Skip to content
Next
Projects
Groups
Snippets
Help
Loading...
Help
Support
Submit feedback
Contribute to GitLab
Switch to GitLab Next
Sign in / Register
Toggle navigation
L
linearalgebra
Project
Project
Details
Activity
Releases
Cycle Analytics
Insights
Repository
Repository
Files
Commits
Branches
Tags
Contributors
Graph
Compare
Charts
Locked Files
Issues
1
Issues
1
List
Boards
Labels
Service Desk
Milestones
Merge Requests
0
Merge Requests
0
CI / CD
CI / CD
Pipelines
Jobs
Schedules
Charts
Security & Compliance
Security & Compliance
Dependency List
Packages
Packages
Container Registry
Wiki
Wiki
Members
Members
Collapse sidebar
Close sidebar
Activity
Graph
Charts
Create a new issue
Jobs
Commits
Issue Boards
Open sidebar
Jim Hefferon
linearalgebra
Commits
c2b2cbd5
Commit
c2b2cbd5
authored
Jan 09, 2012
by
Jim Hefferon
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
spell checks for the first and second chapter
parent
dcc4189f
Changes
12
Hide whitespace changes
Inline
Sidebyside
Showing
12 changed files
with
68 additions
and
68 deletions
+68
68
crystal.tex
crystal.tex
+11
11
dimen.tex
dimen.tex
+1
1
gr1.tex
gr1.tex
+9
9
gr2.tex
gr2.tex
+2
2
gr3.tex
gr3.tex
+4
4
leontif.tex
leontif.tex
+2
2
network.tex
network.tex
+6
6
ppivot.tex
ppivot.tex
+2
2
voting.tex
voting.tex
+8
8
vs1.tex
vs1.tex
+4
4
vs2.tex
vs2.tex
+5
5
vs3.tex
vs3.tex
+14
14
No files found.
crystal.tex
View file @
c2b2cbd5
...
...
@@ 13,9 +13,9 @@ Remarkably, the explanation for the cubical external shape is the simplest
one that we could imagine:~the internal shape, the way the atoms lie,
is also cubical.
The internal structure is pictured below.
Salt is sodium cloride, and the
small spheres shown are sodium while the big ones are cloride.
To simplify the view, it only shows the sodiums and clorides on the front,
Salt is sodium c
h
loride, and the
small spheres shown are sodium while the big ones are c
h
loride.
To simplify the view, it only shows the sodiums and c
h
lorides on the front,
top, and right.
\begin{center}
\includegraphics
{
ch2.8
}
...
...
@@ 33,7 +33,7 @@ There we have a square repeated many times.
The distance between the corners of the square
cell is about
$
3
.
34
$
~
\AA
ngstroms
(an
\AA
ngstrom is
$
10
^{

10
}$
~meters).
Obviously that unit is unwield
l
y.
Obviously that unit is unwieldy.
Instead we can
take as a unit the length of each square's side.
That is, we naturally adopt this basis.
...
...
@@ 56,7 +56,7 @@ much weaker than the bonds inside the planes, which explains why
pencils write
\Dash
the graphite can be sheared so that the planes slide
off and are left on the paper.
We can get a conv
ien
ent unit of length by
We can get a conv
eni
ent unit of length by
decomposing the hexagonal ring into three regions that are rotations
of this
\definend
{
unit cell
}
.
\index
{
crystals!unit cell
}
\begin{center}
%graphite
...
...
@@ 106,7 +106,7 @@ The examples here show that
the structures of crystals is complicated enough to need
some organized system to give the locations of the atoms and how they
are chemically bound.
One tool for that organization is a conv
ien
ent basis.
One tool for that organization is a conv
eni
ent basis.
This application of bases is simple but it shows a
natural science context where
the idea arises naturally.
...
...
@@ 199,19 +199,19 @@ the idea arises naturally.
\item
This illustrates how the
dimensions of a unit cell could be computed from the shape
in which a substance crystalizes
in which a substance crystal
l
izes
(
\cite
{
Ebbing
}
, p.~462).
\begin{exparts}
\partsitem
Recall that there are
$
6
.
