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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; ... ...
 ... ... @@ -9,8 +9,8 @@ give a sense of how they arise. The first example is from Physics.\index{Physics problem} Suppose that we have three objects, one with a mass known to be 2~kg. We are asked to find the unknown masses. one with a mass known to be 2~kg and we want to find the unknown masses. Suppose further that experimentation with a meter stick produces these two balances. \begin{center} ... ... @@ -130,7 +130,7 @@ the system. We don't need guesswork or good luck; there is an algorithm that always works. This algorithm is called This algorithm is \definend{Gauss' method}\index{Gauss' method}% \index{system of linear equations!Gauss' method} (or \definend{Gaussian elimination}\index{Gaussian elimination}% ... ... @@ -229,7 +229,7 @@ can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding $$-1$$ times the row to itself has the effect of multiplying the row by $$0$$. Finally, swapping a row with itself is disallowed Finally, we disallow swapping a row with itself to make some results in the fourth chapter easier to state and remember, and also because it's pointless. ... ... @@ -306,7 +306,7 @@ When writing out the calculations, we will abbreviate row $$i$$' by $$\rho_i$$'. For instance, we will denote a row combination operation by $$k\rho_i+\rho_j$$, with the row that is changed written second. with the row that changes written second. To save writing we will often combine addition steps when they use the same $\rho_i$; see the next example. ... ... @@ -373,7 +373,7 @@ chapter, Gauss' method gives this. \end{linsys} \end{eqnarray*} So $$c=4$$, and back-substitution gives that $$h=1$$. (The Chemistry problem is solved later.) (We will solve the Chemistry problem later.) \end{example} \begin{example} ... ... @@ -464,7 +464,7 @@ Back-substitution gives $$w=1$$, $$z=2$$ , $$y=-1$$, and $$x=-1$$. The row rescaling operation is not needed, strictly speaking, to solve linear systems. It is included here because we will use it But we will use it later in this chapter as part of a variation on Gauss' method, the Gauss-Jordan method. ... ... @@ -550,7 +550,7 @@ that is not what causes the inconsistency\Dash \nearbyexample{ex:MoreEqsThanUnks} has more equations than unknowns and yet is consistent. Nor is having more equations than unknowns necessary for inconsistency, as is illustrated by this inconsistent system with the inconsistency, as we see with this inconsistent system that has the same number of equations as unknowns. \begin{eqnarray*} \begin{linsys}{2} ... ... @@ -594,19 +594,19 @@ because we do not get a contradictory equation. \end{example} Don't be fooled by the final system in that example. A $0=0$' equation does not signal that a system has many solutions. A $0=0$' equation it not the signal that a system has many solutions. \begin{example} \label{ex:NoZerosInfManySols} The absence of a $$0=0$$' does not keep a system from having many different solutions. This system is in echelon form has no $0=0$', but has infinitely many solutions. \begin{equation*} \begin{linsys}{3} x &+ &y &+ &z &= &0 \\ & &y &+ &z &= &0 \end{linsys} \end{equation*} has no $0=0$', but has infinitely many solutions. Some solutions are:~$(0,1,-1)$, $(0,1/2,-1/2)$, $(0,0,0)$, and $(0,-\pi,\pi)$. There are infinitely many solutions because ... ... @@ -618,7 +618,7 @@ many solutions. \nearbyexample{ex:MoreEqsThanUnks} shows that. So does this system, which does not have any solutions at all despite that when it is brought to echelon form it has a $0=0$' row. in echelon form it has a $0=0$' row. \begin{eqnarray*} \begin{linsys}{3} 2x & & &- &2z &= &6 \\ ... ... @@ -689,7 +689,7 @@ a no response by showing that no solution exists.} \end{linsys}\end{exparts*} \begin{answer} 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 ... ... @@ -848,15 +848,16 @@ a no response by showing that no solution exists.} \end{exparts} \end{answer} \recommended \item There are methods for solving linear systems other We can solve linear systems by methods other than Gauss' method. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. That step is repeated until there is an equation with only one Then we repeat that step until there is an equation with only one variable. From that, the first number in the solution is derived, and then back-substitution can be done. From that we get the first number in the solution and then we get the rest with back-substitution. This method takes longer than Gauss' method, since it involves more arithmetic operations, and is also more likely to lead to errors. ... ... @@ -1023,7 +1024,7 @@ a no response by showing that no solution exists.} a balance the reaction problem\Dash have infinitely many solutions? \begin{answer} 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). ... ... @@ -1097,7 +1098,7 @@ a no response by showing that no solution exists.