Commit 924664ea authored by Jim Hefferon's avatar Jim Hefferon

jc3 edits

parent 1f04eaf8
This diff is collapsed.
......@@ -26456,12 +26456,12 @@ ans = 0.017398
\begin{equation*}
\Re^2
\supset\set{\colvec{0 \\ p}\suchthat p\in\C}
\supset\set{\zero\,}
\supset\set{\zero}
=\cdots
\end{equation*}
and the chain of nullspaces
\begin{equation*}
\set{\zero\,}
\set{\zero}
\subset\set{\colvec{q \\ 0}\suchthat q\in\C}
\subset\Re^2
=\cdots
......@@ -26481,11 +26481,11 @@ ans = 0.017398
\begin{equation*}
\polyspace_2=\polyspace_2=\cdots
\qquad
\set{\zero\,}=\set{\zero\,}=\cdots
\set{\zero}=\set{\zero}=\cdots
\end{equation*}
Thus, obviously,
generalized spaces are $\genrangespace{t_2}=\polyspace_2$
and $\gennullspace{t_2}=\set{\zero\,}$.
and $\gennullspace{t_2}=\set{\zero}$.
\partsitem We have
\begin{equation*}
\colvec{a \\ b \\ c}
......@@ -26503,7 +26503,7 @@ ans = 0.017398
\end{equation*}
and the chain of nullspaces
\begin{equation*}
\set{\zero\,}
\set{\zero}
\subset\set{\colvec{0 \\ 0 \\ r}\suchthat r\in \C}
\subset\set{\colvec{0 \\ q \\ r}\suchthat q,r\in \C}
=\cdots
......@@ -26573,14 +26573,14 @@ ans = 0.017398
\end{ans}
\subsection{Subsection Five.III.2: Strings}
\begin{ans}{Five.III.2.17}
\begin{ans}{Five.III.2.18}
Three.
It is at least three because $\ell^2(\,(1,1,1)\,)=(0,0,1)\neq \zero$.
It is at most three because
$(x,y,z)\mapsto (0,x,y)\mapsto (0,0,x)\mapsto (0,0,0)$.
\end{ans}
\begin{ans}{Five.III.2.18}
\begin{ans}{Five.III.2.19}
\begin{exparts}
\partsitem The domain has dimension four.
The map's action is that any vector in the space
......@@ -26647,7 +26647,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.III.2.19}
\begin{ans}{Five.III.2.20}
By \nearbylemma{le:RangeAndNullChains} the nullity has grown as
large as possible by the $n$-th iteration where $n$ is the dimension
of the domain.
......@@ -26689,7 +26689,7 @@ ans = 0.017398
is to note that they are nonsingular.
\end{ans}
\begin{ans}{Five.III.2.20}
\begin{ans}{Five.III.2.21}
The table os calculations
\begin{center}
\begin{tabular}{c|cc}
......@@ -26744,7 +26744,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.III.2.21}
\begin{ans}{Five.III.2.22}
\begin{exparts}
\partsitem The canonical form has a $\nbyn{3}$ block and a
$\nbyn{2}$ block
......@@ -26803,7 +26803,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.III.2.22}
\begin{ans}{Five.III.2.23}
\begin{exparts*}
\partsitem The calculation
\begin{center}
......@@ -26911,7 +26911,7 @@ ans = 0.017398
\end{exparts*}
\end{ans}
\begin{ans}{Five.III.2.23}
\begin{ans}{Five.III.2.24}
A couple of examples
\begin{equation*}
\begin{mat}[r]
......@@ -26975,7 +26975,7 @@ ans = 0.017398
See \nearbyexercise{exer:IndNilLftShift}.
\end{ans}
\begin{ans}{Five.III.2.24}
\begin{ans}{Five.III.2.25}
Yes.
Generalize the last sentence in \nearbyexample{ex:NilMatNotCanon}.
As to the index, that same last sentence shows that the index of the new
......@@ -26990,7 +26990,7 @@ ans = 0.017398
This is \nearbyexercise{exer:MatNilIffMapNil} below.
\end{ans}
\begin{ans}{Five.III.2.25}
\begin{ans}{Five.III.2.26}
Observe that a canonical form nilpotent matrix has only
zero eigenvalues; e.g., the determinant of this lower-triangular matrix
\begin{equation*}
......@@ -27010,19 +27010,19 @@ ans = 0.017398
$\lambda$ is zero.
