Commit 4c31b009 authored by Jim Hefferon's avatar Jim Hefferon

adjust cover slightly

parent 49ea36fa
...@@ -35,9 +35,15 @@ defaultpen(p); ...@@ -35,9 +35,15 @@ defaultpen(p);
// From version 3 of the book, 2016 // From version 3 of the book, 2016
// // From https://color.adobe.com/Ice-cream-chill-color-theme-6611012/edit/?copy=true&base=2&rule=Custom&selected=4&name=Copy%20of%20Ice%20cream%20chill&mode=rgb&rgbvalues=0.611765,0.329412,0.290196,0.705882,0.745098,0.572549,0.992157,0.894118,0.733333,0.788235,0.803922,0.733333,0.886275,0.67451,0.435294&swatchOrder=0,1,2,3,4 // // From https://color.adobe.com/Ice-cream-chill-color-theme-6611012/edit/?copy=true&base=2&rule=Custom&selected=4&name=Copy%20of%20Ice%20cream%20chill&mode=rgb&rgbvalues=0.611765,0.329412,0.290196,0.705882,0.745098,0.572549,0.992157,0.894118,0.733333,0.788235,0.803922,0.733333,0.886275,0.67451,0.435294&swatchOrder=0,1,2,3,4
pen darkcolor = rgb(226/255, 172/255, 111/255); // pen darkcolor = rgb(180/255, 190/255, 146/255);
pen lightcolor = rgb(180/255, 190/255, 146/255); // pen boldcolor = rgb(156/255, 84/255, 74/255);
pen boldcolor = rgb(156/255, 84/255, 74/255); // pen lightcolor = rgb(226/255, 172/255, 111/255);
// pen bgcolor = rgb(201/255, 205/255, 187/255);
// pen flourishcolor = rgb(253/255, 228/255, 187/255)
pen darkcolor = rgb(180/255, 190/255, 146/255); // green
pen boldcolor = rgb(191/255, 103/255, 91/255);
pen lightcolor = rgb(226/255, 172/255, 111/255);
pen bgcolor = rgb(201/255, 205/255, 187/255);
pen flourishcolor = rgb(253/255, 228/255, 187/255); pen flourishcolor = rgb(253/255, 228/255, 187/255);
real XLIMIT_POS=1; real XLIMIT_POS=1;
......
...@@ -29935,49 +29935,6 @@ octave:6> gplot z ...@@ -29935,49 +29935,6 @@ octave:6> gplot z
\end{ans} \end{ans}
\begin{ans}{Five.II.3.24} \begin{ans}{Five.II.3.24}
\begin{exparts}
\item The characteristic polynomial factors as
$x^2-20x+75=(x-5)(x-15)$,
so the eigenvalues are $\lambda_1=5$ and~$\lambda_2=15$.
These are the associated eigenspaces.
\begin{equation*}
V_5=\set{k\colvec{1 \\ 2}\suchthat k\in\C}
\quad
V_{15}=\set{k\colvec{-2 \\ 1}\suchthat k\in\C}
\end{equation*}
For each eigenvalue, both the algebraic and geometric multiplicities
are~$1$.
\item The characteristic polynomial $x^3 - 6x^2 + 32$ factors into
$(x+2)(x-4)^2$.
The eigenvectors are $\lambda_1=-2$ and $\lambda_2=4$.
Here are the associated eigenspaces.
\begin{equation*}
V_{-2}=\set{k\colvec{1 \\ 1 \\ 2}\suchthat k\in\C}
\quad
V_4=\set{k_1\colvec{1 \\ 1 \\ 0}
+k_2\colvec{1 \\ 0 \\ -1} \suchthat k_1,k_2\in\C}
\end{equation*}
For $\lambda_1=-2$ the algebraic and geometric multiplicities are
both~$1$.
For $\lambda_2=4$ the algebraic and geometric multiplicities are
both~$2$.
\item The characteristic polynomial $x^3 - 5x^2 + 8x - 4$ factors into
$(x-1)(x-2)^2$.
The eigenvalues are $\lambda_1=1$ and $\lambda_2=2$.
Here are the associated eigenspaces.
\begin{equation*}
V_1=\set{k\colvec{-6 \\ 3 \\ 1}\suchthat k\in \C}
\quad
V_2=\set{k\colvec{1 \\ 0 \\ 0}\suchthat k\in\C}
\end{equation*}
For $\lambda_1=1$ the algebraic and geometric multiplicities
are both~$1$.
