Commit 7b8c49f1 by Jim Hefferon

### gr3 edits

parent 1223db30
No preview for this file type
 ... ... @@ -6496,9 +6496,9 @@ \partsitem Yes this is a basis. The relation \begin{equation*} c_1\colvec{1 \\ 2 \\ 3} +c_2\colvec{3 \\ 2 \\ 1} +c_3\colvec{0 \\ 0 \\ 1} c_1\colvec[r]{1 \\ 2 \\ 3} +c_2\colvec[r]{3 \\ 2 \\ 1} +c_3\colvec[r]{0 \\ 0 \\ 1} =\colvec{x \\ y \\ z} \end{equation*} gives ... ... @@ -6522,8 +6522,8 @@ \partsitem This is not a basis. Setting it up as in the prior item \begin{equation*} c_1\colvec{1 \\ 2 \\ 3} +c_2\colvec{3 \\ 2 \\ 1} c_1\colvec[r]{1 \\ 2 \\ 3} +c_2\colvec[r]{3 \\ 2 \\ 1} =\colvec{x \\ y \\ z} \end{equation*} gives a linear system whose solution ... ... @@ -6594,9 +6594,9 @@ \begin{exparts} \partsitem We solve \begin{equation*} c_1\colvec{1 \\ 1} +c_2\colvec{-1 \\ 1} =\colvec{1 \\ 2} c_1\colvec[r]{1 \\ 1} +c_2\colvec[r]{-1 \\ 1} =\colvec[r]{1 \\ 2} \end{equation*} with \begin{equation*} ... ... @@ -6613,7 +6613,7 @@ and conclude that $$c_2=1/2$$ and so $$c_1=3/2$$. Thus, the representation is this. \begin{equation*} \rep{\colvec{1 \\ 2}}{B}=\colvec{3/2 \\ 1/2}_B \rep{\colvec[r]{1 \\ 2}}{B}=\colvec[r]{3/2 \\ 1/2}_B \end{equation*} \partsitem The relationship $c_1\cdot(1)+c_2\cdot(1+x)+c_3\cdot(1+x+x^2)+c_4\cdot(1+x+x^2+x^3) ... ... @@ -6621,10 +6621,10 @@ is easily solved by eye to give that$c_4=1$,$c_3=0$,$c_2=-1$, and$c_1=0$. \begin{equation*} \rep{x^2+x^3}{D}=\colvec{0 \\ -1 \\ 0 \\ 1}_D \rep{x^2+x^3}{D}=\colvec[r]{0 \\ -1 \\ 0 \\ 1}_D \end{equation*} \partsitem $$\rep{\colvec{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4} =\colvec{0 \\ -1 \\ 0 \\ 1}_{\stdbasis_4}$$ \partsitem $$\rep[r]{\colvec{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4} =\colvec[r]{0 \\ -1 \\ 0 \\ 1}_{\stdbasis_4}$$ \end{exparts} \end{ans} ... ... @@ -6655,15 +6655,15 @@ \begin{equation*} \set{\colvec{4x_2-3x_3+x_4 \\ x_2 \\ x_3 \\ x_4} \suchthat x_2,x_3,x_4\in\Re} =\set{x_2\colvec{4 \\ 1 \\ 0 \\ 0} +x_3\colvec{-3 \\ 0 \\ 1 \\ 0} +x_4\colvec{1 \\ 0 \\ 0 \\ 1} \suchthat x_2,x_3,x_4\in\Re} =\set{x_2\colvec[r]{4 \\ 1 \\ 0 \\ 0} +x_3\colvec[r]{-3 \\ 0 \\ 1 \\ 0} +x_4\colvec[r]{1 \\ 0 \\ 0 \\ 1} \suchthat x_2,x_3,x_4\in\Re} \end{equation*} and so the obvious candidate for the basis is this. \begin{equation*} \sequence{\colvec{4 \\ 1 \\ 0 \\ 0}, \colvec{-3 \\ 0 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 0 \\ 1} } \sequence{\colvec[r]{4 \\ 1 \\ 0 \\ 0}, \colvec[r]{-3 \\ 0 \\ 1 \\ 0}, \colvec[r]{1 \\ 0 \\ 0 \\ 1} } \end{equation*} We've shown that this spans the space, and showing it is also linearly independent is routine. ... ... @@ -7000,10 +7000,10 @@ Here is a subset of$\Re^2$that is not a basis, and two different linear combinations of its elements that sum to the same vector. \begin{equation*} \set{\colvec{1 \\ 2},\colvec{2 \\ 4}} \set{\colvec[r]{1 \\ 2},\colvec[r]{2 \\ 4}} \qquad 2\cdot\colvec{1 \\ 2}+0\cdot\colvec{2 \\ 4} =0\cdot\colvec{1 \\ 2}+1\cdot\colvec{2 \\ 4} 2\cdot\colvec[r]{1 \\ 2}+0\cdot\colvec[r]{2 \\ 4} =0\cdot\colvec[r]{1 \\ 2}+1\cdot\colvec[r]{2 \\ 4} \end{equation*} Thus, when a subset is not a basis, it can be the case that its linear combinations are not unique. ... ... @@ -7012,13 +7012,13 @@ combinations must be not unique. For instance, this set \begin{equation*} \set{\colvec{1 \\ 2}} \set{\colvec[r]{1 \\ 2}} \end{equation*} does have the property that \begin{equation*} c_1\cdot\colvec{1 \\ 2} c_1\cdot\colvec[r]{1 \\ 2} = c_2\cdot\colvec{1 \\ 2} c_2\cdot\colvec[r]{1 \\ 2} \end{equation*} implies that$c_1=c_2$. The idea here is that this subset fails to be a basis because it fails ... ... @@ -7107,7 +7107,7 @@ +c_2\colvec{x_2 \\ y_2 \\ z_2} +c_3\colvec{x_3 \\ y_3 \\ z_3} +c_4\colvec{x_4 \\ y_4 \\ z_4} =\colvec{0 \\ 0 \\ 0} =\colvec[r]{0 \\ 0 \\ 0} \end{equation*} gives rise to a linear system \begin{equation*} ... ... @@ -7170,12 +7170,12 @@ \end{equation*} and so a natural candidate for a basis is this. \begin{equation*} \sequence{\colvec{0 \\ 1 \\ 0},\colvec{0 \\ 0 \\ 1}} \sequence{\colvec[r]{0 \\ 1 \\ 0},\colvec[r]{0 \\ 0 \\ 1}} \end{equation*} To check linear independence we set up \begin{equation*} c_1\colvec{0 \\ 1 \\ 0}+c_2\colvec{0 \\ 0 \\ 1} =\colvec{1 \\ 0 \\ 0} c_1\colvec[r]{0 \\ 1 \\ 0}+c_2\colvec[r]{0 \\ 0 \\ 1} =\colvec[r]{1 \\ 0 \\ 0} \end{equation*} (the vector on the right is the zero object in this space). That yields the linear system ... ... @@ -7191,12 +7191,12 @@ \end{ans} \subsection{Subsection Two.III.2: Dimension} \begin{ans}{Two.III.2.14} \begin{ans}{Two.III.2.15} One basis is $$\sequence{1,x,x^2}$$, and so the dimension is three. \end{ans} \begin{ans}{Two.III.2.15} \begin{ans}{Two.III.2.16} The solution set is \begin{equation*} \set{\colvec{4x_2-3x_3+x_4 \\ x_2 \\ x_3 \\ x_4} ... ... @@ -7204,15 +7204,15 @@ \end{equation*} so a natural basis is this \begin{equation*} \sequence{\colvec{4 \\ 1 \\ 0 \\ 0}, \colvec{-3 \\ 0 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 0 \\ 1} } \sequence{\colvec[r]{4 \\ 1 \\ 0 \\ 0}, \colvec[r]{-3 \\ 0 \\ 1 \\ 0}, \colvec[r]{1 \\ 0 \\ 0 \\ 1} } \end{equation*} (checking linear independence is easy). Thus the dimension is three. \end{ans} \begin{ans}{Two.III.2.16} \begin{ans}{Two.III.2.17} For this space \begin{equation*} \set{\begin{mat} ... ... @@ -7252,7 +7252,7 @@ The dimension is four. \end{ans} \begin{ans}{Two.III.2.17} \begin{ans}{Two.III.2.18} \begin{exparts} \partsitem As in the prior exercise, the space$\matspace_{\nbyn{2}}$of matrices without restriction has this basis ... ... @@ -7345,7 +7345,7 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.18} \begin{ans}{Two.III.2.19} The bases for these spaces are developed in the answer set of the prior subsection. \begin{exparts} ... ... @@ -7361,7 +7361,7 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.19} \begin{ans}{Two.III.2.20} First recall that$\cos2\theta=\cos^2\theta-\sin^2\theta$, and so deletion of$\cos2\theta$from this set leaves the span unchanged. What's left, the set ... ... @@ -7375,7 +7375,7 @@ That shows that the span is a dimension three vector