} Gauss' method works by combining the equations in a system to make new equations. \begin{exparts} \partsitem Can the equation $$3x-2y=5$$ be derived by a sequence of \partsitem Can we derive the equation $$3x-2y=5$$ by a sequence of Gaussian reduction steps from the equations in this system? \begin{equation*} \begin{linsys}{2} ... ... @@ -1105,7 +1106,7 @@ a no response by showing that no solution exists.} 4x &- &y &= &6 \end{linsys} \end{equation*} \partsitem Can the equation $$5x-3y=2$$ be derived by a sequence of \partsitem Can we derive the equation $$5x-3y=2$$ with a sequence of Gaussian reduction steps from the equations in this system? \begin{equation*} \begin{linsys}{2} ... ... @@ -1113,7 +1114,7 @@ a no response by showing that no solution exists.} 3x &+ &y &= &4 \end{linsys} \end{equation*} \partsitem Can the equation $$6x-9y+5z=-2$$ be derived \partsitem Can we derive $$6x-9y+5z=-2$$ by a sequence of Gaussian reduction steps from the equations in the system? \begin{equation*} ... ... @@ -1128,7 +1129,7 @@ a no response by showing that no solution exists.} \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. ... ... @@ -1345,7 +1346,7 @@ a no response by showing that no solution exists.} 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*} ... ... @@ -1390,7 +1391,7 @@ a no response by showing that no solution exists.} \recommended \item Are any of the operations used in Gauss' method redundant? That is, can any of the operations be made from a combination That is, can we make any of the operations from a combination of the others? \begin{answer} Yes. ... ... @@ -1409,7 +1410,7 @@ a no response by showing that no solution exists.}S_1\rightarrow S_2$then there is a row operation to go back$S_2\rightarrow S_1$. \begin{answer} Swapping rows is reversed by swapping back. Reverse a row swap by swapping back. \begin{eqnarray*} \begin{linsys}{3} a_{1,1}x_1 &+ &\cdots &+ &a_{1,n}x_n &= &d_1 \\ ... ... @@ -1664,7 +1665,7 @@ a no response by showing that no solution exists.} \subsection{Describing the Solution Set} A linear system with a unique solution has a solution set with one element. A linear system with no solution has a solution set that is empty. % In these cases the solution set is easy to describe. In these cases the solution set is easy to describe. Solution sets are a challenge to describe only when they contain many elements. \begin{example} ... ... @@ -1692,9 +1693,9 @@ not all of the variables are leading variables. \nearbytheorem{th:GaussMethod} shows that a triple$(x,y,z)$satisfies the first system if and only if it satisfies the third. Thus the solution set Thus we can describe the solution set$\set{(x,y,z)\suchthat\text{$2x+z=3$ and $x-y-z=1$ and $3x-y=4$}}$can also be described as as$\set{(x,y,z)\suchthat\text{$2x+z=3$ and $-y-3z/2=-1/2$}}$. However, this second description is not optimal. It has two equations instead of three ... ... @@ -1707,7 +1708,7 @@ The second equation gives$y=(1/2)-(3/2)z$and the first equation gives$x=(3/2)-(1/2)z$. Thus the solution set can be described as Thus the solution set is this. \begin{equation*} \set{ (x,y,z)= ((3/2)-(1/2)z,(1/2)-(3/2)z,z)\suchthat z\in\Re} ... ... @@ -1819,7 +1820,7 @@ In the prior example$y$and~$w$are free because in the echelon form system they do not lead while they are parameters because of how they are used in the solution set description. we use them to describe the set of solutions. Had we instead rewritten the second equation as$w=2/3-(1/3)z$then the free variables would still be$y$and~$w$but the parameters ... ... @@ -1854,7 +1855,7 @@ This is another system with infinitely many solutions. The leading variables are $$x$$, $$y$$, and $$w$$. The variable $$z$$ is free. Notice that, although there are infinitely many solutions, the value of the variable$w$is fixed at$-1$. solutions, the value of$w$doesn't vary but is constant at$-1$. To parametrize, write $$w$$ in terms of $$z$$ with $$w=-1+0z$$. Then $$y=(1/4)z$$. Substitute for $$y$$ in the first ... ... @@ -1894,7 +1895,7 @@ has$2$~rows and$3$~columns and so is a $$\nbym{2}{3}$$ matrix. (read that aloud as two-by-three''); the number of rows is always first. Entries are named by the corresponding lower-case letter We denote entries with the corresponding lower-case letter so that$a_{i,j}$is the number in row~$i$and column~$j$of the array. The entry in the second row and first column is $$a_{2,1}=3$$. ... ... @@ -1977,7 +1978,7 @@ We will write them vertically, in one-column matrices. For instance, the top line says that $$x=2-2z+2w$$ and the second line says that $$y= -1+z-w$$. The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. picture the solution sets. \begin{definition} A \definend{vector}\index{vector} ... ... @@ -2061,7 +2062,7 @@ matrix.