For $\lambda_2=2$ the algebraic multiplicity is~$2$ but the
geometric multiplicity is~$1$.
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.25}
The characteristic equation The characteristic equation
\begin{equation*} \begin{equation*}
0= 0=
...@@ -30045,7 +30002,7 @@ octave:6> gplot z ...@@ -30045,7 +30002,7 @@ octave:6> gplot z
\end{equation*} \end{equation*}
\end{ans} \end{ans}
\begin{ans}{Five.II.3.26} \begin{ans}{Five.II.3.25}
The characteristic equation is The characteristic equation is
\begin{equation*} \begin{equation*}
0= 0=
...@@ -30083,7 +30040,7 @@ octave:6> gplot z ...@@ -30083,7 +30040,7 @@ octave:6> gplot z
\end{equation*} \end{equation*}
\end{ans} \end{ans}
\begin{ans}{Five.II.3.27} \begin{ans}{Five.II.3.26}
\begin{exparts} \begin{exparts}
\partsitem The characteristic equation is \partsitem The characteristic equation is
\begin{equation*} \begin{equation*}
...@@ -30285,6 +30242,49 @@ octave:6> gplot z ...@@ -30285,6 +30242,49 @@ octave:6> gplot z
\end{equation*} \end{equation*}
\end{exparts} \end{exparts}
\end{ans}
\begin{ans}{Five.II.3.27}
\begin{exparts}
\item The characteristic polynomial factors as
$x^2-20x+75=(x-5)(x-15)$,
so the eigenvalues are $\lambda_1=5$ and~$\lambda_2=15$.
These are the associated eigenspaces.
\begin{equation*}
V_5=\set{k\colvec{1 \\ 2}\suchthat k\in\C}
\quad
V_{15}=\set{k\colvec{-2 \\ 1}\suchthat k\in\C}
\end{equation*}
For each eigenvalue, both the algebraic and geometric multiplicities
are~$1$.
\item The characteristic polynomial $x^3 - 6x^2 + 32$ factors into
$(x+2)(x-4)^2$.
The eigenvectors are $\lambda_1=-2$ and $\lambda_2=4$.
Here are the associated eigenspaces.
\begin{equation*}
V_{-2}=\set{k\colvec{1 \\ 1 \\ 2}\suchthat k\in\C}
\quad
V_4=\set{k_1\colvec{1 \\ 1 \\ 0}
+k_2\colvec{1 \\ 0 \\ -1} \suchthat k_1,k_2\in\C}
\end{equation*}
For $\lambda_1=-2$ the algebraic and geometric multiplicities are
both~$1$.
For $\lambda_2=4$ the algebraic and geometric multiplicities are
both~$2$.
\item The characteristic polynomial $x^3 - 5x^2 + 8x - 4$ factors into
$(x-1)(x-2)^2$.
The eigenvalues are $\lambda_1=1$ and $\lambda_2=2$.
Here are the associated eigenspaces.
\begin{equation*}
V_1=\set{k\colvec{-6 \\ 3 \\ 1}\suchthat k\in \C}
\quad
V_2=\set{k\colvec{1 \\ 0 \\ 0}\suchthat k\in\C}
\end{equation*}
For $\lambda_1=1$ the algebraic and geometric multiplicities
are both~$1$.
For $\lambda_2=2$ the algebraic multiplicity is~$2$ but the
geometric multiplicity is~$1$.
\end{exparts}
\end{ans} \end{ans}
\begin{ans}{Five.II.3.28} \begin{ans}{Five.II.3.28}
With respect to the natural basis $B=\sequence{1,x,x^2}$ With respect to the natural basis $B=\sequence{1,x,x^2}$
...@@ -11,8 +11,10 @@ ...@@ -11,8 +11,10 @@
% \definecolor{coverflourishcolor}{HTML}{AB1A25} % % \definecolor{coverflourishcolor}{HTML}{AB1A25} %
% Cover for version 3, 2016: % Cover for version 3, 2016:
% % From https://color.adobe.com/Ice-cream-chill-color-theme-6611012/edit/?copy=true&base=2&rule=Custom&selected=3&name=Copy%20of%20Ice%20cream%20chill&mode=rgb&rgbvalues=0.611765,0.329412,0.290196,0.705882,0.745098,0.572549,0.992157,0.894118,0.733333,0.788235,0.803922,0.733333,0.886275,0.67451,0.435294&swatchOrder=0,1,2,3,4 % % From https://color.adobe.com/Ice-cream-chill-color-theme-6611012/edit/?copy=true&base=2&rule=Custom&selected=3&name=Copy%20of%20Ice%20cream%20chill&mode=rgb&rgbvalues=0.611765,0.329412,0.290196,0.705882,0.745098,0.572549,0.992157,0.894118,0.733333,0.788235,0.803922,0.733333,0.886275,0.67451,0.435294&swatchOrder=0,1,2,3,4
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...@@ -25,10 +27,11 @@ ...@@ -25,10 +27,11 @@
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......