space. \end{ans} \begin{ans}{Two.III.2.20} \begin{ans}{Two.III.2.21} Here is a basis \begin{equation*} \sequence{(1+0i,0+0i,\dots,0+0i),\, ... ... @@ -7384,7 +7384,7 @@ and so the dimension is $$2\cdot 47=94$$. \end{ans} \begin{ans}{Two.III.2.21} \begin{ans}{Two.III.2.22} A basis is \begin{equation*} \sequence{ ... ... @@ -7408,7 +7408,7 @@ and thus the dimension is $$3\cdot 5=15$$. \end{ans} \begin{ans}{Two.III.2.22} \begin{ans}{Two.III.2.23} In a four-dimensional space a set of four vectors is linearly independent if and only if it spans the space. The form of these vectors makes linear independence easy to show ... ... @@ -7416,7 +7416,7 @@ third components, etc.). \end{ans} \begin{ans}{Two.III.2.23} \begin{ans}{Two.III.2.24} \begin{exparts} \partsitem The diagram for$\polyspace_2$has four levels. The top level has the only three-dimensional subspace, ... ... @@ -7432,7 +7432,7 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.24} \begin{ans}{Two.III.2.25} \begin{exparts*} \partsitem One \partsitem Two ... ... @@ -7440,7 +7440,7 @@ \end{exparts*} \end{ans} \begin{ans}{Two.III.2.25} \begin{ans}{Two.III.2.26} We need only produce an infinite linearly independent set. One is $$\sequence{f_1,f_2,\ldots}$$ where $$\map{f_i}{\Re}{\Re}$$ is ... ... @@ -7453,7 +7453,7 @@ the function that has value$1$only at$x=i$. \end{ans} \begin{ans}{Two.III.2.26} \begin{ans}{Two.III.2.27} A function is a set of ordered pairs$(x,f(x))$. So there is only one function with an empty domain, namely the empty set. ... ... @@ -7461,11 +7461,11 @@ and has dimension zero. \end{ans} \begin{ans}{Two.III.2.27} \begin{ans}{Two.III.2.28} Apply \nearbycorollary{cor:NoLiSetGreatDim}. \end{ans} \begin{ans}{Two.III.2.28} \begin{ans}{Two.III.2.29} A plane has the form$\set{\vec{p}+t_1\vec{v}_1+t_2\vec{v}_2\suchthat t_1,t_2\in\Re}$. (The first chapter also calls this a `$2$-flat', and contains ... ... @@ -7488,7 +7488,7 @@ any sequence made from the set's elements is a basis). \end{ans} \begin{ans}{Two.III.2.29} \begin{ans}{Two.III.2.30} Let the space$V$be finite dimensional. Let$S$be a subspace of$V$. \begin{exparts} ... ... @@ -7503,12 +7503,12 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.30} \begin{ans}{Two.III.2.31} It ensures that we exhaust the $$\vec{\beta}$$'s. That is, it justifies the first sentence of the last paragraph. \end{ans} \begin{ans}{Two.III.2.31} \begin{ans}{Two.III.2.32} Let $$B_U$$ be a basis for $$U$$ and let $$B_W$$ be a basis for $$W$$. The set $$B_U\union B_W$$ is linearly dependent as it is a ... ... @@ -7523,7 +7523,7 @@ intersection. \end{ans} \begin{ans}{Two.III.2.32} \begin{ans}{Two.III.2.33} First, note that a set is a basis for some space if and only if it is linearly independent, because in that case it is a basis for its own span. ... ... @@ -7554,9 +7554,9 @@ It is not, however, a basis for the intersection of the spaces. For instance, these are bases for $$\Re^2$$: \begin{equation*} B_1=\sequence{\colvec{1 \\ 0},\colvec{0 \\ 1}} B_1=\sequence{\colvec[r]{1 \\ 0},\colvec[r]{0 \\ 