Commit 5cc77136 by Jim Hefferon

### change first vector space example to have restrictions

parent 934f6247
 ... ... @@ -8,4 +8,4 @@ See http://joshua.smcvt.edu/linearalgebra for more information and contact details. Jim Hefferon 2012-June-28 \ No newline at end of file 2016-Nov-30
 ... ... @@ -679,7 +679,7 @@ Because the $i$-th row is unchanged, the operation $-k\rho_i+\rho_j$ returns the $j$-th row to its original state. (Observe that the $$i=j$$ conditino on the $k\rho_i+\rho_j$ (Observe that the $$i=j$$ condition on the $k\rho_i+\rho_j$ is needed, or else this could happen \begin{equation*} \begin{linsys}{2} ... ... @@ -8793,7 +8793,7 @@ sage: solve(system, x,y) \end{ans} \begin{ans}{Two.III.1.39} We have (using these odball operations with care) We have (using these oddball operations with care) \begin{multline*} \set{\colvec{1-y-z \\ y \\ z}\suchthat y,z\in\Re} =\set{\colvec{-y+1 \\ y \\ 0} ... ... @@ -11142,7 +11142,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $T>R>D$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $R>T>D$ \\ ... ... @@ -11150,7 +11150,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $D>T>R$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $T>D>R$ \\ ... ... @@ -11158,7 +11158,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $R>D>T$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \end{tabular} \end{center} ... ... @@ -11283,7 +11283,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $T>R>D$ \\ \textit{-cancelled} \textit{-canceled} \end{tabular} \\ \hline \begin{tabular}{l} $R>T>D$ \\ ... ... @@ -11291,7 +11291,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $D>T>R$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $T>D>R$ \\ ... ... @@ -11299,7 +11299,7 @@ sage: solve(system, x,y) \end{tabular} &\begin{tabular}{l} $R>D>T$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \end{tabular} \end{center} ... ... @@ -16069,7 +16069,7 @@ sage: solve(system, x,y) (iii)~Setting $a=b=c=0$ in the calculation gives infinitely many solutions. Paramatrizing using the free variable~$z$ leads to this description Parametrizing using the free variable~$z$ leads to this description of the nullspace. \begin{equation*} \nullspace{h}=\set{\colvec{x \\ y \\ z}\suchthat ... ... @@ -19917,7 +19917,7 @@ sage: solve(system, x,y) \quad \rep{\identity(x^2)}{B}=\colvec{-1 \\ 0 \\ 1} \end{equation*} These calcuations give $Q$. These calculations give $Q$. \begin{multline*} \rep{\identity( \begin{mat} ... ... @@ -19984,7 +19984,7 @@ sage: solve(system, x,y) 0 &1 \end{mat} \end{equation*} These calcuations give $Q=\rep{\identity}{D,\hat{D}}$. These calculations give $Q=\rep{\identity}{D,\hat{D}}$. \begin{equation*} \rep{\identity(\colvec{0 \\ 1})}{\hat{D}}=\colvec{0 \\ 1} \quad ... ... @@ -20380,7 +20380,7 @@ sage: solve(system, x,y) \begin{equation*} \begin{CD} V_{\wrt{B_1}} @>t>T> V_{\wrt{B_1}} \\ @V\text{\scriptsize$\identity$} VV @V\text{scriptsize$\identity$} VV \\ @V\text{\scriptsize$\identity$} VV @V\text{\scriptsize$\identity$} VV \\ V_{\wrt{B_2}} @>t>\hat{T}> V_{\wrt{B_2}} \end{CD} \end{equation*} ... ... @@ -23137,7 +23137,7 @@ sage: p.save('bridges.pdf') \end{center} We can easily verify that the resulting vector has unit length and forms an angle with the $x$-axis of $-\pi/12$, which is indeed a rotation clockwose of $\pi/4$ radians a rotation clockwise of $\pi/4$ radians since $(\pi/6)-(\pi/4)=-\pi/12$. \end{exparts} ... ... @@ -30455,7 +30455,7 @@ octave:6> gplot z \partsitem Computing the eigenvalues gives $\lambda_1=3$, $\lambda_2=0$, and $\lambda_3=-4$. Thus the diagonalzation of the matrix is this. Thus the diagonalization of the matrix is this. \begin{equation*} \begin{mat} 3 &0 &0 \\
 ... ... @@ -1074,7 +1074,7 @@ write it as a function of the rows We want a formula to determine whether an $\nbyn{n}$ matrix is nonsingular. We will not begin by stating such a formula. Instead we will begin by considering, for eacn~$n$, Instead we will begin by considering, for each~$n$, the function that such a formula calculates. We will define this function by a list of properties. We will then prove that a function with these properties exists and is unique, ... ...
