.10} \end{center} Sometimes, to emphasize this property vectors have of not being anchored, they are referred to as \definend{free}\index{vector!free} vectors. we can refer to them as \definend{free}\index{vector!free} vectors. Thus, these free vectors are equal as each is a displacement of one over and two up. \begin{center} ... ... @@ -246,7 +246,7 @@ canonical representation ends at that point. \Re^n= \set{\colvec{v_1 \\ \vdotswithin{v_1} \\ v_n}\suchthat v_1,\ldots,v_n\in\Re} \end{equation*} And, addition and scalar multiplication are done component-wise. And, we do addition and scalar multiplication component-wise. Having considered points, we now turn to the lines. In $\Re^2$, the line through $$(1,2)$$ and $$(3,1)$$ ... ... @@ -279,7 +279,8 @@ the line through $$(1,2,1)$$ and $$(2,3,2)$$ is the set of and lines in even higher-dimensional spaces work in the same way. In $\Re^3$, a line uses one parameter so that there is freedom to move back and forth a line uses one parameter so that a particle on that line is free to move back and forth in one dimension, and a plane involves two parameters. For example, the plane through the points ... ... @@ -306,7 +307,7 @@ For example, the plane through the points \colvec[r]{1 \\ 0 \\ 5} \end{equation*} are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described As with the line, note that we describe some points in this plane with negative $t$'s or negative $s$'s or both. In algebra and calculus we often use a description of planes involving ... ... @@ -314,14 +315,14 @@ a single equation as the condition that describes the relationship among the first, second, and third coordinates of points in a plane. \newsavebox{\jhscratchbox} \savebox{\jhscratchbox}{\includegraphics{ch1.18}} \newlength{\jhscratchlength}\newlength{\jhscratchheight} \settowidth{\jhscratchlength}{\usebox{\jhscratchbox}} \begin{center} \usebox{\jhscratchbox} %\includegraphics{ch1.18} \end{center} % \newsavebox{\jhscratchbox} % \savebox{\jhscratchbox}{\includegraphics{ch1.18}} % \newlength{\jhscratchlength}\newlength{\jhscratchheight} % \settowidth{\jhscratchlength}{\usebox{\jhscratchbox}} \begin{equation*} % \usebox{\jhscratchbox} \vcenteredhbox{\includegraphics{ch1.18}} \end{equation*} % \begin{equation*} % P=\set{\colvec{x \\ y \\ z}\suchthat 2x+3y-z=4} % \end{equation*} ... ... @@ -329,12 +330,12 @@ The translation from such a description to the vector description that we favor in this book is to think of the condition as a one-equation linear system and parametrize $$x=2-y/2-z/2$$. \begin{equation*} \vcenteredhbox{\includegraphics{ch1.19}} \end{equation*} % \begin{center} % \includegraphics{ch1.19} % \makebox[\jhscratchlength][l]{\includegraphics{ch1.20}} % \end{center} \begin{center} \makebox[\jhscratchlength][l]{\includegraphics{ch1.20}} \end{center} % \begin{equation*} % P=\set{\colvec[r]{2 \\ 0 \\ 0} % +y\cdot\colvec[r]{-1/2 \\ 1 \\ 0} ... ... @@ -540,7 +541,7 @@ namely by any particular solution. -\colvec[r]{-1 \\ 0 \\ -4} =\colvec[r]{3 \\ 0 \\ 7} \end{equation*} that plane can be described in this way. we can describe that plane in this way. \begin{equation*} \set{\colvec[r]{-1 \\ 0 \\ -4} +m\colvec[r]{1 \\ 1 \\ 2} ... ... @@ -754,7 +755,7 @@ namely by any particular solution. The other equality is similar. \end{answer} \item How should $\Re^0$ be defined? How should we define $\Re^0$? \begin{answer} We shall later define it to be a set with one element\Dash an origin''. ... ... @@ -816,7 +817,8 @@ namely by any particular solution. \subsectionoptional{Length and Angle Measures} We've translated the first section's results about solution sets into geometric terms, to better understand those sets. But we must be careful not to be misled by our own terms; labeling subsets of But we must be careful not to be misled by our own terms\Dash labeling subsets of $$\Re^k$$ of the forms $$\set{\vec{p}+t\vec{v}\suchthat t\in\Re}$$ and $$\set{\vec{p}+t\vec{v}+s\vec{w}\suchthat t,s\in\Re}$$ ... ... @@ -1060,8 +1062,8 @@ between two nonzero vectors $$\vec{u},\vec{v}\in\Re^n$$ is \arccos(\,\frac{\vec{u}\dotprod\vec{v}}{ \norm{\vec{u}\,}\,\norm{\vec{v}\,} }\,) \end{equation*} (the angle between the zero vector and any other vector is defined to be a right angle). (by definition, the angle between the zero vector and any other vector is right). \end{definition} \noindent Thus vectors from $$\Re^n$$ are ... ... @@ -1078,7 +1080,7 @@ These vectors are orthogonal. \qquad $\colvec[r]{1 \\ -1}\dotprod\colvec[r]{1 \\ 1}=0$ \end{center} The arrows are shown away from canonical position We've drawn the arrows away from canonical position but nevertheless the vectors are orthogonal. \end{example} ... ... @@ -1181,11 +1183,11 @@ Not every vector in each is orthogonal to all vectors in the other. \colvec[r]{1 \\ 3 \\ -1} \end{equation*} \begin{answer} The set We could describe the set \begin{equation*} \set{\colvec{x \\ y \\ z}\suchthat 1x+3y-1z=0} \end{equation*} can also be described with parameters in this way. with parameters in this way. \begin{equation*} \set{\colvec[r]{-3 \\ 1 \\ 0}y+\colvec[r]{1 \\ 0 \\ 1}z \suchthat y,z\in\Re} ... ... @@ -1245,7 +1247,7 @@ Not every vector in each is orthogonal to all vectors in the other. \partsitem Associate? \partsitem How does it interact with scalar multiplication? \end{exparts} As always, any assertion must be backed by either a proof or an example. As always, you must back any assertion with either a proof or an example. \begin{answer} Assume that $$\vec{u},\vec{v},\vec{w}\in\Re^n$$ have components $$u_1,\ldots,u_n,v_1,\ldots,w_n$$. ... ... @@ -1541,7 +1543,7 @@ Not every vector in each is orthogonal to all vectors in the other. \item Describe the angle between two vectors in $$\Re^1$$. \begin{answer} The angle between $$(a)$$ and $$(b)$$ is found We can find the angle between $$(a)$$ and $$(b)$$ (for $$a,b\neq 0$$) with \begin{equation*} \arccos(\frac{ab}{\sqrt{a^2}\sqrt{b^2}}). ... ... @@ -1659,7 +1661,7 @@ Not every vector in each is orthogonal to all vectors in the other. The $$\vec{z}_1+\vec{z}_2=\zero$$ case is easy. For the rest, by the definition of angle, we will be done if we show this. we will be finished if we show this. \begin{equation*} \frac{\vec{z}_1\dotprod(\vec{z}_1+\vec{z}_2)}{ \norm{\vec{z}_1}\,\norm{\vec{z}_1+\vec{z}_2} } ... ...
