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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 5 "     " }{TEXT 256 0 "" }{TEXT 
257 11 "MATH 2270-1" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "
" }{TEXT 259 39 "Symmetric matrices, conics and quadrics" }}{PARA 258 
"" 0 "" {TEXT -1 16 "December 2, 2005" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 329 "Since this is formally your third Ma
ple project, although it is really just part of your homework, you may
 again work in groups of up to three individuals, and you may do this \+
group work for the entire homework assignment due December 9.  So, eac
h group hands in one assignment, with the group member names prominent
ly displayed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 260 14 "Conic sections" }{TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 133 "restart:with(linalg):with(plots):#for computation
s and pictures\nwith(student):#to do algebra computations like complet
ing the square\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 128 "Let's do the problem we did in class on Wednesday, using
 Maple. First, recall what we did by hand:  We started with the equati
on" }}{PARA 262 "" 0 "" {XPPEDIT 18 0 "2*x^2+2*y^2+5*x*y = 1;" "6#/,(*
&\"\"#\"\"\")%\"xGF&F'F'*&F&F')%\"yGF&F'F'*(\"\"&F'F)F'F,F'F'F'" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 57 "We rewrote this equation
 using a 2 by 2 symmetric matrix:" }}{PARA 263 "" 0 "" {XPPEDIT 18 0 "
matrix([[x, y]])*matrix([[2, 5/2], [5/2, 2]])*matrix([[x], [y]]) = 1;
" "6#/*(-%'matrixG6#7#7$%\"xG%\"yG\"\"\"-F&6#7$7$\"\"##\"\"&F17$F2F1F,
-F&6#7$7#F*7#F+F,F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 137 "B
y hand, we computed the eigenvalues and eigenvectors of this matrix.  \+
Since we're currently using Maple, we can verify Wednesday's work:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "A:=matrix(2,2,[2,5/2,5/2,2])
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "eigenvectors(A); #comp
ute eigenvalues, algebraic multiplicities,\n  #and bases for each eige
nspace" }}}{PARA 0 "" 0 "" {TEXT -1 98 "As I can show you now on the b
lackboard, and/or show you again in class on Monday, eigenvectors to" 
}}{PARA 0 "" 0 "" {TEXT -1 269 "symmetric matrices with different eige
nvalues are always perpendicular.  So, for this matrix A, I can define
 an orthonormal eigenbasis, and I'll choose to make it positively orie
nted.  These two orthonormal basis vectors are the columns of the chan
ge of basis matrix S:" }}{PARA 264 "" 0 "" {XPPEDIT 18 0 "S := matrix(
[[1/2*2^(1/2), -1/2*2^(1/2)], [1/2*2^(1/2), 1/2*2^(1/2)]]);" "6#>%\"SG
-%'matrixG6#7$7$,$*&\"\"#!\"\"F,#\"\"\"F,F/,$*&F,F-F,F.F-7$F*F*" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 15 "Precisely, if  " }
{XPPEDIT 18 0 "matrix([[u], [v]]);" "6#-%'matrixG6#7$7#%\"uG7#%\"vG" }
{TEXT -1 70 " are the coordinates of a point with respect to our new b
asis, and if " }{XPPEDIT 18 0 "matrix([[x], [y]]);" "6#-%'matrixG6#7$7
#%\"xG7#%\"yG" }{TEXT -1 62 " are the coordinates with respect to our \+
standard basis, then " }}{PARA 267 "" 0 "" {XPPEDIT 18 0 "matrix([[x],
 [y]]) = [S]*matrix([[u], [v]]);" "6#/-%'matrixG6#7$7#%\"xG7#%\"yG*&7#
%\"SG\"\"\"-F%6#7$7#%\"uG7#%\"vGF/" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 96 "Then, from our discussion of diagonalization we know that
 for the diagonal matrix of eigenvalues" }}{PARA 266 "" 0 "" {XPPEDIT 
18 0 "Lambda := matrix([[9/2, 0], [0, -1/2]]);" "6#>%'LambdaG-%'matrix
G6#7$7$#\"\"*\"\"#\"\"!7$F-#!\"\"F," }{TEXT -1 1 "," }}{PARA 265 "" 0 
"" {XPPEDIT 18 0 "Lambda = [S]^T*[A]*[S];" "6#/%'LambdaG*()7#%\"SG%\"T
G\"\"\"7#%\"AGF*F'F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 93 "P