 ... ... @@ -1666,7 +1666,7 @@ a no response by showing that no solution exists.} Because the $i$-th row is unchanged, the operation $-k\rho_i+\rho_j$ returns the $j$-th row to its original state. (Observe that the $$i=j$$ conditino on the $k\rho_i+\rho_j$ (Observe that the $$i=j$$ condition on the $k\rho_i+\rho_j$ is needed, or else this could happen \begin{equation*} \begin{linsys}{2} ... ... @@ -2298,7 +2298,7 @@ operation extends beyond vectors to apply to any matrix.\index{matrix!scalar multiplication}\index{scalar multiplication!matrix} We write scalar multiplication either as $$r\cdot\vec{v}$$ or $$\vec{v}\cdot r$$, and soemtimes even omit the $\cdot$' symbol:~$r\vec{v}$. $$\vec{v}\cdot r$$, and sometimes even omit the $\cdot$' symbol:~$r\vec{v}$. (Do not refer to scalar multiplication as scalar product' because that name is for a different operation.) ... ...
 ... ... @@ -1260,7 +1260,7 @@ Not every vector in each is orthogonal to all vectors in the other. \end{answer} \item Describe the set of vectors in $$\Re^3$$ orthogonal to the one with entires $1$, $3$, and~$-1$. with entries $1$, $3$, and~$-1$. % \begin{equation*} % \colvec[r]{1 \\ 3 \\ -1} % \end{equation*} ... ...
 ... ... @@ -625,7 +625,7 @@ The Chain Rule multiplies the matrices. \end{center} We can easily verify that the resulting vector has unit length and forms an angle with the $x$-axis of $-\pi/12$, which is indeed a rotation clockwose of $\pi/4$ radians a rotation clockwise of $\pi/4$ radians since $(\pi/6)-(\pi/4)=-\pi/12$. \end{exparts} \end{answer} ... ...
 ... ... @@ -584,7 +584,7 @@ if and only if they have the same rank). B=\sequence{\colvec{1 \\ 0},\colvec{1 \\ 1}},\hspace{0.7em} D=\sequence{\colvec{2 \\ 0},\colvec{0 \\ -2}} \end{equation*} We will convert to the matrix representing~$t$ with resepct to $D,D$. We will convert to the matrix representing~$t$ with respect to $D,D$. \begin{exparts} \partsitem Draw the arrow diagram. \partsitem Give the matrix that represents the left and right ... ... @@ -1290,7 +1290,7 @@ Thus if $N$ were to be similar to a diagonal matrix then that matrix would have have at least one nonzero entry on its diagonal. The square of $N$ is the zero matrix. This imples that for any map~$n$ represented by $$N$$ This implies that for any map~$n$ represented by $$N$$ (with respect to some $B,B$) the composition $$\composed{n}{n}$$ is the zero map. This in turn implies that ... ... @@ -2974,7 +2974,7 @@ In contrast, \nearbyexample{ex:AlgMultDifferentThanGeoMult} shows that the eigenvalue $\lambda=2$ has algebraic multiplicity~$2$ but geometric multiplicity~$1$. For every transformation, each eigenvalue has geometric multiplicy greater than or equal to~$1$ by \nearbylemma{le:EigSpaceIsSubSp}. multiplicity greater than or equal to~$1$ by \nearbylemma{le:EigSpaceIsSubSp}. (And, an eigenvalue must have geometric multiplicity less than or equal to its algebraic multiplicity, although proving this is beyond our scope.) ... ... @@ -3905,7 +3905,7 @@ In the next section we study matrices that cannot be diagonalized. \partsitem Computing the eigenvalues gives $\lambda_1=3$, $\lambda_2=0$, and $\lambda_3=-4$. Thus the diagonalzation of the matrix is this. Thus the diagonalization of the matrix is this. \begin{equation*} \begin{mat} 3 &0 &0 \\ ... ...