022
\times
10
^{
23
}$
atoms in a mole
(this is Av
a
gadro's number).
(this is Av
o
gadro's number).
From that, and the fact that platinum has a mass of
$
195
.
08
$
grams
per mole, calculate the mass of each atom.
\partsitem
Platinum crystalizes in a facecentered cubic lattice
\partsitem
Platinum crystal
l
izes in a facecentered cubic lattice
with atoms at each lattice point,
that is, it looks like the middle picture given above for
the diamond crystal.
Find the number of platinums per unit cell
(hint:~sum the fractions of platinums that are inside of a single
Find the number of platinum
'
s per unit cell
(hint:~sum the fractions of platinum
'
s that are inside of a single
cell).
\partsitem
From that, find the mass of a unit cell.
\partsitem
Platinum crystal has a
...
...
dimen.tex
View file @
c2b2cbd5
...
...
@@ 486,7 +486,7 @@ modeling is in \cite{Giordano82}.
\item
\cite
{
Einstein1911
}
conjectured that the infrared characteristic frequencies of a solid
may be determined by the same forces between atoms as determine
the solid's ord
a
nary elastic behavior.
the solid's ord
i
nary elastic behavior.
The relevant quantities are these.
\begin{center}
\begin{tabular}
{
rl
}
...
...
gr1.tex
View file @
c2b2cbd5
...
...
@@ 312,7 +312,7 @@ often combine addition steps when they use the same $\rho_i$; see the
next example.
\begin{example}
Gauss' method system
m
atically applies the row operations to solve a system.
Gauss' method systematically applies the row operations to solve a system.
Here is a typical case.
\begin{equation*}
\begin{linsys}
{
3
}
...
...
@@ 1299,7 +1299,7 @@ a no response by showing that no solution exists.}
\text
{
and
\
}
a
_{
m,1
}
s
_
1+a
_{
m,2
}
s
_
2+
\cdots
+a
_{
m,n
}
s
_
n
&
=d
_
m
\end{align*}
(this is straightforward cancel
l
ing on both sides of the
$
i
$
th equation),
(this is straightforward canceling on both sides of the
$
i
$
th equation),
which says that
\(
(
s
_
1
,
\ldots
,s
_
n
)
\)
solves
\begin{equation*}
\begin{linsys}
{
4
}
...
...
@@ 1607,7 +1607,7 @@ a no response by showing that no solution exists.}
The answer to this question would have been the same had we known only
that
{
\em
at least
\/
}
14 commissioners preferred
$
B
$
over
$
C
$
.
The seemingly paradoxical nature of the commissioner
s
's preferences
The seemingly paradoxical nature of the commissioner's preferences
(
$
A
$
is preferred to
$
B
$
, and
$
B
$
is preferred to
$
C
$
, and
$
C
$
is
preferred to
$
A
$
), an example of ``nontransitive dominance'', is not
uncommon when individual choices are pooled.
...
...
@@ 1928,7 +1928,7 @@ with this matrix.
\end{amat}
\end{equation*}
The vertical bar just reminds a reader of the difference between the
coefficients on the system
s
's left hand side and the constants on the right.
coefficients on the system's left hand side and the constants on the right.
With a bar, this is an
\definend
{
augmented
\/
}
\index
{
matrix!augmented
}
\index
{
augmented matrix
}
matrix.
In this notation
...
...
@@ 2694,7 +2694,7 @@ We will know exactly what can and cannot happen in a reduction.
&
&
30s
&

&
8a
&
=
&
16
\,
000
\end{linsys}
\end{eqnarray*}
To describe the solution set we can param
a
trize using
$
a
$
.
To describe the solution set we can param
e
trize using
$
a
$
.
\begin{equation*}
\set
{
\colvec
{
c
\\
s
\\
a
}
=
\colvec
{
20
\,
000/30
\\
16
\,
000/30
\\
0
}
...
...
@@ 2957,7 +2957,7 @@ We will know exactly what can and cannot happen in a reduction.