\end{ans}
\begin{ans}{Five.III.2.26}
\begin{ans}{Five.III.2.27}
No, by \nearbylemma{le:RangeAndNullChains} for a map on a
two-dimensional space, the nullity has grown
as large as possible by the second iteration.
\end{ans}
\begin{ans}{Five.III.2.27}
\begin{ans}{Five.III.2.28}
The index of nilpotency of a transformation can be zero only when the
vector starting the string
must be $\zero$, that is, only when $V$ is a trivial space.
\end{ans}
\begin{ans}{Five.III.2.28}
\begin{ans}{Five.III.2.29}
\begin{exparts}
\partsitem Any member $\vec{w}$ of the span can be written as
a linear combination
......@@ -27057,7 +27057,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.III.2.29}
\begin{ans}{Five.III.2.30}
We must check that
\( B\union\hat{C}\union\set{\vec{v}_1,\dots,\vec{v}_j} \) is linearly
independent where \( B \) is a \( t \)-string basis for
......@@ -27086,7 +27086,7 @@ ans = 0.017398
the remaining coefficients are zero also.
\end{ans}
\begin{ans}{Five.III.2.30}
\begin{ans}{Five.III.2.31}
For any basis $B$,
a transformation~$n$ is nilpotent if and only if
$N=\rep{n}{B,B}$ is a nilpotent matrix.
......@@ -27095,7 +27095,7 @@ ans = 0.017398
is the zero matrix.
\end{ans}
\begin{ans}{Five.III.2.31}
\begin{ans}{Five.III.2.32}
It can be of any size greater than or equal to one.
To have a transformation that is nilpotent of index four,
whose cube has rangespace of dimension~$k$, take a vector space,
......@@ -27122,7 +27122,7 @@ ans = 0.017398
would not be four.
\end{ans}
\begin{ans}{Five.III.2.32}
\begin{ans}{Five.III.2.33}
These two have only zero for eigenvalues
\begin{equation*}
\begin{mat}
......@@ -27139,7 +27139,7 @@ ans = 0.017398
representatives, namely, themselves).
\end{ans}
\begin{ans}{Five.III.2.33}
\begin{ans}{Five.III.2.34}
A simple reordering of the string basis will do.
For instance, a map that is assoicated with this string basis
\begin{equation*}
......@@ -27165,14 +27165,14 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.III.2.34}
\begin{ans}{Five.III.2.35}
Let $\map{t}{V}{V}$ be the transformation.
If \( \rank (t)=\nullity (t) \) then the equation
\( \rank(t)+\nullity(t)=\dim (V) \) shows that
\( \dim (V) \) is even.
\end{ans}
\begin{ans}{Five.III.2.35}
\begin{ans}{Five.III.2.36}
For the matrices to be nilpotent they must be square.
For them to commute they must be the same size.
Thus their product and sum are defined.
......@@ -27188,7 +27188,7 @@ ans = 0.017398
The sum is similar; use the Binomial Theorem.
\end{ans}
\begin{ans}{Five.III.2.36}
\begin{ans}{Five.III.2.37}
Some experimentation gives the idea for the proof.
Expansion of the second power
\begin{equation*}
......@@ -27217,7 +27217,7 @@ ans = 0.017398
has a power higher than $k$, and so the term gives the zero matrix.