1}} \quad\text{and}\quad B_2=\sequence{\colvec{2 \\ 0},\colvec{0 \\ 2}} B_2=\sequence[r]{\colvec{2 \\ 0},\colvec[r]{0 \\ 2}} \end{equation*} and $$\Re^2\intersection\Re^2=\Re^2$$, but $$B_1\intersection B_2$$ is empty. ... ... @@ -7564,9 +7564,9 @@ for a subset of the intersection of the spaces. \partsitem The$\cup$of bases need not be a basis: in $$\Re^2$$ \begin{equation*} B_1=\sequence{\colvec{1 \\ 0},\colvec{1 \\ 1}} B_1=\sequence{\colvec[r]{1 \\ 0},\colvec[r]{1 \\ 1}} \quad\text{and}\quad B_2=\sequence{\colvec{1 \\ 0},\colvec{0 \\ 2}} B_2=\sequence{\colvec[r]{1 \\ 0},\colvec[r]{0 \\ 2}} \end{equation*} $$B_1\union B_2$$ is not linearly independent. A necessary and sufficient condition for a$\cup$of two bases ... ... @@ -7583,7 +7583,7 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.33} \begin{ans}{Two.III.2.34} \begin{exparts} \partsitem A basis for $$U$$ is a linearly independent set in $$W$$ ... ... @@ -7608,7 +7608,7 @@ \end{exparts} \end{ans} \begin{ans}{Two.III.2.34} \begin{ans}{Two.III.2.35} The possibilities for the dimension of$V$are$0$,$1$,$n-1$, and$n$. ... ... @@ -7633,7 +7633,7 @@ Therefore, \begin{equation*} \frac{1}{y-x}\bigl(\sigma_1(\vec{v})-\sigma_2(\vec{v})\bigr)= \colvec{-1 \\ 1 \\ 0 \\ \vdots \\ 0} \colvec[r]{-1 \\ 1 \\ 0 \\ \vdotswithin{-1} \\ 0} \end{equation*} is in$V$. That is,$\vec{e}_2-\vec{e}_1\in V$, where$\vec{e}_1$,$\vec{e}_2$, ... ... @@ -7677,7 +7677,7 @@ 3 &8 \end{mat} \) \partsitem $$\rowvec{0 &0 &0}$$ \partsitem $$\colvec{-1 \\ -2}$$ \partsitem $$\colvec[r]{-1 \\ -2}$$ \end{exparts*} \end{ans} ... ... @@ -7726,9 +7726,9 @@ \partsitem No. To see if there are $$c_1,c_2\in\Re$$ such that \begin{equation*} c_1\colvec{1 \\ 1} +c_2\colvec{1 \\ 1} =\colvec{1 \\ 3} c_1\colvec[r]{1 \\ 1} +c_2\colvec[r]{1 \\ 1} =\colvec[r]{1 \\ 3} \end{equation*} we can use Gauss' method on the resulting linear system. \begin{equation*} ... ... @@ -7746,10 +7746,10 @@ \partsitem Yes. From this relationship \begin{equation*} c_1\colvec{1 \\ 2 \\ 1} +c_2\colvec{3 \\ 0 \\ -3} +c_3\colvec{1 \\ 4 \\ 3} =\colvec{1 \\ 0 \\ 0} c_1\colvec[r]{1 \\ 2 \\ 1} +c_2\colvec[r]{3 \\ 0 \\ -3} +c_3\colvec[r]{1 \\ 4 \\ 3} =\colvec[r]{1 \\ 0 \\ 0} \end{equation*} we get a linear system that, when Gauss' method is applied, \begin{equation*} ... ... @@ -7908,8 +7908,8 @@ \end{equation*} and then transposing back gives this basis. \begin{equation*} \sequence{\colvec{1 \\ 2 \\ 1}, \colvec{0 \\ -5 \\ -4} } \sequence{\colvec[r]{1 \\ 2 \\ 1}, \colvec[r]{0 \\ -5 \\ -4} } \end{equation*} \partsitem Notice first that the surrounding space is given as$\polyspace_3$, not$\polyspace_2$. ... ... @@ -8042,7 +8042,7 @@ all multiples of $$\rowvec{1 &3}$$ while the column space consists of multiples of \begin{equation*} \colvec{1 \\ 2} \colvec[r]{1 \\ 2} \end{equation*} so we also cannot argue that the two spaces must be simply transposes of each other. ... ... @@ -8107,9 +8107,9 @@ of coefficients. To find a basis for the column space, \begin{equation*} \set{c_1\colvec{3 \\ 1 \\ 2} +c_2\colvec{2 \\ 0 \\ 