Those creatures are found in three colors: red, green and blue.
There are
$
13
$
~ red
arbuzoids,
$
15
$
~blue ones, and
$
17
$
~green. When
two differently colo
u
red arbuzoids meet, they
two differently colored arbuzoids meet, they
both change to the third color.
The question is, can it ever happen that all
...
...
@@ 2965,7 +2965,7 @@ We will know exactly what can and cannot happen in a reduction.
\begin{answer}
\answerasgiven
My solution was to define the numbers of arbuzoids
as
$
3
$
dimen
t
ional vectors, and express all possible
as
$
3
$
dimen
s
ional vectors, and express all possible
elementary transitions as such vectors, too:
\begin{center}
\begin{tabular}
{
rr
}
...
...
@@ 3100,7 +3100,7 @@ We will know exactly what can and cannot happen in a reduction.
It is known that it may contain one or more of the metals aluminum,
copper, silver, or lead.
When weighed successively under standard conditions in water, benzene,
alcohol, and glycerin
e
its respective weights are
\(
6588
\)
,
\(
6688
\)
,
alcohol, and glycerin its respective weights are
\(
6588
\)
,
\(
6688
\)
,
\(
6778
\)
, and
\(
6328
\)
grams.
How much, if any, of the forenamed metals does it contain if the
specific gravities of the designated substances are taken to be as follows?
...
...
@@ 3109,7 +3109,7 @@ We will know exactly what can and cannot happen in a reduction.
Aluminum
&\(
2
.
7
\)
&
\makebox
[3em]
{
\mbox
{}
\hfill\mbox
{}}
&
Alcohol
&
0.81
\\
Copper
&\(
8
.
9
\)
&
&
Benzene
&\(
0
.
90
\)
\\
Gold
&\(
19
.
3
\)
&
&
Glycerin
e
&\(
1
.
26
\)
\\
Gold
&\(
19
.
3
\)
&
&
Glycerin
&\(
1
.
26
\)
\\
Lead
&\(
11
.
3
\)
&
&
Water
&\(
1
.
00
\)
\\
Silver
&\(
10
.
8
\)
\end{tabular}
...
...
gr2.tex
View file @
c2b2cbd5
...
...
@@ 375,7 +375,7 @@ is a threedimensional linear surface.
Again, the intuition is that a line permits motion in one direction,
a plane permits motion in
combinations of two directions, etc.
(When the dimen
a
ion of the linear surface is one less than the dimension
(When the dimen
s
ion of the linear surface is one less than the dimension
of the space, that is, when we have an
$
n

1
$
flat in
$
\Re
^
n
$
,
then the surface is called a
...
...
@@ 876,7 +876,7 @@ while the right side gives this.
(u
_
1
^
2+u
_
2
^
2+u
_
3
^
2)+(v
_
1
^
2+v
_
2
^
2+v
_
3
^
2)
2
\,\norm
{
\vec
{
u
}
\,
}
\,\norm
{
\vec
{
v
}
\,
}
\cos\theta
\end{equation*}
Cancel
l
ing squares and dividing by
$
2
$
gives
Canceling squares and dividing by
$
2
$
gives
the formula that we want.
\begin{equation*}
\theta
...
...
gr3.tex
View file @
c2b2cbd5
...
...
@@ 269,11 +269,11 @@ Again, the point of view that we are developing, buttressed now by this lemma,
is that the term `reduces to' is misleading:~where
\(
A
\longrightarrow
B
\)
, we shouldn't think of
\(
B
\)
as
``after''
\(
A
\)
or ``simpler than''
$
A
$
.
Instead we should think of them as interreducible or interrelated.
Instead we should think of them as inter

reducible or interrelated.
Below is a picture of the idea.
The matrices from the start of this section and their
reduced echelon form version are shown in a cluster.
They are all interreducible.
They are all inter

reducible.
\begin{center}
\includegraphics
{
ch1.28
}
\end{center}
...
...
@@ 311,7 +311,7 @@ gives a reduction from \( A \) to \( C \).