 ... ... @@ -4,7 +4,7 @@ \section{Reduced Echelon Form} After developing the mechanics of Gauss' method, we observed that it can be done in more than one way. One example is that from this matrix For example, from this matrix \begin{equation*} \begin{mat}[r] 2 &2 \\ ... ... @@ -114,7 +114,7 @@ by combining upwards. The answer is $$x=1$$, $$y=1$$, and $$z=2$$. \end{example} Using one entry to clear out the rest of a column is called \definend{pivoting}\index{pivoting} on that entry. \definend{pivoting}\index{pivoting} on that entry. Note that the row combination operations in the first stage move left to right, from column one to column three, ... ... @@ -176,7 +176,7 @@ variable is the leading variable in some row. And, we can read off when the system has an infinite solution set because there is no contradiction and at least one variable is free. If the echelon form is reduced then we can read off not just the size of the In reduced echelon form we can read off not just the size of the solution set but also its description. We have no trouble describing the solution set when it is empty, of course. \nearbyexample{exm:GJRedReadOffSol} and~\ref{exm:GJRedReadOffSolTwo} ... ... @@ -261,8 +261,8 @@ is undone by subtracting the same multiple of row $$i$$ from row $$j$$. \;\grstep{-k\rho_i+\rho_j} A \end{equation*} (The $i\neq j$ conditions is needed. See \nearbyexercise{exer:INotJMakesRowOpsRev}.) (We need the $i\neq j$ condition; see \nearbyexercise{exer:INotJMakesRowOpsRev}.) \end{proof} Again, the point of view that we are developing, buttressed now by this lemma, ... ... @@ -271,9 +271,9 @@ is that the term reduces to' is misleading:~where after'' $$A$$ or simpler than'' $A$. Instead we should think of them as inter-reducible or interrelated. Below is a picture of the idea. The matrices from the start of this section and their reduced echelon form version are shown in a cluster. They are all inter-reducible. It shows the matrices from the start of this section and their reduced echelon form version in a cluster as inter-reducible. \begin{center} \includegraphics{ch1.28} \end{center} ... ... @@ -320,14 +320,14 @@ The diagram below shows the collection of all matrices as a box. Inside that box, each matrix lies in some class. Matrices are in the same class if and only if they are interreducible. The classes are disjoint\Dash no matrix is in two distinct classes. The collection of matrices has been partitioned into We have partitioned the collection of matrices into \definend{row equivalence classes}.\appendrefs{partitions and class representatives}\index{partition!row equivalence classes} \begin{center} \includegraphics{ch1.27} \end{center} \noindent One of the classes in this partition is the cluster of matrices from the start of this section that is shown above, cluster of matrices from the start of this section shown above, expanded to include all of the nonsingular $\nbyn{2}$ matrices. The next subsection proves that the reduced echelon form of a matrix is ... ... @@ -681,7 +681,7 @@ class. +\colvec[r]{-1 \\ -1 \\ 0 \\ 1}w \suchthat z,w\in\Re} \end{equation*} (of course, the zero vector could be omitted from the description). (of course, we could omit the zero vector from the description). \partsitem The Jordan'' half \begin{equation*} \grstep{-(1/7)\rho_2} ... ... @@ -901,7 +901,7 @@ class. For symmetric, we assume $A$ has the same sum of entries as~$B$ and obviously then $B$ has the same sum of entries as~$A$. Transitivity is no harder\Dash if $A$ has the same sum of entries as $B$ and $B$ has the same sum of entries as $C$ then clearly as $B$ and $B$ has the same sum of entries as $C$ then $A$ has the same as $C$. \end{answer} \item \cite{Cleary} ... ... @@ -947,7 +947,7 @@ class. and show that the $-1\cdot\rho_1+\rho_1$ operation is not reversed by $1\cdot\rho_1+\rho_1$. \partsitem Expand the proof of that lemma to make explicit exactly where the $i\neq j$ condition on combining is used. it uses the $i\neq j$ condition on combining. \end{exparts} \begin{answer} \begin{exparts} ... ... @@ -1042,7 +1042,8 @@ the situations that we were investigating. Here, where we are investigating row equivalence, we know that the set of all matrices breaks into the row equivalence classes. When we finish the proof here, we will have a way to understand each of those classes\Dash its matrices can be thought of as derived by row operations from the classes\Dash we can think of its matrices as derived by row operations from the unique reduced echelon form matrix in that class. Put in more operational terms, ... ... @@ -1113,7 +1114,7 @@ which is a linear combination of the $x$'s. \end{proof} In this subsection we will use the convention that, where a matrix is named with an upper case roman letter, that, where an upper case roman letter names the matrix then the matching lower-case greek letter names the rows. \begin{equation*} A= ... ... @@ -1435,7 +1436,7 @@ another way and get $y$ and $w$ free. We end this section with a recap. In Gauss' method we start with a matrix and then derive a sequence of other matrices. We defined two matrices to be related if one can be derived from the other. We defined two matrices to be related if we can derive one from the other. That relation is an equivalence relation, %\appendrefs{equivalence relation} called row equivalence, and so partitions the set of all matrices into row equivalence classes. ... ... @@ -2045,7 +2046,7 @@ and the class of matrices row equivalent to this % \end{answer} \recommended \item The Linear Combination Lemma says which equations can be gotten from Gaussian reduction from a given linear system. Gaussian reduction of a given linear system. \begin{enumerate} \item Produce an equation not implied by this system. \begin{equation*} ... ...