utting all of this together, points whose standard coordinates satisfy
 the original equation" }}{PARA 268 "" 0 "" {XPPEDIT 18 0 "matrix([[x,
 y]])*[A]*matrix([[x], [y]]) = 1;" "6#/*(-%'matrixG6#7#7$%\"xG%\"yG\"
\"\"7#%\"AGF,-F&6#7$7#F*7#F+F,F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 60 "have coordinates in the new basis which satisfy the equat
ion" }}{PARA 269 "" 0 "" {XPPEDIT 18 0 "matrix([[u, v]])*[S]^T*[A]*[S]
*matrix([[u], [v]]) = 1;" "6#/*,-%'matrixG6#7#7$%\"uG%\"vG\"\"\")7#%\"
SG%\"TGF,7#%\"AGF,F.F,-F&6#7$7#F*7#F+F,F," }{TEXT -1 1 "," }}{PARA 
270 "" 0 "" {XPPEDIT 18 0 "matrix([[u, v]])*[Lambda]*matrix([[u], [v]]
) = 1;" "6#/*(-%'matrixG6#7#7$%\"uG%\"vG\"\"\"7#%'LambdaGF,-F&6#7$7#F*
7#F+F,F," }{TEXT -1 1 "," }}{PARA 271 "" 0 "" {XPPEDIT 18 0 "9/2*u^2-1
/2*v^2 = 1;" "6#/,&*(\"\"*\"\"\"\"\"#!\"\"%\"uGF(F'*&F(F)%\"vGF(F)F'" 
}{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 133 "We recognize this as a
 hyperbola opening along the u-axis, i.e. along the axis of our first \+
basis vector.  In fact, here's a picture:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "implicitplot(2*x^2+2*
y^2+5*x*y= 1,\n       x=-3..3,y=-3..3, grid=[100,100],\n       color=`
black`);\n" }}}{PARA 0 "" 0 "" {TEXT -1 33 "There are three things to \+
notice:" }}{PARA 0 "" 0 "" {TEXT -1 481 "(A)  Just knowing the eigenva
lues of the matrix A allowed us to deduce that the curve had to be a r
otated hyperbola!  In general, the eigenvalue information of a quadrat
ic equation in two variables will almost tell you enough to deduce whi
ch conic it describes.  If you have a quadratic equation in three vari
ables, the eigenvalue information will limit the quadric surface possi
bilities, but you will often need to do more algebra to deduce exactly
 which creature you've captured." }}{PARA 0 "" 0 "" {TEXT -1 567 "(B) \+
 The spectral theorem, which we will talk about on Monday, says that s
ymmetric matrices always have orthonnormal eigenbases, so the recipe w
e followed above in a particular case, will let us diagonalize any sym
metric matrix (and the corresponding quadratic expression which is cal
led a quadratic form),  finding new coordinates in which these \"quadr
atic forms\" and equations become diagonal (have no cross terms).  The
re are important applications of this procedure, beyond the work with \+
conics and quartics which we play with in this Maple file and the home
work." }}{PARA 0 "" 0 "" {TEXT -1 149 "(C)  We could be taking a lot m
ore advantage of the fact that we're using Maple, to automate this pro
cess of identifying conics and quadric surfaces." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "     Here's how to make M
aple help you." }}{PARA 0 "" 0 "" {TEXT -1 148 "     Let's do #30 page
 423 of the Kolman text handout: We wish to identify the conic of poin
ts whose standard coordinates [x,y] satisfy the equation" }}{PARA 260 
"" 0 "" {XPPEDIT 18 0 "33*sqrt(2)*x-31*sqrt(2)*y+8*x^2-16*x*y+8*y^2+70
 = 0;" "6#/,.*&-%%sqrtG6#\"\"#\"\"\"%\"xGF*\"#L*(\"#JF*F&F*%\"yGF*!\"
\"*&\"\")F*)F+F)F*F**(\"#;F*F+F*F/F*F0*&F2F*)F/F)F*F*\"#qF*\"\"!" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 120 "Notice there are linear
 terms so this is a rotated AND translated conic!  We will write this \+
equation in matrix form, as" }}{PARA 261 "" 0 "" {XPPEDIT 18 0 "x^T*A*
x+B*x+C;" "6#,(*()%\"xG%\"TG\"\"\"%\"AGF(F&F(F(*&%\"BGF(F&F(F(%\"CGF(
" }{TEXT -1 4 " = 0" }}{PARA 0 "" 0 "" {TEXT -1 17 "where the vector \+
" }{TEXT 263 1 "x" }{TEXT -1 26 " is the coordinate vector " }
{XPPEDIT 18 0 "matrix([[x], [y]]);" "6#-%'matrixG6#7$7#%\"xG7#%\"yG" }
{TEXT -1 297 ", A is a symmetric two by two matrix, B is a one by two \+