 ... ... @@ -3032,7 +3032,7 @@ And, we shall see how to find the matrix that represents a map's inverse. (iii)~Setting $a=b=c=0$ in the calculation gives infinitely many solutions. Paramatrizing using the free variable~$z$ leads to this description Parametrizing using the free variable~$z$ leads to this description of the nullspace. \begin{equation*} \nullspace{h}=\set{\colvec{x \\ y \\ z}\suchthat ... ... @@ -3243,7 +3243,7 @@ And, we shall see how to find the matrix that represents a map's inverse. (where $$\map{h,\hat{h}}{V}{W}$$) then the dimension of the range space of $$h$$ equals the dimension of the range space of $$\hat{h}$$. Must these equal-dimensioned range spaces actually be the same? Must these equal-dimensional range spaces actually be the same? \begin{answer} No, the range spaces may differ. \nearbyexample{ex:CngBasesChgMap} shows this. ... ...
 ... ... @@ -2061,7 +2061,7 @@ Thus we can view any linear map as a projection. \quad \rep{\identity(x^2)}{B}=\colvec{-1 \\ 0 \\ 1} \end{equation*} These calcuations give $Q$. These calculations give $Q$. \begin{multline*} \rep{\identity( \begin{mat} ... ... @@ -2136,7 +2136,7 @@ Thus we can view any linear map as a projection. 0 &1 \end{mat} \end{equation*} These calcuations give $Q=\rep{\identity}{D,\hat{D}}$. These calculations give $Q=\rep{\identity}{D,\hat{D}}$. \begin{equation*} \rep{\identity(\colvec{0 \\ 1})}{\hat{D}}=\colvec{0 \\ 1} \quad ... ... @@ -2611,7 +2611,7 @@ Thus we can view any linear map as a projection. \begin{equation*} \begin{CD} V_{\wrt{B_1}} @>t>T> V_{\wrt{B_1}} \\ @V\text{\scriptsize$\identity$} VV @V\text{scriptsize$\identity$} VV \\ @V\text{\scriptsize$\identity$} VV @V\text{\scriptsize$\identity$} VV \\ V_{\wrt{B_2}} @>t>\hat{T}> V_{\wrt{B_2}} \end{CD} \end{equation*} ... ...
 ... ... @@ -65,34 +65,20 @@ With that as motivation the second chapter does vector spaces over the reals. In the schedule below this happens at the start of the third week. % Another example of the emphasis here on motivation and naturalness % is that the chapter on linear maps % does not begin with the definition of homomorphism. % Instead it begins with the definition of isomorphism, which % is natural\Dash students themselves % observe that some spaces are the same'' as others. % After that, % the next section takes the reasonable step of % isolating the operation-preservation idea % to define homomorphism. % This loses some mathematical slickness % but it is a good trade because it gives to students % a large gain in sensibility. A student progresses most in mathematics while doing exercises. A student progresses most in mathematics by doing exercises. The problem sets start with routine checks and range up to reasonably involved proofs. % Since instructors often assign about a dozen exercises I have aimed to typically put two dozen in each set, thereby giving a selection. In particular there is a good selection of the medium-difficult problems In particular there is a good number of the medium-difficult problems that stretch a learner, but not too far. At the high end, there are even a few that are puzzles At the high end, there are a few that are puzzles taken from various journals, competitions, or problems collections, which are marked with a \puzzlemark' (as part of the fun I have tried to keet the original wording). (as part of the fun I have worked to keep the original wording). That is, as with the rest of the book, the exercises are aimed to both build an ability at, ... ... @@ -145,9 +131,9 @@ See the web page. \noindent{\bf Acknowledgments.} A lesson of software development is that complex projects have bugs, and need a process for bug fixes. and need a process to fix them. I am grateful for reports from both instructors and students. I periodically issue revisions, and acknowledge in the book's source I periodically issue revisions and acknowledge in the book's source all of the reports that I use. My current contact information is on the web page above. ... ... @@ -179,11 +165,11 @@ While an experienced instructor knows what subjects and pace suit their class, this semester's timetable (graciously shared by George Ashline) may help you plan a sensible rate. It presumes Section~One.II, the elements of vectors. It presumes that you have already studied the material of Section~One.II, the elements of vectors. \begin{center} % George Ashline's \begin{tabular}{r|*{2}{p{\colwidth}}l} \textit{week} \multicolumn{1}{r}{\textit{week}} &\textit{Monday} &\textit{Wednesday} &\textit{Friday} \\ \hline ... ... @@ -203,19 +189,20 @@ presumes Section~One.II, the elements of vectors. 14 &Five.II.1, 2 &Five.II.2 &Five.II.3 \end{tabular} \end{center} As enrichment, you might pick one or two topics that appeal to you from the end of each chapter, such as the ones on As enrichment, you could pick one or two extra things that appeal to you, from the lab manual or from the Topics from the end of each chapter. I like the Topics on Voting Paradoxes, Geometry of Linear Maps, and Coupled Oscillators, or from the lab manual. (When I teach with this schedule as a target, I find that I often have room for a couple of days of extras.) Geometry of Linear Maps, and Coupled Oscillators. % (When I teach with this schedule as a target, I often have room % for a couple of days of extras.) You'll get more from these if you have access to software for calculations. I recommend \textit{Sage}, freely available from \url{http://sagemath.org}. In the table of contents I have marked some subsections as optional if I have marked a few subsections as optional if some instructors will pass over them in favor of spending more time elsewhere. Note that ... ...
 ... ... @@ -307,7 +307,7 @@ We shall write where $a$ is an integer that is positive if the remaining lists are on the left, where $a$ is negative if the lists are on the right, and zero if the cancellation was perfect. Similiarly we have integers $b$ and $c$ for the second and third rows, Similarly we have integers $b$ and $c$ for the second and third rows, which can each be positive, negative, or zero. Then the election is determined by this sum. ... ... @@ -517,7 +517,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $T>R>D$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $R>T>D$ \\ ... ... @@ -525,7 +525,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $D>T>R$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $T>D>R$ \\ ... ... @@ -533,7 +533,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $R>D>T$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \end{tabular} \end{center} ... ... @@ -676,7 +676,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $T>R>D$ \\ \textit{-cancelled} \textit{-canceled} \end{tabular} \\ \hline \begin{tabular}{l} $R>T>D$ \\ ... ... @@ -684,7 +684,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $D>T>R$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \begin{tabular}{l} $T>D>R$ \\ ... ... @@ -692,7 +692,7 @@ for his kind and illuminating discussions.)} \end{tabular} &\begin{tabular}{l} $R>D>T$ \\ \textit{-cancelled-} \textit{-canceled-} \end{tabular} \\ \hline \end{tabular} \end{center} ... ...
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 ... ... @@ -1360,7 +1360,7 @@ That will start in the next subsection. \end{equation*} Find a basis. \begin{answer} We have (using these odball operations with care) We have (using these oddball operations with care) \begin{multline*} \set{\colvec{1-y-z \\ y \\ z}\suchthat y,z\in\Re} =\set{\colvec{-y+1 \\ y \\ 0} ... ... @@ -1685,7 +1685,7 @@ In that section we could not show that these are We can show it now. The prior corollary proves that There are no, say, five-dimensional subspaces of three-space. Further, by nearbydefinition{df:Dimension} the dimension of every space Further, by \nearbydefinition{df:Dimension} the dimension of every space is a whole number so there are no subspaces of $\Re^3$ that are somehow $1.5$-dimensional, between lines and planes. Thus the list of subspaces that we gave is exhaustive; ... ...
 ... ... @@ -44,9 +44,9 @@ Write~$x(t)$ for the vertical motion over time and $\theta(t)$ for the rotational motion. Fix the coordinate system so that in rest position $x=0$ and~$\theta=0$, so that positive $x$'s are up, and so that positive $\theta$'s are counterclockwise when vewed from above. are counterclockwise when viewed from above. We start by modelling the motion of a mass on a We start by modeling the motion of a mass on a spring constrained to have no twist. This is simpler because there is only one motion, one degree of freedom. Put the mass in rest position and ... ... @@ -114,7 +114,7 @@ will make the spring longer. If at the same moment the vertical motion is that the spring is getting shorter, then superimposing the two could result in their almost cancelling. could result in their almost canceling. The bob ends up not moving vertically much at all, just twisting. With a properly adjusted device this could last for a number of seconds. ... ... @@ -410,7 +410,7 @@ See \cite{BergMarshall}. But we are setting it to zero so that doesn't matter. The roots are the same as in the Topic body. \end{answer} \item Buld a Wilberforce pendulum out of a Slinky Jr and a soup can. \item Build a Wilberforce pendulum out of a Slinky Jr and a soup can. You can drill holes in the can for bolts, either two or four of them, that you can use to adjust the moment of inertia of the can so the periods of vertical and rotational ... ...
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