2} +c_3\colvec{4 \\ -1 \\ 5} \set{c_1\colvec[r]{3 \\ 1 \\ 2} +c_2\colvec[r]{2 \\ 0 \\ 2} +c_3\colvec[r]{4 \\ -1 \\ 5} \suchthat c_1,c_2,c_3\in\Re} \end{equation*} we take the three vectors from the spanning set, transpose, reduce, ... ... @@ -8129,8 +8129,8 @@ \end{equation*} and transpose back to get this. \begin{equation*} \sequence{ \colvec{3 \\ 1 \\ 2}, \colvec{0 \\ -2/3 \\ 2/3} } \sequence{ \colvec[r]{3 \\ 1 \\ 2}, \colvec[r]{0 \\ -2/3 \\ 2/3} } \end{equation*} \end{ans} ... ... @@ -8140,22 +8140,22 @@ \trans{(rA+sB)} &=\trans{\begin{mat} ra_{1,1}+sb_{1,1} &\ldots &ra_{1,n}+sb_{1,n} \\ \vdots & &\vdots \\ &\vdots \\ ra_{m,1}+sb_{m,1} &\ldots &ra_{m,n}+sb_{m,n} \end{mat} } \\ &=\begin{mat} ra_{1,1}+sb_{1,1} &\ldots &ra_{m,1}+sb_{m,1} \\ \vdots \\ &\vdots \\ ra_{1,n}+sb_{1,n} &\ldots &ra_{m,n}+sb_{m,n} \end{mat} \\ &=\begin{mat} ra_{1,1} &\ldots &ra_{m,1} \\ \vdots \\ &\vdots \\ ra_{1,n} &\ldots &ra_{m,n} \end{mat} +\begin{mat} sb_{1,1} &\ldots &sb_{m,1} \\ \vdots \\ &\vdots \\ sb_{1,n} &\ldots &sb_{m,n} \end{mat} \\ &=r\trans{A}+s\trans{B} ... ... @@ -8290,9 +8290,9 @@ we can apply it here: \begin{equation*} \colspace{A}=\spanof{\set{ \colvec{1 \\ 4}, \colvec{2 \\ 5}, \colvec{3 \\ 6} } } \colvec[r]{1 \\ 4}, \colvec[r]{2 \\ 5}, \colvec[r]{3 \\ 6} } } \end{equation*} while \begin{equation*} ... ... @@ -8462,11 +8462,11 @@ To see that any vector in the plane is a combination of vectors from these parts, consider this relationship. \begin{equation*} \colvec{a \\ b}=c_1\colvec{1 \\ 1}+c_2\colvec{1 \\ 1.1} \colvec{a \\ b}=c_1\colvec[r]{1 \\ 1}+c_2\colvec[r]{1 \\ 1.1} \end{equation*} We could now simply note that the set \begin{equation*} \set{\colvec{1 \\ 1},\colvec{1 \\ 1.1}} \set{\colvec[r]{1 \\ 1},\colvec[r]{1 \\ 1.1}} \end{equation*} is a basis for the space (because it is clearly linearly independent, and has size two in$\Re^2$), and thus ther is one and ... ... @@ -8495,7 +8495,7 @@ \partsitem Yes. Each vector in the plane is a sum in this way \begin{equation*} \colvec{x \\ y}=\colvec{x \\ y}+\colvec{0 \\ 0} \colvec{x \\ y}=\colvec{x \\ y}+\colvec[r]{0 \\ 0} \end{equation*} and the intersection of the two subspaces is trivial. \partsitem No. No preview for this file type  ... ... @@ -505,7 +505,7 @@ as in \nearbyexample{ex:RealVecSpaces}. sequence, or that it means a function from$\N$to$\Re$.) \end{example} \begin{example} \begin{example} \label{ex:PolysOfAllFiniteDegrees} The set of polynomials with real coefficients \begin{equation*} \set{ a_0+a_1x+\cdots+a_nx^n\suchthat n\in\N ... ...  ... ... @@ -62,7 +62,7 @@ and it spans $$\Re^2$$. \begin{example} This basis for $$\Re^2$$ \begin{equation*} \sequence{\colvec{1 \\ 1},\colvec{2 \\ 4}} \sequence{\colvec[r]{1 \\ 1},\colvec[r]{2 \\ 4}} \end{equation*} differs from the prior one because the vectors are in a different order. The verification that it is a basis is just as in the prior example. ... ... @@ -72,7 +72,7 @@ The verification that it is a basis is just as in the prior example. The space $$\Re^2$$ has many bases. Another one