\end{ans}
\begin{ans}{Five.III.2.37}
\begin{ans}{Five.III.2.38}
Use the geometric series:
$
I-N^{k+1}=(I-N)(N^k+N^{k-1}+\cdots+I)
......@@ -300,7 +300,7 @@ beginfig(8) % graph of rank falling and nullity rising
% label.lrt(btex \small power etex,z2);
% label.ulft(btex \small rank etex,z4);
% point of interest
z10=(10w,4v);
z10=(10w,3.75v);
% ticks
pickup updown_tick;
for i = 1 upto 3:
......@@ -314,7 +314,7 @@ beginfig(8) % graph of rank falling and nullity rising
for i = 17 upto 21:
drawdot (i*w,0v);
endfor
label.bot(btex \scriptsize $k$ etex,(18w,0v));
% label.bot(btex \scriptsize $k$ etex,(18w,0v));
label.bot(btex \scriptsize $n$ etex,(20w,0v));
% label.bot(btex \scriptsize $k+1$ etex,(14w,0v));
pickup sidetoside_tick;
......@@ -334,20 +334,33 @@ beginfig(8) % graph of rank falling and nullity rising
pickup pensquare scaled line_width_light;
draw (x10,y10+0.5v)--(x10,y4-0.5v) withcolor .8white; % vert line above dot
draw (x10-.5whisker_width,y4-0.5v)--(x10+.5whisker_width,y4-0.5v) withcolor .8white; % whisker at top
draw (x10-.5whisker_width,y10+0.5v)--(x10+.5whisker_width,y10+0.5v) withcolor .8white; % whisker at top
draw (x10-.5whisker_width,y10+0.5v)--(x10+.5whisker_width,y10+0.5v) withcolor .8white; % whisker at bot
label.rt(btex \scriptsize $\nullity (t^j)$ etex,(x10,.5[y10,y4]));
draw (x10,y10-0.5v)--(x10,1v) withcolor .8white; % vert line below dot
draw (x10-.5whisker_width,y10-0.5v)--(x10+.5whisker_width,y10-0.5v) withcolor .8white; % whisker at top
draw (x10-.5whisker_width,1v)--(x10+.5whisker_width,1v) withcolor .8white; % whisker at bot
label.lft(btex \scriptsize $\rank (t^j)$ etex,(x10,.5[y10,1v]));
label.lft(btex \scriptsize $\rank (t^j)$ etex,(x10,.55[y10,1v]));
pickup pencircle scaled line_width_light;
drawpoint((1w,9v));
drawpoint((2w,7v));
drawpoint((3w,6v));
drawpoint((17w,2.25v));
drawpoint((17w,2.2v));
drawpoint((18w,2v));
drawpoint((19w,2v));
drawpoint((20w,2v));
% drawpoint((21w,2v));
label(btex \scriptsize $\ldots$ etex,(21w,2v));
% Generalized spaces
z20=(20w,2v); % were to anchor the line for generalized spaces
pickup pensquare scaled line_width_light;
draw (x20,y20+0.5v)--(x20,y4-0.5v) withcolor .8white; % vert line above dot
draw (x20-.5whisker_width,y4-0.5v)--(x20+.5whisker_width,y4-0.5v) withcolor .8white; % whisker at top
draw (x20-.5whisker_width,y20+0.5v)--(x20+.5whisker_width,y20+0.5v) withcolor .8white; % whisker at bot
label.rt(btex \scriptsize $\dim(\gennullspace{t})$ etex,(x20,.5[y20,y4]));
draw (x20,y20-0.5v)--(x20,1v) withcolor .8white; % vert line below dot
draw (x20-.5whisker_width,y20-0.5v)--(x20+.5whisker_width,y20-0.5v) withcolor .8white; % whisker at top
draw (x20-.5whisker_width,1v)--(x20+.5whisker_width,1v) withcolor .8white; % whisker at bot
label.rt(btex \scriptsize $\dim(\genrangespace{t})$ etex,(x20,.8[y20,1v]));
endfig;
......
This diff is collapsed.
......@@ -2202,7 +2202,7 @@ Thus the characteristic equation of \( t \) on
And thus $q_j=0$ for all $j\neq i$.
Now consider the restriction of \( t \) to \( \genrangespace{t-\lambda_i} \).
By Note~II.\ref{note:RestONeToOne}, the map
By Lemma~III.\ref{lem:RestONeToOne}, the map
\( t-\lambda_i \) is one-to-one on
\( \genrangespace{t-\lambda_i} \) and so \( \lambda_i \) is not an
eigenvalue of \( t \) on that subspace.
......
This diff is collapsed.
......@@ -268,6 +268,15 @@
\put(0,-3){\makebox[0pt]{#1}}
\end{picture}}}
% In tables of matrices, add a bit of extra space above and below each line.
% Note that the included material is set in math mode
\newlength{\extramatrixvspace}
\setlength{\extramatrixvspace}{.75ex}
\newcommand{\matrixvenlarge}[1]{\raisebox{-0.5\height}{\vbox{
\vspace*{\extramatrixvspace}
\hbox{$#1$}
\vspace*{\extramatrixvspace}
}}}
% from ltugboat.cls 2000-Apr-27 for making dashes.
......
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