\end{proof}
\begin{definition}
Two matrices that are interreducible by the elementary row operations
Two matrices that are inter

reducible by the elementary row operations
are
\definend
{
row equivalent
}
.
\index
{
matrix!row equivalence
}
%
\index
{
row equivalence
}
\index
{
equivalence relation!row equivalence
}
\end{definition}
...
...
@@ 917,7 +917,7 @@ class.
\end{exparts}
\begin{answer}
To be an equivalence, each relation must be reflexive, symmetric, and
trasitive.
tra
n
sitive.
\begin{exparts}
\item
This relation
is not symmetric because if
$
x
$
has taken
$
4
$
~classes and
$
y
$
...
...
leontif.tex
View file @
c2b2cbd5
...
...
@@ 3,7 +3,7 @@
% 2001Jun09
\topic
{
InputOutput Analysis
}
\index
{
InputOutput Analysis(
}
An economy is an immensely complicated network of interdependence
s
.
An economy is an immensely complicated network of interdependence.
Changes in one part can ripple out to affect other parts.
Economists have struggled to be able to describe, and to make
predictions about, such a complicated object and
...
...
@@ 173,7 +173,7 @@ among more sectors of an economy;
these models are typically solved on a computer.
Naturally also, a single model does not suit every case and
assuring that the assumptions underlying a model
are reasonable for a particular prediction requires the judg
e
ments of experts.
are reasonable for a particular prediction requires the judgments of experts.
With those caveats
however, this model has proven in practice to be a useful and accurate tool for
economic analysis.
...
...
network.tex
View file @
c2b2cbd5
...
...
@@ 157,7 +157,7 @@ To analyze it, we can place the arrows in this way.
\begin{center}
\includegraphics
{
ch1.39
}
\end{center}
Kirchoff's Current Law, applied to the
Kirch
h
off's Current Law, applied to the
top node, the left node, the right node, and the bottom node gives
these.
\begin{align*}
...
...
@@ 231,7 +231,7 @@ For instance, the exercises analyze some networks of streets.
and
$
i
_
3
=
81
/
14
$
.
Of course, the first and second paragraphs yield the same answer.
Esentially, in the first paragraph we solved the linear system
Es
s
entially, in the first paragraph we solved the linear system
by a method less systematic than Gauss' method, solving for some
of the variables and then substituting.
\partsitem
...
...
@@ 358,7 +358,7 @@ For instance, the exercises analyze some networks of streets.
\end{answer}
% \item Prove Th\`evenin's Theorem.
\item
[]
\textit
{
There are networks other than electrical ones, and
we can ask how well Kirchoff's laws apply to them.
we can ask how well Kirch
h
off's laws apply to them.
The remaining questions consider an extension to
networks of streets.
}
\item
...
...
@@ 386,8 +386,8 @@ For instance, the exercises analyze some networks of streets.
We can set up equations to model how the traffic flows.
\begin{exparts}
\partsitem
Adapt Kirchoff's Current Law to this circumstance.
Is it a reasonable model
l
ing assumption?
Adapt Kirch
h
off's Current Law to this circumstance.
Is it a reasonable modeling assumption?
\partsitem
Label the three betweenroad arcs in the circle with a variable.
Using the (adapted) Current Law,
...
...
@@ 511,7 +511,7 @@ For instance, the exercises analyze some networks of streets.
\begin{center}
\includegraphics
{
ch1.51
}
\end{center}
We apply Kirchoff's principle that the flow into the intersection
We apply Kirch
h
off's principle that the flow into the intersection
of Willow and Shelburne must equal the flow out to get
$
i
_
1
+
25
=
i
_
2
+
125
$
.
Doing the intersections from right to left and top to bottom
...
...
ppivot.tex
View file @
c2b2cbd5
...
...
@@ 218,7 +218,7 @@ where experts have worked hard to counter what can go wrong.
\item
Using two decimal places, add
$
253
$
and
$
2
/
3
$
.
\begin{answer}
Sc
eintific notation is convien
ent to express the twoplace restriction.