matrix, C is a scalar.  Remember how to turn the quadratic expression \+
into a matrix product, by using the exact coefficients of the perfect \+
squares in the diagonal entries, and half of the cross-term coefficien
ts in the off-diagonal entries:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 154 "A:=matrix(2,2,[8,-8,-8,8]);\n    #matrix for quadratic form\n
B:=matrix(1,2,[33*sqrt(2),-31*sqrt(2)]);\n    #row matrix for linear t
erm\nC:=70; # constant term" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
291 "fmat:=v->evalm(transpose(v)&*A&*v\n                 + B&*v +C) ;
\n     #this function takes a vector v and computes \n     # transpose
(v)Av + Bv + C ,\n     #which will be a one by one matrix\nf:=v->simpl
ify(fmat(v)[1]); #this extracts the entry of the one\n     #by one mat
rix fmat(v)\n           " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
150 "fmat(vector([x,y])); #should be a one by one matrix\nf(vector([x,
y]));  #should get its entry, which is what we want.\n   #should agree
 with our example" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "eqtn:=
f([x,y])=0;\n   #should be our equation" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 149 "implicitplot(eqtn,x=-5..5,y=-5..5, color=black,\n   \+
 grid=[100,100]); #what conic is this???? (In\n      #any case, you ca
n chuckle at Maple's picture)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 43 "Now let's go about the change of variab
les:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "data:=eigenvects(A);
#get eigenvectors" }}}{PARA 0 "" 0 "" {TEXT -1 111 "You pick things ou
t of the object above systematically, using inidices to work through t
he nesting of brackets:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "d
ata[1];#first piece of data" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
22 "data[1][1];#eigenvalue" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
34 "data[1][2];#algebraic multiplicity" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "data[1][3];#basis for eigenspace" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 33 "data[1][3][1];#actual eigenvector" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 237 "So that's how \+
to extract the eigenvectors.  Remember once more that if you have two \+
eigenvectors from a symmetric matrix with different eigenvalues, they \+
are automatically perpendicular, and we can make a rotated ortho-norma
l eigenbasis!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "v1:=data[1
][3][1];#first eigenvector\nv2:=data[2][3][1];#second eigenvector\nw1:
=v1/norm(v1,2);#normalized\nw2:=v2/norm(v2,2);#normalized" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 164 "S:=augment(w1,w2):#our orthogonal \+
matrix\n  if det(S)<0 then S:=augment(w2,w1):\n  fi:\nevalm(S); #I put
 it the \"if\" clause to make the\n#new basis vectors right-handed" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "f(evalm(S&*[u,v]));#do the
 change of variables...\n#here [u,v] are the new-basis coordinates of \+
a point\n#and S is the change of coordinates matrix from new basis\n#t
o standard basis." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "simpli
fy(%);#simplify it!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "comp
letesquare(%,u); #complete the square in u" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "completesquare(%,v);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "Eqtn:=%=0;\n  #this is the new equation\n  " }}}
{PARA 0 "" 0 "" {TEXT -1 179 "     This is a translated parabola!  Cou
ld you have predicted this just from the eigenvalues of the matrix?  W
hat eigenvalue property would indicate the possibility of an ellipse?