...@@ -2858,14 +2858,13 @@ It is a nontrivial subspace. ...@@ -2858,14 +2858,13 @@ It is a nontrivial subspace.
\begin{proof} \begin{proof}
%<*pf:EigSpaceIsSubSp> %<*pf:EigSpaceIsSubSp>
To show that an eigenspace is a subspace, Notice first that
notice first that $V_{\lambda}$ is not empty; it contains the zero
$V_{\lambda}$ contains the zero
vector since $t(\zero)=\zero$, which equals vector since $t(\zero)=\zero$, which equals
$\lambda\cdot \zero$. $\lambda\cdot \zero$.
So the eigenspace is a nonempty subset of the space. To show that an eigenspace is a subspace,
What remains is to check closure of this set under linear combinations. what remains is to check closure of this set under linear combinations.
Take \( \vec{\zeta}_1,\ldots,\vec{\zeta}_n\in V_{\lambda} \) and then verify Take \( \vec{\zeta}_1,\ldots,\vec{\zeta}_n\in V_{\lambda} \) and then
\begin{align*} \begin{align*}
t(\lincombo{c}{\vec{\zeta}}) t(\lincombo{c}{\vec{\zeta}})
&=c_1t(\vec{\zeta}_1)+\dots+c_nt(\vec{\zeta}_n) \\ &=c_1t(\vec{\zeta}_1)+\dots+c_nt(\vec{\zeta}_n) \\
...@@ -3290,71 +3289,6 @@ In the next section we study matrices that cannot be diagonalized. ...@@ -3290,71 +3289,6 @@ In the next section we study matrices that cannot be diagonalized.
\end{equation*} \end{equation*}
\end{exparts} \end{exparts}
\end{answer} \end{answer}
\item % TODO: credit http://algebra.math.ust.hk/eigen/05_multiplicity/lecture2.shtml
For each matrix, find the characteristic equation and the
eigenvalues, and associated eigenspaces.
Also find the algebraic and geometric multiplicities.
\begin{exparts*}
\partsitem
$\begin{mat}
13 &-4 \\
-4 &7
\end{mat}$
\partsitem
$\begin{mat}
1 &3 &-3 \\
-3 &7 &-3 \\
-6 &6 &-2
\end{mat}$
\partsitem
$\begin{mat}
2 &3 &-3 \\
0 &2 &-3 \\
0 &0 &1
\end{mat}$
\end{exparts*}
\begin{answer}
\begin{exparts}
\item The characteristic polynomial factors as
$x^2-20x+75=(x-5)(x-15)$,
so the eigenvalues are $\lambda_1=5$ and~$\lambda_2=15$.
These are the associated eigenspaces.
\begin{equation*}
V_5=\set{k\colvec{1 \\ 2}\suchthat k\in\C}
\quad
V_{15}=\set{k\colvec{-2 \\ 1}\suchthat k\in\C}
\end{equation*}
For each eigenvalue, both the algebraic and geometric multiplicities
are~$1$.
\item The characteristic polynomial $x^3 - 6x^2 + 32$ factors into
$(x+2)(x-4)^2$.
The eigenvectors are $\lambda_1=-2$ and $\lambda_2=4$.
Here are the associated eigenspaces.
\begin{equation*}
V_{-2}=\set{k\colvec{1 \\ 1 \\ 2}\suchthat k\in\C}
\quad
V_4=\set{k_1\colvec{1 \\ 1 \\ 0}
+k_2\colvec{1 \\ 0 \\ -1} \suchthat k_1,k_2\in\C}
\end{equation*}
For $\lambda_1=-2$ the algebraic and geometric multiplicities are
both~$1$.
For $\lambda_2=4$ the algebraic and geometric multiplicities are
both~$2$.
\item The characteristic polynomial $x^3 - 5x^2 + 8x - 4$ factors into
$(x-1)(x-2)^2$.