Sc
ientific notation is conveni
ent to express the twoplace restriction.
We have
$
.
25
\times
10
^{
2
}
+
.
67
\times
10
^{
0
}
=
.
25
\times
10
^{
2
}$
.
The
$
2
/
3
$
has no apparent effect.
\end{answer}
...
...
@@ 325,7 +325,7 @@ where experts have worked hard to counter what can go wrong.
\begin{exparts}
\partsitem
Solve the system by hand.
Notice that the
$
\varepsilon
$
's divide out only because there is
an exact cancelation of the integer parts on the right side
an exact cancel
l
ation of the integer parts on the right side
as well as on the left.
\partsitem
Solve the system by hand, rounding to two decimal
places, and with
$
\varepsilon
=
0
.
001
$
.
...
...
voting.tex
View file @
c2b2cbd5
...
...
@@ 116,7 +116,7 @@ Mathematicians also study voting paradoxes simply because they are
interesting.
One interesting aspect is that
the group's overall majority cycle occurs despite that
each single voter
s
's preference list is
each single voter's preference list is
\definend
{
rational
}
\index
{
voting paradox!rational preference order
}
, in
a straightline order.
That is, the majority cycle seems to arise in the aggregate
...
...
@@ 125,7 +125,7 @@ However we can use linear algebra to argue that
a tendency toward cyclic preference is actually present
in each voter's list and
that it surfaces when there is more adding of the tendency
than cancel
l
ing.
than canceling.
For this,
abbreviating the choices as
$
D
$
,
$
R
$
, and
$
T
$
,
...
...
@@ 149,7 +149,7 @@ for instance, the Political Science mock election
yields the circular group preference shown earlier.
Of course, taking linear combinations is linear algebra.
The graphical cycle notation is suggestive but inconv
ien
ent so we
The graphical cycle notation is suggestive but inconv
eni
ent so we
use column vectors by starting at the
$
D
$
and
taking the numbers from the cycle in counterclockwise order.
Thus, the mock election and a single
$
D>R>T
$
vote are represented in this way.
...
...
@@ 326,7 +326,7 @@ voting paradox can happen only when the
tendencies toward cyclic preference reinforce each other.
For the proof, assume that
opposite preference orders have been cancel
l
ed and we are left with one set
opposite preference orders have been canceled and we are left with one set
of preference lists from each of the three rows.
Consider the sum of these three
(here, the numbers
$
a
$
,
$
b
$
, and
$
c
$
could be positive, negative, or zero).
...
...
@@ 384,8 +384,8 @@ for his kind and illuminating discussions.)}
Suppose that this voter ranks each candidate
on each of three criteria.
\begin{exparts}
\partsitem
Draw up a table with the rows label
l
ed `Democrat',
`Republican', and `Third', and the columns label
l
ed
\partsitem
Draw up a table with the rows labeled `Democrat',
`Republican', and `Third', and the columns labeled
`character', `experience', and `policies'.
Inside each column, rank some
candidate as most preferred,
...
...
@@ 548,7 +548,7 @@ for his kind and illuminating discussions.)}
\end{center}
All three come from the same side of the table (the left),
as the result from this Topic says must happen.
Tallying the election can now proceed, using the cancel
l
ed numbers
Tallying the election can now proceed, using the canceled numbers
\begin{equation*}
% \psset{xunit=4pt,yunit=4pt,runit=4pt}
3
\cdot
\votinggraphic
{
31
}
...
...
@@ 701,7 +701,7 @@ for his kind and illuminating discussions.)}
\end{tabular}
\\
\hline
\end{tabular}
\end{center}
The election, using the cancel
l
ed numbers, is this.
The election, using the canceled numbers, is this.
\begin{equation*}
% \psset{xunit=4pt,yunit=4pt,runit=4pt}
3
\cdot\votinggraphic
{
35
}
...
...
vs1.tex
View file @
c2b2cbd5
...
...
@@ 403,7 +403,7 @@ $
Although this space is not a subset of any
\(
\Re
^
n
\)
,
there is a sense in which we can think of
$
\polyspace
_
3
$
as ``the same'' as
\(
\Re
^
4
\)
.
