Commit 666d4189 authored by Jim Hefferon's avatar Jim Hefferon

edits for vs2

parent 56301810
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......@@ -5577,7 +5577,7 @@
\end{ans}
\subsection{Subsection Two.II.1: Definition and Examples}
\begin{ans}{Two.II.1.18}
\begin{ans}{Two.II.1.19}
For each of these, when the subset is independent it must be proved, and
when the subset is dependent an example of a dependence must be given.
\begin{exparts}
......@@ -5682,7 +5682,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.19}
\begin{ans}{Two.II.1.20}
In the cases of independence, that must be proved.
Otherwise, a specific dependence must be produced.
(Of course, dependences other than the ones exhibited here are possible.)
......@@ -5749,7 +5749,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.20}
\begin{ans}{Two.II.1.21}
Let $Z$ be the zero function $Z(x)=0$, which is the additive identity in
the vector space under discussion.
\begin{exparts}
......@@ -5786,7 +5786,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.21}
\begin{ans}{Two.II.1.22}
In each case, that the set is independent must be proved, and that it is
dependent must be shown by exihibiting a specific dependence.
\begin{exparts}
......@@ -5831,25 +5831,25 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.22}
\begin{ans}{Two.II.1.23}
No, that equation is not a linear relationship.
In fact this set is independent, as the system arising from taking
\( x \) to be \( 0 \), \( \pi/6 \) and \( \pi/4 \) shows.
\end{ans}
\begin{ans}{Two.II.1.23}
\begin{ans}{Two.II.1.24}
To emphasize that the equation
\( 1\cdot\vec{s}+(-1)\cdot\vec{s}=\zero \)
does not make the set dependent.
\end{ans}
\begin{ans}{Two.II.1.24}
\begin{ans}{Two.II.1.25}
We have already showed this: the Linear Combination
Lemma and its corollary state that in an echelon form matrix,
no nonzero row is a linear combination of the others.
\end{ans}
\begin{ans}{Two.II.1.25}
\begin{ans}{Two.II.1.26}
\begin{exparts}
\partsitem Assume that the set
\( \set{\vec{u},\vec{v},\vec{w}} \) is linearly
......@@ -5884,7 +5884,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.26}
\begin{ans}{Two.II.1.27}
\begin{exparts}
\partsitem A singleton set $\set{\vec{v}}$ is linearly independent
if and only if $\vec{v}\neq\zero$.
......@@ -5917,11 +5917,11 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.27}
\begin{ans}{Two.II.1.28}
This set is linearly dependent set because it contains the zero vector.
\end{ans}
\begin{ans}{Two.II.1.28}
\begin{ans}{Two.II.1.29}
The `if' half is given by \nearbylemma{le:SubsetPreserveLI}.
The converse (the `only if' statement) does not hold.
An example is to consider the vector space \( \Re^2 \) and
......@@ -5933,7 +5933,7 @@
\end{equation*}
\end{ans}
\begin{ans}{Two.II.1.29}
\begin{ans}{Two.II.1.30}
\begin{exparts}
\partsitem The linear system arising from
\begin{equation*}
......@@ -6005,7 +6005,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.30}
\begin{ans}{Two.II.1.31}
In this `if and only if' statement, the `if' half is clear\Dash if
the polynomial is the zero polynomial then the function that arises
from the action of the polynomial must be the zero
......@@ -6018,13 +6018,13 @@
polynomial.
\end{ans}
\begin{ans}{Two.II.1.31}
\begin{ans}{Two.II.1.32}
The work in this section suggests that an \( n \)-dimensional
non-degenerate linear surface should be defined as the span of a linearly
independent set of \( n \) vectors.
\end{ans}
\begin{ans}{Two.II.1.32}
\begin{ans}{Two.II.1.33}
\begin{exparts}
\partsitem For any $a_{1,1}$, \ldots, $a_{2,4}$,
\begin{equation*}
......@@ -6063,7 +6063,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.33}
\begin{ans}{Two.II.1.34}
Yes; here is one.
\begin{equation*}
\set{\colvec{1 \\ 0 \\ 0},
......@@ -6073,13 +6073,13 @@
\end{equation*}
\end{ans}
\begin{ans}{Two.II.1.34}
\begin{ans}{Two.II.1.35}
Yes.
The two improper subsets, the entire set and the empty subset, serve as
examples.
\end{ans}
\begin{ans}{Two.II.1.35}
\begin{ans}{Two.II.1.36}
In \( \Re^4 \) the biggest linearly independent set has
four vectors.
There are many examples of such sets, this is one.
......@@ -6123,7 +6123,7 @@
The smallest is \( \set{\zero} \).
\end{ans}
\begin{ans}{Two.II.1.36}
\begin{ans}{Two.II.1.37}
\begin{exparts}
\partsitem The intersection of two linearly independent sets
$S\intersection T$ must be linearly
......@@ -6154,7 +6154,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.37}
\begin{ans}{Two.II.1.38}
\begin{exparts}
\partsitem \nearbylemma{le:LDIffANonTrivLinRel}
requires that the vectors
......@@ -6226,7 +6226,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.38}
\begin{ans}{Two.II.1.39}
\begin{exparts}
\partsitem We do induction on the number of vectors in the finite set
\( S \).
......@@ -6282,7 +6282,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.39}
\begin{ans}{Two.II.1.40}
\begin{exparts}
\partsitem
Assuming first that \( a\neq 0 \),
......@@ -6429,7 +6429,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.40}
\begin{ans}{Two.II.1.41}
Recall that two vectors from \( \Re^n \) are perpendicular if and
only if their dot product is zero.
\begin{exparts}
......@@ -6456,7 +6456,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.41}
\begin{ans}{Two.II.1.42}
\begin{exparts}
\partsitem This check is routine.
\partsitem The summation is infinite (has infinitely many summands).
......@@ -6473,7 +6473,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Two.II.1.42}
\begin{ans}{Two.II.1.43}
It is both `if' and `only if'.
Let \( T \) be a subset of the subspace \( S \) of the vector space
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......@@ -2037,9 +2037,9 @@ type of subspaces, so in fact this picture shows them all.
\put(45,0){\makebox(0,0){\tiny\( \set{\colvec[r]{0 \\ 0 \\ 0} } \)} }
\end{picture}
\end{center}
They are described as spans of sets with a minimal number of members,
The subspaces are described as spans of sets with a minimal number of members
and are shown connected to their supersets.
Note that these subspaces fall naturally into levels\Dash planes on one level,
Note that the subspaces fall naturally into levels\Dash planes on one level,
lines on another,
etc.\Dash according to how many vectors are in a minimal-sized
spanning set.
......
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