The eigenvalues are $\lambda_1=1$ and $\lambda_2=2$.
Here are the associated eigenspaces.
\begin{equation*}
V_1=\set{k\colvec{-6 \\ 3 \\ 1}\suchthat k\in \C}
\quad
V_2=\set{k\colvec{1 \\ 0 \\ 0}\suchthat k\in\C}
\end{equation*}
For $\lambda_1=1$ the algebraic and geometric multiplicities
are both~$1$.
For $\lambda_2=2$ the algebraic multiplicity is~$2$ but the
geometric multiplicity is~$1$.
\end{exparts}
\end{answer}
\item \item
Find the characteristic equation, and the Find the characteristic equation, and the
eigenvalues and associated eigenvectors for this matrix. eigenvalues and associated eigenvectors for this matrix.
...@@ -3697,11 +3631,76 @@ In the next section we study matrices that cannot be diagonalized. ...@@ -3697,11 +3631,76 @@ In the next section we study matrices that cannot be diagonalized.
\end{equation*} \end{equation*}
\end{exparts} \end{exparts}
\end{answer} \end{answer}
\item
For each matrix, find the characteristic polynomial, and the
eigenvalues and associated eigenspaces.
Also find the algebraic and geometric multiplicities.
\begin{exparts*}
\partsitem
$\begin{mat}
13 &-4 \\
-4 &7
\end{mat}$
\partsitem
$\begin{mat}
1 &3 &-3 \\
-3 &7 &-3 \\
-6 &6 &-2
\end{mat}$
\partsitem
$\begin{mat}
2 &3 &-3 \\
0 &2 &-3 \\
0 &0 &1
\end{mat}$
\end{exparts*}
\begin{answer}
\begin{exparts}
\item The characteristic polynomial factors as
$x^2-20x+75=(x-5)(x-15)$,
so the eigenvalues are $\lambda_1=5$ and~$\lambda_2=15$.
These are the associated eigenspaces.
\begin{equation*}
V_5=\set{k\colvec{1 \\ 2}\suchthat k\in\C}
\quad
V_{15}=\set{k\colvec{-2 \\ 1}\suchthat k\in\C}
\end{equation*}
For each eigenvalue, both the algebraic and geometric multiplicities
are~$1$.
\item The characteristic polynomial $x^3 - 6x^2 + 32$ factors into
$(x+2)(x-4)^2$.
The eigenvectors are $\lambda_1=-2$ and $\lambda_2=4$.
Here are the associated eigenspaces.
\begin{equation*}
V_{-2}=\set{k\colvec{1 \\ 1 \\ 2}\suchthat k\in\C}
\quad
V_4=\set{k_1\colvec{1 \\ 1 \\ 0}
+k_2\colvec{1 \\ 0 \\ -1} \suchthat k_1,k_2\in\C}
\end{equation*}
For $\lambda_1=-2$ the algebraic and geometric multiplicities are
both~$1$.
For $\lambda_2=4$ the algebraic and geometric multiplicities are
both~$2$.
\item The characteristic polynomial $x^3 - 5x^2 + 8x - 4$ factors into
$(x-1)(x-2)^2$.
The eigenvalues are $\lambda_1=1$ and $\lambda_2=2$.
Here are the associated eigenspaces.
\begin{equation*}
V_1=\set{k\colvec{-6 \\ 3 \\ 1}\suchthat k\in \C}
\quad
V_2=\set{k\colvec{1 \\ 0 \\ 0}\suchthat k\in\C}
\end{equation*}
For $\lambda_1=1$ the algebraic and geometric multiplicities
are both~$1$.
For $\lambda_2=2$ the algebraic multiplicity is~$2$ but the
geometric multiplicity is~$1$.
\end{exparts}
\end{answer}
\recommended \item \recommended \item
Let \( \map{t}{\polyspace_2}{\polyspace_2} \) be Let \( \map{t}{\polyspace_2}{\polyspace_2} \) be this linear map.
\begin{equation*} \begin{equation*}
a_0+a_1x+a_2x^2\mapsto a_0+a_1x+a_2x^2\mapsto
(5a_0+6a_1+2a_2)-(a_1+8a_2)x+(a_0-2a_2)x^2. (5a_0+6a_1+2a_2)-(a_1+8a_2)x+(a_0-2a_2)x^2
\end{equation*} \end{equation*}
Find its eigenvalues and the associated eigenvectors. Find its eigenvalues and the associated eigenvectors.
\begin{answer} \begin{answer}
......
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