If we identify these two space
s
's elements in this way
If we identify these two space's elements in this way
\begin{equation*}
a
_
0+a
_
1x+a
_
2x
^
2+a
_
3x
^
3
\quad\text
{
corresponds to
}
\quad
...
...
@@ 1067,7 +1067,7 @@ But from now on our primary objects of study will be vector spaces.
\begin{exparts}
\partsitem
No:
\(
1
\cdot
(
0
,
1
)+
1
\cdot
(
0
,
1
)
\neq
(
1
+
1
)
\cdot
(
0
,
1
)
\)
.
\partsitem
No; the same calculation as the prior answer shows
a con
t
ition in the definition of a vector space that is
a con
d
ition in the definition of a vector space that is
violated.
Another example of a violation of the conditions for a
vector space is that
\(
1
\cdot
(
0
,
1
)
\neq
(
0
,
1
)
\)
.
...
...
@@ 2641,7 +2641,7 @@ The next section studies spanning sets that are minimal.
\(
c
\cdot\vec
{
s
}
+
d
\cdot\vec
{
s
}
\,
\)
in
\(
S
\)
.
The check for the third, fourth, and fifth conditions are similar to the
second condition
s
's check just given.
second condition's check just given.
\end{answer}
\item
Show that each vector space has only one trivial subspace.
...
...
@@ 2981,7 +2981,7 @@ The next section studies spanning sets that are minimal.
interacts with the usual set operations.
\begin{exparts}
\partsitem
If
\(
A,B
\)
are subspaces of a vector space, must
their inter
es
ction
their inter
se
ction
\(
A
\intersection
B
\)
be a subspace?
Always? Sometimes? Never?
\partsitem
Must the union
\(
A
\union
B
\)
be a subspace?
...
...
vs2.tex
View file @
c2b2cbd5
...
...
@@ 414,7 +414,7 @@ gives a system
\end{equation*}
with leading variables
$
c
_
1
$
,
$
c
_
2
$
, and
$
c
_
4
$
and
free variables
$
c
_
3
$
and
$
c
_
5
$
.
Param
a
trizing the solution set
Param
e
trizing the solution set
\begin{equation*}
\set
{
\colvec
{
c
_
1
\\
c
_
2
\\
c
_
3
\\
c
_
4
\\
c
_
5
}
=
c
_
3
\colvec
{
1
\\
1
\\
1
\\
0
\\
0
}
...
...
@@ 936,7 +936,7 @@ tells us that a linearly independent set is maximal when it spans the space.
\end{exparts*}
\begin{answer}
In each case, that the set is independent must be proved, and that it is
dependent must be shown by ex
i
hibiting a specific dependence.
dependent must be shown by exhibiting a specific dependence.
\begin{exparts}
\partsitem
This set is dependent.
The familiar relation
$
\sin
^
2
(
x
)+
\cos
^
2
(
x
)=
1
$
shows that
...
...
@@ 965,7 +965,7 @@ tells us that a linearly independent set is maximal when it spans the space.
$
c
_
1
\cdot
(
1
+
x
)
^
2
+
c
_
2
\cdot
(
x
^
2
+
2
x
)=
3
$
is satisfied by
$
c
_
1
=
3
$
and
$
c
_
2
=
3
$
.
\partsitem
This set is dependent.
The easiest way to see that is to recall the trig
i
nometric
The easiest way to see that is to recall the trig
o
nometric
relationship
$
\cos
^
2
(
x
)
\sin
^
2
(
x
)=
\cos
(
2
x
)
$
.
(
\textit
{
Remark.
}
A person who doesn't recall this, and tries some
$
x
$
's,
...
...
@@ 1279,7 +1279,7 @@ tells us that a linearly independent set is maximal when it spans the space.
With three vectors from
$
\Re
^
2
$
, the argument from the prior item
still applies, with the slight change that Gauss' method now only
leaves at least one variable free (but that still gives infintely many
leaves at least one variable free (but that still gives infin
i
tely many
solutions).