is this. \begin{equation*} \sequence{ \colvec{1 \\ 0},\colvec{0 \\ 1} } \sequence{ \colvec[r]{1 \\ 0},\colvec[r]{0 \\ 1} } \end{equation*} The verification is easy. \end{example} ... ... @@ -81,10 +81,10 @@ The verification is easy. For any $$\Re^n$$, \begin{equation*} \stdbasis_n=\sequence{ \colvec{1 \\ 0 \\ \vdots \\ 0}, \colvec{0 \\ 1 \\ \vdots \\ 0}, \colvec[r]{1 \\ 0 \\ \vdotswithin{0} \\ 0}, \colvec[r]{0 \\ 1 \\ \vdotswithin{0} \\ 0}, \dots,\, \colvec{0 \\ 0 \\ \vdots \\ 1}} \colvec[r]{0 \\ 0 \\ \vdotswithin{0} \\ 1}} \end{equation*} is the \definend{standard}\index{standard basis}% \index{basis!standard} (or \definend{natural}) basis. ... ... @@ -92,12 +92,12 @@ We denote these vectors by $$\vec{e}_1,\dots,\vec{e}_n$$. \end{definition} \noindent (Calculus books refer to$\Re^2$'s standard basis vectors Calculus books refer to$\Re^2$'s standard basis vectors $$\vec{\imath}$$ and $$\vec{\jmath}$$ instead of$\vec{e}_1$and$\vec{e}_2$, and they refer to $$\Re^3$$'s standard basis vectors $$\vec{\imath}$$, $$\vec{\jmath}$$, and $$\vec{k}$$ instead of$\vec{e}_1$,$\vec{e}_2$, and$\vec{e}_3$.) instead of$\vec{e}_1$,$\vec{e}_2$, and$\vec{e}_3$. Note that $$\vec{e}_1$$ means something different in a discussion of $$\Re^3$$ than it means in a discussion of $$\Re^2$$. ... ... @@ -150,8 +150,8 @@ such as this one \end{equation*} by parametrizing. \begin{equation*} \set{\colvec{-1 \\ 1 \\ 0 \\ 0}y +\colvec{1 \\ 0 \\ -1 \\ 1}w \set{\colvec[r]{-1 \\ 1 \\ 0 \\ 0}y +\colvec[r]{1 \\ 0 \\ -1 \\ 1}w \suchthat y,w\in\Re } \end{equation*} Thus the vector space of solutions is ... ... @@ -301,13 +301,13 @@ In $$\polyspace_3$$, with respect to the basis $$B=\sequence{1,2x,2x^2,2x^3}$$, the representation of $$x+x^2$$ is \begin{equation*} \rep{x+x^2}{B}=\colvec{0 \\ 1/2 \\ 1/2 \\ 0}_B \rep{x+x^2}{B}=\colvec[r]{0 \\ 1/2 \\ 1/2 \\ 0}_B \end{equation*} (note that the coordinates are scalars, not vectors). With respect to a different basis $$D=\sequence{1+x,1-x,x+x^2,x+x^3}$$, the representation \begin{equation*} \rep{x+x^2}{D}=\colvec{0 \\ 0 \\ 1 \\ 0}_D \rep{x+x^2}{D}=\colvec[r]{0 \\ 0 \\ 1 \\ 0}_D \end{equation*} is different. \end{example} ... ... @@ -327,20 +327,20 @@ refers to the plane vector that, when in canonical position, ends at To find the coordinates of that vector with respect to the basis \begin{equation*} B=\sequence{ \colvec{1 \\ 1}, \colvec{0 \\ 2} } \colvec[r]{1 \\ 1}, \colvec[r]{0 \\ 2} } \end{equation*} we solve \begin{equation*} c_1\colvec{1 \\ 1} +c_2\colvec{0 \\ 2} c_1\colvec[r]{1 \\ 1} +c_2\colvec[r]{0 \\ 2} = \colvec{3 \\ 2} \colvec[r]{3 \\ 2} \end{equation*} to get that$c_1=3$and$c_2=1/2$. Then we have this. \begin{equation*} \rep{\vec{v}}{B}=\colvec{3 \\ -1/2} \rep{\vec{v}}{B}=\colvec[r]{3 \\ -1/2} \end{equation*} Here, although we've ommited the subscript $$B$$ from the column, the fact that the right side is a representation is clear from the context. ... ... @@ -372,20 +372,20 @@ We will see that in the next subsection. Decide if each is a basis for $$\Re^3$$. \begin{exparts*} \partsitem $$\sequence{ \colvec{1 \\ 2 \\ 3}, \colvec{3 \\ 2 \\ 1}, \colvec{0 \\ 0 \\ 1}}$$ \colvec[r]{1 \\ 2 \\ 3}, \colvec[r]{3 \\ 2 \\ 1}, \colvec[r]{0 \\ 0 \\ 1}} \) \partsitem $$\sequence{ \colvec{1 \\ 2 \\ 3}, \colvec{3 \\ 2 \\ 1}}$$ \colvec[r]{1 \\ 2 \\ 3}, \colvec[r]{3 \\ 2 \\ 1}} \) \partsitem $$\sequence{ \colvec{0 \\ 2 \\ -1}, \colvec{1 \\ 1 \\ 1}, \colvec{2 \\ 5 \\ 0}}$$ \colvec[r]{0 \\ 2 \\ -1}, \colvec[r]{1 \\ 1 \\ 1}, \colvec[r]{2 \\ 5 \\ 0}} \) \partsitem $$\sequence{ \colvec{0 \\ 2 \\ -1}, \colvec{1 \\ 1 \\ 1}, \colvec{1 \\ 3 \\ 0}}$$ \colvec[r]{0 \\ 2 \\ -1}, \colvec[r]{1 \\ 1 \\ 1}, \colvec[r]{1 \\ 3 \\ 0}} \) \end{exparts*} \begin{answer} By \nearbytheorem{th:BasisIffUniqueRepWRT}, each is a basis if and only ... ... @@ -395,9 +395,9 @@ We will see that in the next subsection. \partsitem Yes this is a basis. The relation \begin{equation*} c_1\colvec{1 \\ 2 \\ 3} +c_2\colvec{3 \\ 2 \\ 1} +c_3\colvec{0 \\ 0 \\ 1} c_1\colvec[r]{1 \\ 2 \\ 3} +c_2\colvec[r]{3 \\ 2 \\ 1} +c_3\colvec[r]{0 \\ 0 \\ 1} =\colvec{x \\ y \\ z} \end{equation*} gives ... ... @@ -421,8 +421,8 @@ We will see that in the next subsection. \partsitem This is not a basis. Setting it up as in the prior item \begin{equation*} c_1\colvec{1 \\ 2 \\ 3} +c_2\colvec{3 \\ 2 \\ 1} c_1\colvec[r]{1 \\ 2 \\ 3} +c_2\colvec[r]{3 \\ 2 \\ 1} =\colvec{x \\ y \\ z} \end{equation*} gives a linear system whose solution ... ... @@ -491,20 +491,20 @@ We will see that in the next subsection. \recommended \item Represent the vector with respect to the basis. \begin{exparts} \partsitem $$\colvec{1 \\ 2}$$, $$B=\sequence{\colvec{1 \\ 1},\colvec{-1 \\ 1}}\subseteq\Re^2$$ \partsitem $$\colvec[r]{1 \\ 2}$$, $$B=\sequence{\colvec[r]{1 \\ 1},\colvec[r]{-1 \\ 1}}\subseteq\Re^2$$ \partsitem $$x^2+x^3$$, $$D=\sequence{1,1+x,1+x+x^2,1+x+x^2+x^3}\subseteq\polyspace_3$$ \partsitem $$\colvec{0 \\ -1 \\ 0 \\ 1}$$, \partsitem $$\colvec[r]{0 \\ -1 \\ 0 \\ 1}$$, $$\stdbasis_4\subseteq\Re^4$$ \end{exparts} \begin{answer} \begin{exparts} \partsitem We solve \begin{equation*} c_1\colvec{1 \\ 1} +c_2\colvec{-1 \\ 1} =\colvec{1 \\ 2} c_1\colvec[r]{1 \\ 1} +c_2\colvec[r]{-1 \\ 1} =\colvec[r]{1 \\ 2} \end{equation*} with \begin{equation*} ... ... @@ -521,7 +521,7 @@ We will see that in the next subsection. and conclude that $$c_2=1/2$$ and so $$c_1=3/2$$. Thus, the representation is this. \begin{equation*} \rep{\colvec{1 \\ 2}}{B}=\colvec{3/2 \\ 1/2}_B \rep{\colvec[r]{1 \\ 2}}{B}=\colvec[r]{3/2 \\ 1/2}_B \end{equation*} \partsitem The relationship$c_1\cdot(1)+c_2\cdot(1+x)+c_3\cdot(1+x+x^2)+c_4\cdot(1+x+x^2+x^3) ... ... @@ -529,10 +529,10 @@ We will see that in the next subsection. is easily solved by eye to give that $c_4=1$, $c_3=0$, $c_2=-1$, and $c_1=0$. \begin{equation*} \rep{x^2+x^3}{D}=\colvec{0 \\ -1 \\ 0 \\ 1}_D \rep{x^2+x^3}{D}=\colvec[r]{0 \\ -1 \\ 0 \\ 1}_D \end{equation*} \partsitem $$\rep{\colvec{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4} =\colvec{0 \\ -1 \\ 0 \\ 1}_{\stdbasis_4}$$ \partsitem $$\rep[r]{\colvec{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4} =\colvec[r]{0 \\ -1 \\ 0 \\ 1}_{\stdbasis_4}$$ \end{exparts} \end{answer} \item ... ... @@ -574,15 +574,15 @@ We will see that in the next subsection. \begin{equation*} \set{\colvec{4x_2-3x_3+x_4 \\ x_2 \\ x_3 \\ x_4} \suchthat x_2,x_3,x_4\in\Re} =\set{x_2\colvec{4 \\ 1 \\ 0 \\ 0} +x_3\colvec{-3 \\ 0 \\ 1 \\ 0} +x_4\colvec{1 \\ 0 \\ 0 \\ 1} \suchthat x_2,x_3,x_4\in\Re} =\set{x_2\colvec[r]{4 \\ 1 \\ 0 \\ 0} +x_3\colvec[r]{-3 \\ 0 \\ 1 \\ 0} +x_4\colvec[r]{1 \\ 0 \\ 0 \\ 1} \suchthat x_2,x_3,x_4\in\Re} \end{equation*} and so the obvious candidate for the basis is this. \begin{equation*} \sequence{\colvec{4 \\ 1 \\ 0 \\ 0}, \colvec{-3 \\ 0 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 0 \\ 1} } \sequence{\colvec[r]{4 \\ 1 \\ 0 \\ 0}, \colvec[r]{-3 \\ 0 \\ 1 \\ 0}, \colvec[r]{1 \\ 0 \\ 0 \\ 1} } \end{equation*} We've shown that this spans the space, and showing it is also linearly independent is routine. ... ... @@ -924,9 +924,9 @@ We will see that in the next subsection. Find one vector $\vec{v}$ that will make each into a basis for the space. \begin{exparts*} \partsitem $\sequence{\colvec{1 \\ 1},\vec{v}}$ in $\Re^2$ \partsitem $\sequence{\colvec{1 \\ 1 \\ 0}, \colvec{0 \\ 1 \\ 0},\vec{v}}$ in $\Re^3$ \partsitem $\sequence{\colvec[r]{1 \\ 1},\vec{v}}$ in $\Re^2$ \partsitem $\sequence{\colvec[r]{1 \\ 1 \\ 0}, \colvec[r]{0 \\ 1 \\ 0},\vec{v}}$ in $\Re^3$ \partsitem $\sequence{x,1+x^2,\vec{v}}$ in $\polyspace_2$ \end{exparts*} \begin{answer} ... ... @@ -992,10 +992,10 @@ We will see that in the next subsection. Here is a subset of $\Re^2$ that is not a basis, and two different linear combinations of its elements that sum to the same vector. \begin{equation*} \set{\colvec{1 \\ 2},\colvec{2 \\ 4}} \set{\colvec[r]{1 \\ 2},\colvec[r]{2 \\ 4}} \qquad 2\cdot\colvec{1 \\ 2}+0\cdot\colvec{2 \\ 4} =0\cdot\colvec{1 \\ 2}+1\cdot\colvec{2 \\ 4} 2\cdot\colvec[r]{1 \\ 2}+0\cdot\colvec[r]{2 \\ 4} =0\cdot\colvec[r]{1 \\ 2}+1\cdot\colvec[r]{2 \\ 4} \end{equation*} Thus, when a subset is not a basis, it can be the case that its linear combinations are not unique. ... ... @@ -1004,13 +1004,13 @@ We will see that in the next subsection. combinations must be not unique. For instance, this set \begin{equation*} \set{\colvec{1 \\ 2}} \set{\colvec[r]{1 \\ 2}} \end{equation*} does have the property that \begin{equation*} c_1\cdot\colvec{1 \\ 2} c_1\cdot\colvec[r]{1 \\ 2} = c_2\cdot\colvec{1 \\ 2} c_2\cdot\colvec[r]{1 \\ 2} \end{equation*} implies that $c_1=c_2$. The idea here is that this subset fails to be a basis because it fails ... ... @@ -1124,7 +1124,7 @@ We will see that in the next subsection. +c_2\colvec{x_2 \\ y_2 \\ z_2} +c_3\colvec{x_3 \\ y_3 \\ z_3} +c_4\colvec{x_4 \\ y_4 \\ z_4} =\colvec{0 \\ 0 \\ 0} =\colvec[r]{0 \\ 0 \\ 0} \end{equation*} gives rise to a linear system \begin{equation*} ... ... @@ -1199,12 +1199,12 @@ We will see that in the next subsection. \end{equation*} and so a natural candidate for a basis is this. \begin{equation*}