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 ... ... @@ -116,7 +116,7 @@ picture themselves doing the same type of work. \noindent{\bf Applications and computers.} %\smallskip The point of view taken here, that students should think of linear algebra as about vector spaces Linear Algebra as about vector spaces and linear maps, is not taken to the complete exclusion of others. Applications and computing are interesting and vital aspects of the subject. ... ... @@ -126,11 +126,11 @@ They are brief enough that an instructor can do one in a day's class or can assign them as independent or small-group projects. Most simply give a reader a taste of the subject, discuss how linear algebra comes in, a taste of the subject, discuss how Linear Algebra comes in, point to some further reading, and give a few exercises. Whether they figure formally in a course or not these help readers see for themselves that linear algebra is a tool that a professional must have. readers see for themselves that Linear Algebra is a tool that a professional must master. ... ... @@ -147,7 +147,7 @@ That page also contains this book's latest version, answers to the exercises, and the \LaTeX\ source. A text is a large and complex project and one of the lessons of software development is that errors are bound to find their way in. development is that such a project is bound to have errors. I welcome bug reports (contributions are acknowledged in the source). I save them and periodically issue revisions. ... ...
 ... ... @@ -204,6 +204,36 @@ Two additional excellent expositions are =\alpha\cdot 1+(1-\alpha)\cdot 1$, which is one. \end{answer} \item For this web of pages, the importance of each page should be equal. Check that for$\alpha=0.85\$. \begin{center} \includegraphics{ch5.11} \end{center} \begin{answer} This \textit{Sage} session gives equal values. \begin{lstlisting} sage: H=matrix(QQ,[[0,0,0,1], [1,0,0,0], [0,1,0,0], [0,0,1,0]]) sage: S=matrix(QQ,[[1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4], [1/4,1/4,1/4,1/4]]) sage: alpha=0.85 sage: G=alpha*H+(1-alpha)*S sage: I=matrix(QQ,[[1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1]]) sage: N=G-I sage: 1200*N [-1155.00000000000 45.0000000000000 45.0000000000000 1065.00000000000] [ 1065.00000000000 -1155.00000000000 45.0000000000000 45.0000000000000] [ 45.0000000000000 1065.00000000000 -1155.00000000000 45.0000000000000] [ 45.0000000000000 45.0000000000000 1065.00000000000 -1155.00000000000] sage: M=matrix(QQ,[[-1155,45,45,1065], [1065,-1155,45,45], [45,1065,-1155,45], [45,45,1065,-1155]]) sage: M.echelon_form() [ 1 0 0 -1] [ 0 1 0 -1] [ 0 0 1 -1] [ 0 0 0 0] sage: v=vector([1,1,1,1]) sage: (v/v.norm()).n() (0.500000000000000, 0.500000000000000, 0.500000000000000, 0.500000000000000) \end{lstlisting} \end{answer} \item \cite{BryanLeise} Give the importance ranking for this web of pages. \begin{center} ... ...
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