\partsitem
The prior item shows that no threeelement subset of
$
\Re
^
2
$
is independent.
...
...
@@ 1781,7 +1781,7 @@ tells us that a linearly independent set is maximal when it spans the space.
(also called
\definend
{
pairwise perpendicular
}
):~if
\(
i
\neq
j
\)
then
\(
\vec
{
v
}_
i
\)
is perpendicular to
\(
\vec
{
v
}_
j
\)
.
Mimicing the proof of the first item above shows that such a set of
Mimic
k
ing the proof of the first item above shows that such a set of
nonzero vectors is linearly independent.
\end{exparts}
\end{answer}
...
...
vs3.tex
View file @
c2b2cbd5
...
...
@@ 114,7 +114,7 @@ This is a natural basis.
Another, more generic, basis is
\(
\sequence
{
\cos\theta

\sin\theta
,
2
\cos\theta
+
3
\sin\theta
}
\)
.
Verfication that these two are bases is
Ver
i
fication that these two are bases is
\nearbyexercise
{
exer:VerifBasesCosPlusSin
}
.
\end{example}
...
...
@@ 162,7 +162,7 @@ basis comprised of the above two elements.
\end{example}
\begin{example}
Paramet
e
rization helps find bases for other vector spaces, not just
Parametrization helps find bases for other vector spaces, not just
for solution sets of homogeneous systems.
To find a basis for this subspace of
$
\matspace
_{
\nbyn
{
2
}}$
\begin{equation*}
...
...
@@ 342,7 +342,7 @@ Then we have this.
\begin{equation*}
\rep
{
\vec
{
v
}}{
B
}
=
\colvec
[r]
{
3
\\
1/2
}
\end{equation*}
Here, although we've om
mi
ted the subscript
\(
B
\)
from the column,
Here, although we've om
it
ted the subscript
\(
B
\)
from the column,
the fact that the right side is a representation is clear from the context.
The advantage the notation and the term `coordinates' is that they
...
...
@@ 930,7 +930,7 @@ We will see that in the next subsection.
\partsitem
$
\sequence
{
x,
1
+
x
^
2
,
\vec
{
v
}}$
in
$
\polyspace
_
2
$
\end{exparts*}
\begin{answer}
Each forms a linearly independent set if
$
\vec
{
v
}$
is om
m
itted.
Each forms a linearly independent set if
$
\vec
{
v
}$
is omitted.
To preserve linear independence, we must expand the span of each.
That is, we must determine the span of each (leaving
$
\vec
{
v
}$
out),
and then pick a
$
\vec
{
v
}$
lying outside of that span.
...
...
@@ 2111,7 +2111,7 @@ that the two senses of `minimal' are equivalent.
members than does
\(
B
_
U
\)
and so equals
\(
B
_
U
\)
.
Since
\(
U
\)
and
\(
W
\)
have the same bases, they are equal.
\partsitem
Let
\(
W
\)
be the space of finitedegree polynomials and
let
\(
U
\)
be the subspace of polynom
ai
ls that have only
let
\(
U
\)
be the subspace of polynom
ia
ls that have only
evenpowered terms
\(
\set
{
a
_
0
+
a
_
1
x
^
2
+
a
_
2
x
^
4
+
\dots
+
a
_
nx
^{
2
n
}
\suchthat
a
_
0
,
\ldots
,a
_
n
\in\Re
}
\)
.
...
...
@@ 4049,7 +4049,7 @@ The dimension of a direct sum is the sum of the dimensions of its summands.
\begin{proof}
In
\nearbylemma
{
le:UniqDecIffBasisDec
}
,
the number of basis vectors in the concatenation equals the sum of
the number of vectors in the subbases that make up the concatenation.
the number of vectors in the sub

bases that make up the concatenation.
\end{proof}
The special case of two subspaces is worth mentioning separately.
...
...
@@ 4602,7 +4602,7 @@ needed to do the Jordan Form construction in the fifth chapter.
\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
independent, and has size two in
$
\Re
^
2
$
), and thus ther
e
is one and
only one solution to the above equation, implying that all
decompositions are unique.
Alternatively, we can solve
...
...
@@ 4833,7 +4833,7 @@ needed to do the Jordan Form construction in the fifth chapter.
\item
\label
{
exer:ThreeSubsPairwseNonTriv
}
(Refer to
\nearbyexample
{
exam:DirSumThree
}
.
This exercise
shows that the requirement that pa
ri
wise intersections be trivial
shows that the requirement that pa
ir
wise intersections be trivial
is genuinely stronger than the requirement only that the intersection of
all of the subspaces be trivial.)
Give a vector space and three subspaces
$
W
_
1
$
,
$
W
_
2
$
, and
$
W
_
3
$
...
...
@@ 4889,7 +4889,7 @@ needed to do the Jordan Form construction in the fifth chapter.
\(
\sequence
{
\vec
{
\omega
}_
1
,
\dots
,
\vec
{
\omega
}_
k,
\vec
{
\beta
}_{
k
+
1
}
,
\dots
,
\vec
{
\beta
}_
n
}
\)
for the whole
space.
Then the complemen of the original subspace has this for a basis:
Then the complemen
t
of the original subspace has this for a basis:
\(
\sequence
{
\vec
{
\beta
}_{
k
+
1
}
,
\dots
,
\vec
{
\beta
}_
n
}
\)
.
\end{answer}
\recommended
\item
...
...
@@ 5071,7 +5071,7 @@ needed to do the Jordan Form construction in the fifth chapter.
\partsitem
Give a symmetric
$
\nbyn
{
2
}$
matrix and an antisymmetric
$
\nbyn
{
2
}$
matrix.
(
\textit
{
Remark.
}
For the second one, be careful about the entries on the diag
i
onal.)
For the second one, be careful about the entries on the diagonal.)
\partsitem
What is the relationship between a square symmetric matrix and
its transpose?
Between a square antisymmetric matrix and its transpose?
...
...
@@ 5260,9 +5260,9 @@ needed to do the Jordan Form construction in the fifth chapter.
\partsitem
Must there be an identity element,
a subspace
\(
I
\)
such that
\(
I
+
W
=
W
+
I
=
W
\)
for all subspaces
\(
W
\)
?
\partsitem
Does leftcancelation hold:~if
\partsitem
Does leftcancel
l
ation hold:~if
\(
W
_
1
+
W
_
2
=
W
_
1
+
W
_
3
\)
then
\(
W
_
2
=
W
_
3
\)
?
Right cancelation?
Right cancel
l
ation?
\end{exparts}
\begin{answer}
\begin{exparts}
...
...
@@ 5279,7 +5279,7 @@ needed to do the Jordan Form construction in the fifth chapter.
\(
W
\)
.
\partsitem
In each vector space, the identity element with respect
to subspace addition is the trivial subspace.
\partsitem
Neither of left or right cancelation needs to hold.
\partsitem
Neither of left or right cancel
l
ation needs to hold.
For an example, in
\(
\Re
^
3
\)
take
\(
W
_
1
\)
to be the
\(
xy
\)
plane, take
\(
W
_
2
\)
to be the
\(
x
\)
axis,
and take
\(
W
_
3
\)
to be the
\(
y
\)
axis.
...
...
@@ 5295,7 +5295,7 @@ needed to do the Jordan Form construction in the fifth chapter.
W
_
1
\directsum
(
W
_
2
\directsum
W
_
3
)
\)
.
\partsitem
Show that
\(
\Re
^
3
\)
is the direct sum of the three axes
(the relevance here is that by the previous item,
we needn't specify which two of the three
e
axes are combined first).
we needn't specify which two of the three axes are combined first).
\partsitem
Does the direct sum operation leftcancel:~does
\(
W
_
1
\directsum
W
_
2
=
W
_
1
\directsum
W
_
3
\)
imply
\(
W
_
2
=
W
_
3
\)
?
Does it rightcancel?
...
...
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment