Commit 7b8c49f1 authored by Jim Hefferon's avatar 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*}