Commit 924664ea by Jim Hefferon

### jc3 edits

parent 1f04eaf8
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 ... ... @@ -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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