\index{matrix!scalar multiplication}\index{scalar multiplication!matrix} We write scalar multiplication in either order, as $$r\cdot\vec{v}$$ or $$\vec{v}\cdot r$$, or without the $\cdot$' symbol:~$r\vec{v}$. (Do not refer to scalar multiplication as scalar product' because that name is used for a different operation.) as scalar product' because we use that name for a different operation.) \begin{example} \begin{equation*} ... ... @@ -2127,7 +2128,7 @@ We write that in vector form. Note how well vector notation sets off the coefficients of each parameter. For instance, the third row of the vector form shows plainly that if $$u$$ is held fixed then $$z$$ increases three times as fast as $$w$$. fixed then $$z$$ increases three times as fast as $$w$$. Another thing shown plainly is that setting both $$w$$ and $$u$$ to zero gives that this vector \begin{equation*} ... ... @@ -2200,7 +2201,7 @@ would tell us about the size of solution sets. It will also help us understand the geometry of the solution sets. Many questions arise from our observation that Gauss' method can be done in Many questions arise from our observation that we can do Gauss' method in more than one way (for instance, when swapping rows we may have a choice of more than one row). \nearbytheorem{th:GaussMethod} says that we must get the same solution set ... ... @@ -3203,7 +3204,7 @@ particular solution and so there are no sums of that form. \begin{theorem} \label{th:GenEqPartPlusHomo} Any linear system's solution set can be described as solution set has the form \begin{equation*} \set{\vec{p}+c_1\vec{\beta}_1+\,\cdots\,+c_k\vec{\beta}_k \suchthat c_1,\,\ldots\,,c_k\in\Re} ... ... @@ -3273,8 +3274,8 @@ with the reduction of the associated homogeneous system. \end{linsys} \end{eqnarray*} Obviously the two reductions go in the same way. We can study how linear systems are reduced by instead studying how the associated homogeneous systems are reduced. We can study how to reduce a linear systems by instead studying how to reduce the associated homogeneous system. \end{example} Studying the associated homogeneous system has a great advantage over ... ... @@ -3434,8 +3435,7 @@ etc. \begin{proof} Apply Gauss' method to get to echelon form. We may get some $$0=0$$' equations that we can drop from the system We may get some $$0=0$$' equations (if the entire system consists of such equations then the theorem is trivially true) but ... ... @@ -3511,12 +3511,13 @@ In particular, we need this in the case where a homogeneous system has a unique solution. Then the homogeneous case fits the pattern of the other solution sets: in the proof above, the solution set is derived by taking the $$c$$'s to be the free variables we derive the solution set by taking the $$c$$'s to be the free variables and if there is a unique solution then there are no free variables. The proof incidentally shows, as discussed after \nearbyexample{ex:Parametrize1}, that solution sets can always be parametrized using the free variables. as discussed after \nearbyexample{ex:Parametrize1}, that we can always parametrize solution sets using the free variables. The next lemma finishes the proof of \nearbytheorem{th:GenEqPartPlusHomo} by considering the particular solution part of the ... ... @@ -3830,7 +3831,7 @@ has either no solutions or else has infinitely many, as with these. The word singular means departing from general expectation'' (people often, naively, expect that systems with the same number of variables as equations will have a unique solution). Thus, it can be thought of as connoting Thus, we can think of it as connoting troublesome,'' or at least not ideal.'' (That singular' applies those systems that do not have one solution is ironic, but it is the standard term.) ... ... @@ -4182,8 +4183,8 @@ of Gauss' method itself in the rest of this chapter. \end{exparts} \end{answer} \recommended \item \nearbylemma{th:GenEqPartHomo} says that any particular solution may be used for$\vec{p}$. \nearbylemma{th:GenEqPartHomo} says that we can use any particular solution for$\vec{p}$. Find, if possible, a general solution to this system \begin{equation*} \begin{linsys}{4} ... ... @@ -4445,8 +4446,8 @@ of Gauss' method itself in the rest of this chapter. \nearbylemma{le:HomoSltnSpanVecs}, what happens if there are no non-$$0=0$$' equations? \begin{answer} 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} ... ... @@ -4516,7 +4517,7 @@ of Gauss' method itself in the rest of this chapter. 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. ... ... @@ -4735,13 +4736,13 @@ we fix$x$and$y$, we can solve for appropriate$m$,$n$, and$p$: & & & &p &= &y\hfill \end{linsys} \end{equation*} shows that that any shows that we can express any \begin{equation*} \vec{v}= \colvec[r]{1 \\ 2 \\ 0}x+ \colvec[r]{1 \\ 0 \\ 1}y \end{equation*} can be expressed as a member of $$R$$ with as a member of $$R$$ with $$m=x$$, $$n=2x$$, and $$p=y$$: \begin{equation*} \vec{v}= ... ... @@ -4951,7 +4952,8 @@ is in $$R$$ but not in $$P$$. =\colvec[r]{1 \\ 8} +\colvec[r]{0 \\ -1}(6-3t) \end{equation*} and so any vector in the form for $$S_1$$ can be stated in the form and so we can state any vector in the form for $$S_1$$ can also in the form needed for inclusion in $$S_2$$. For $$S_2\subseteq S_1$$, we look for $$t$$ so that ... ... @@ -5051,7 +5053,7 @@ is in $$R$$ but not in $$P$$. =\colvec[r]{1 \\ 3 \\ 1}(2m) +\colvec[r]{2 \\ 1 \\ 5}(m-2n) \end{equation*} and so any member of $$S_2$$ can be expressed in the form needed for and so we can express any member of $$S_2$$ in the form needed for $$S_1$$. \partsitem These sets are equal. ... ...  ... ... @@ -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} } ... ...