" }}{PARA 259 "" 1 "" {TEXT -1 454 "     Let's collect all of our comm
ands into one template .  (People who like to program could also turn \+
this into a procedure, which would let you input an equation, and whic
h would output the conic information and a plot.) You can see from whe
re the colons and semicolons are that the output will be  the eigenval
ues,  the transition matrix,  a plot, and an equation just short of st
andard form.  (You can improve plots by picking a denser grid size.)  \+
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "restart;with(linalg):wit
h(plots):with(student):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 816 
"\nA:=matrix(2,2,[8,-8,-8,8]);\nB:=matrix(1,2,[33*sqrt(2),-31*sqrt(2)]
);  \nC:=70;\nfmat:=v->evalm(transpose(v)&*A&*v\n                 + B&
*v + C):\nf:=v->simplify(fmat(v)[1]):\neigenvals(A); #show the eigenva
lues\ndata:=eigenvectors(A):\nv1:=data[1][3][1]:#first eigenvector\nv2
:=data[2][3][1]:#second eigenvector\nw1:=v1/norm(v1,2):#normalized\nw2
:=v2/norm(v2,2):#normalized\nP:=augment(w1,w2):#our orthogonal matrix,
\n  #unless we want to switch columns\n  if det(P)<0 then\n     P:=aug
ment(w2,w1):\n  fi:\nevalm(P);\nimplicitplot(f([x,y])=0,x=-5..5,y=-5..
5,\n     grid=[100,100],color=`black`);\n     #increase grid size for \+
better picture, but too big\n     #takes too long \nf(evalm(P&*[u,v]))
:#do the change of variables\nsimplify(%):#simplify it!\ncompletesquar
e(%,u): #complete the square in u\ncompletesquare(%,v):#and v\nEqtn:=%
=0;\n" }}}{PARA 0 "" 0 "" {TEXT -1 294 "You can check your book proble
ms about conic sections using such a template, in the case of two dist
inct eigenvalues.  (If both eigenvalues are the same, then A is simila
r to a multiple of the identity matrix, so IS a multiple of the identi
ty matrix, so you won't need any fancy computer help.)" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 17 "Quadric Surfaces:" 
}}{PARA 0 "" 0 "" {TEXT -1 17 "Bonus Question : " }{TEXT 262 23 "worth
 10 midterm points" }{TEXT -1 651 " (for each group member!)  Create a
 template or procedure which will diagonalize, graph, and put into sta
ndard form, the solution set to any quadratic equation in 3 variables \+
- i.e. analogous to the template above which works for conic sections.
  You need only create a template or procedure which works when there \+
are 3 distinct eigenvalues.  (If you successfully tackle the  case of \+
higher geometric multiplicity you will need several conditionals to al
low for the structure of the eigenvectors array.  Also, in that case y
ou will need to use the Maple Gram-Schmidt command or your own, to get
 an orthonormal basis of the offending eigenspace basis." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 490 "     To make 3-d \+
plots you can use implicitplot3d.  If you make your grid too fine the \+
plot will take a long time to make, so try the default grid at first a
nd adjust if necessary.  You might also have to adjust your limits in \+
the plot to get a better picture.  You can manipulate 3d plots with yo
ur mouse.  There are a lot of interesting plot options which you write
 into the command or access from the plotting toolbar.  One that helps
 me see things is to use the ``boxed'' axes option.  " }}{PARA 0 "" 0 
"" {TEXT -1 359 "     The 3d plotting routines have been known to cras
h Maple, so save your file often.  Sometimes what really happens is th
at a dialog box gets opened behind one of your windows, where you can'
t see it, so that it will appear that Maple has frozen.  If this happe
ns, minimize your maple window (the black dot a the upper left corner)
, and then re-display it. " }}{PARA 0 "" 0 "" {TEXT -1 143 "     Here'
s a plot to play with, #1 on page 430 of Kolman  You should be able to
 move it around to make it look like a one-sheeted hyperboloid." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "implicitplot3d(x^2 + y^2 +2
*z^2 - 2*x*y -4*x*z -4*y*z + 4*x = 8,\n              x=-5..5,y=-5..5,z
=-5..5, axes=`boxed`);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "How is that pi
cture related to this calculation:" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "A:=matrix(3,3,[1,-1,-2,-1,1,-2,-2,-2,2]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvalues(A);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 98 "Could you \+
find a rotated basis in 3-space which would put this quadric surface i
nto diagonal form?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eigenv
ects(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "14 \+
0 0" 29 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }