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Math 125,
Fall 2005
N. Kerzman
QUADRATIC FORMS AND SYMMETRIC MATRICES.
EIGENVALUES, EIGENVECTORS, DIAGONALIZATION, AND
REDUCTION OF QUADRICS TO PRINCIPAL AXES. A ROAD MAP
INVOLVING MATLAB.
We use MatLab notation without warning.
Sometimes we use A*B for matrix multiplication (in MatLab style)
and sometimes we just write AB (in math style).
All matrices have real entries.
All symmetric matrices A in this write-up are 3 by 3 unless
otherwise stated. Statements easily extend to any n by n
symmetric A.
References:
Linear Algebra by Otto Bretscher, Prentice Hall. A nice
undergraduate textbook.
Matrix
Theory , Vols. I, II, by F. Gantmacher, Chelsea. A
thorough graduate level classic and reference. Very useful.
-------1. Motivation. Quadratic functions in two variables.
The goal is to plot and understand the plane curve C
given by x 2  2 y 2  xy  1 . Think of C as the level set of
q( x, y )  x 2  2 y 2  xy where q( x, y )  1 .
Thanks to MatLab, it is easy
to plot C using meshgrid and the contour commands. Try it and see
what you get.
Now you want to better understand what you plotted. A
wonderful and far-reaching train of thought starts at this point.
Introduce the column vector [x;y], its transpose [x;y]’ (which is
the row vector [x,y]), and a newly created matrix
 1 1 / 2
A
 . You can rewrite
1 / 2 2 
q( x, y )  [ x; y ]'* A * [ x, y ] , or skipping the matrix product symbol *, as
is done in mathematics—but not in MatLab!—
q( x, y )  [ x; y ]' A [ x, y ]
(1.1)
Verify that (1) is true. A is a 2 by 2 matrix. It is
symmetric, i.e., A(i,j)=A(j,i). A was concocted by putting in
its main diagonal the coefficients of the “pure” quadratic terms
in q(x,y), and in the off-diagonal entries the coefficient of the
“mixed” term xy divided by 2.
The same is true for any function q( x, y )  ax 2  by 2  cxy
 a
c / 2
The symmetric matrix A  
 ,
c / 2 b 
constructed like the previous A, does the trick. You can write
q(x,y) via (1), exactly as in the example. Please check it!
Functions q( x, y )  ax 2  by 2  cxy are called quadratic forms , or
just quadratic functions in two variables.
with a,b,c real.
Notice the interplay:
PING: Any quadratic form q( x, y ) gives rise to
matrix A such that (1) is true.
a 2 by 2 symmetric
PONG: Any symmetric 2 by 2 matrix A creates a quadratic form
q(x,y) defined by (1).
-------2. Quadratic forms in three variables.
Getting more ambitious, let’s
surface S in 3-d space given by
Change of notation.
plot and understand now
the
9 x 2  7 y 2  3z 2  2 xy  4 xz  6 yz  1
S is the level surface of the quadratic function in three
variables q( x, y, z )  9 x 2  7 y 2  3z 2  2 xy  4 xz  6 yz for the level
q( x, y , z )  1 . MatLab plots S using meshgrid for 3-d arrays and
isosurface. Try it: Success! Now we’ll try to understand S.
We change to a more systematic notation because what we are
up to works just as well in 2,3,4, 5, or in any number of
variables. Here is what we do: instead of x,y,z, we write
x1 , x2 , x3 and reserve the name x for the column vector
x  [ x1 ; x2 ; x3 ] .
A new 3 by 3 matrix A is created as before: In the
main diagonal are 9, 7, and 3, i.e., the coefficients of the
“pure” quadratic terms. In the off-diagonal position
A(i,j)=coefficient of x i x j divided by 2. I.e.,
 9 1 2 
A   1 7  3


 2  3 3 
A is a 3 by 3 symmetric matrix.
In the new notation, the same easy verifications done in
the two-variables case show that
q( x1 , x2 , x3 )  x' A x
(2.1)
What was said in our example extends word for word to any
quadratic form
q( x 1 , x 2 , x 3 )  a 11 ( x 1 ) 2  a 22 ( x 2 ) 2  a 33 ( x 3 ) 2  a 12 x 1 x 2  a 13 x 1 x 3  a 23 x 2 x 3
The symmetric matrix A arising from q has A(i,i)= A(i, i )  aii
and A(i, j )  aij / 2 if i  j
The same PING-PONG still goes on:
q( x1 , x2 , x3 ) produces a
symmetric A for which (2.1) holds. And any symmetric A generates
a q( x1 , x2 , x3 ) defined by (2.1).
-------3. When A is diagonal,
1. the surface S given by q( x, y , z )  1 is very easy to
undertand, and
2. the a ii are precisely the eigenvalues i
of A
Reasons:
1. If all A(i,j)=0 off the main diagonal, then, in the old
notation, the equation of the surface S is
q( x, y, z )  a11 x 2  a 22 y 2  a33 z 2  1 . Our study of quadrics started with
examples of this type.
E.g, if all the aii  0 , S is an ellipsoid.
If two of the aii  0 and one is <0, S is a hyperboloid of one
sheet. While if two are <0 and one is >0 we’ll get a hyperboloid
of two sheets. (And if all three are <0 what do we get? And if
some aii  0 ? Answer please!)
2. Basic linear algebra says that  is an eigenvalue of A
exactly when p( )  det( I  A)  0 . Write the matrix
I  A .
You will see that p( )  (  a11 )(   a22 )(   a33 ) . This makes
it obvious which are the roots of p( ) , i.e., the eigenvalues of
A.
The eigenvectors of a diagonal A
J, K, along the coordinate axes.
are simply the famous I,
1 
E.g., v1  0 , etc.
 
0
Please
check it.
--------
4. Eigenvalues and eigenvectors of a symmetric matrix A
THEOREM
For any SYMMETRIC 3 by 3 matrix A,



the eigenvalues are necessarily real.
For different eigenvalues 1  2 of A, the corresponding
eigenvectors v1 and v2 are automatically orthogonal (
i.e., v1  v2 in the sense of space geometry or of
multivariable calculus).
There are ALWAYS three eigenvectors v1, v2, v3 of A that
are orthogonal to each other and normalized of length=1;
v1 corresponds to 1 , etc.
The theorem is valid because A is symmetric. It does not
matter whether the eigenvalues of A are repeated or not.
Our
references have the proofs-- but you don’t need to study them
for this class.
--------
5. The heart of the matter.
A crucial change of variables.
Let V be the 3 by 3 matrix whose columns are the v1, v2, v3
above.
Then
(5.1)
V’*V=I
is an immediate consequence of the definition of matrix product
and of the ortonormality of v1,v2,v3. Please make sure you
verify and understand this. Hence V has inv(V) and
`
(5.2) inv(V)=V’
Any square matrix V satisfying (5.2) is called orthogonal.
We get another ping-pong important for our business.
PING: start with any three vectors v1,v2,v3 ; place them as
columns of a matrix V. If v1,v2,v3 are orthonormal, then V
satisfies (5.1), i.e., V is orthogonal.
PONG: A way to check whether three arbitrary vectors are
orthonormal is to check whether the matrix V that has them as
columns is orthogonal, i.e., you should check whether V satisfies
(5.1)or not. This is very easy to do with MatLab.
Next: Define R as the diagonal 3 by 3 matrix
1 0
(5.3) R=  0 2

 0 0
0
0

3 
A direct consequence of the matrix multiplication
definition and of what it means to be an eigenvector is
(5.4) A*V=V*R
or, equivalently,
(5.5) A=V*R*inv(V)=V*R*V’
Please make sure you verify and understand (5.4) and (5.5).
NOW DO A CHANGE OF VARIABLES.
Write
(5.6) q( x1 , x2 , x3 )  x' A x  x'VRV ' x
and define a column vector
(5.7)
y V' x
 y1 
Here y   y 2 
 
 y 3 
appears in our story for the first time.
Magic now sets in. Remember that for any two matrices A and B
(A*B)’=B’*A’. Hence
y '  x'V
(5.8)
and you can rewrite (5.6) as
(5.9) q( x1 , x2 , x3 )  x' A x  y' Ry
BUT R IS DIAGONAL AND SECTION 3 ABOVE APPLIES. THE SURFACE
S SATISFYING q( x1 , x2 , x3 )  x' A x  1 IS HENCE VERY EASY TO ANALYZE IN
TERMS OF y .
NAMELY, S IS THE SET OF POINTS
x  ( x1 , x2 , x3 ) IN
SPACE SUCH THAT y ' Ry  1 , i.e.,S IS DEFINED BY
(5.10)
1 y12  2 y2 2  3 y33  1
WHERE y IS THE VECTOR ASSOCIATED WITH x VIA (5.7) .
(5.10) has no mixed terms. Hence it will be an ellipsoid
if all three eigenvalues are >0, an…, etc. in other cases.
-------6. The geometric meaning of y
associated to x  [ x1 ; x2 ; x3 ]
this y mean?
is fundamental:
via (5.7), i.e.,
y is
y  V ' x . What does
What are these three numbers y1 , y2 , y3 ?
Now, x1 , x2 , x3 are the components of a point P in space.
According to what the word “component” means,
(5.11)
P  x1 I  x2 J  x3 K
where I, J, K are the famous unit vectors along the coordinate
system axes. We are talking about the basic system of reference
used from the beginning to do analytic geometry.
By (5.7),
(5.12)
x V * y
The definition of matrix multiplication shows immediately that
(5.13)
 x1 
 x  = y v1  y v 2  y v3
2
3
 2 1
 x3 
 x1 
Since  x 2   x1 I  x2 J  x3 K , we get,
 
 x3 
(5.14) x1 I  x2 J  x3 K  y1 v1  y2 v 2  y3 v3
CONCLUSION: ANY POINT P WITH COMPONENTS x1 , x2 , x3 IN THE
ORIGINAL SYSTEM OF REFERENCE BASED ON I, J, K HAS
COMPONENTS y1 , y2 , y3 IN THE SYSTEM OF REFERENCE DETERMINED BY
v1,v2,v3.
-------THE SURFACE S IS MADE UP OF THE POINTS P WHOSE COMPONENTS
y1 , y2 , y3 IN THE EIGENVECTORS v1,v2,v3 REFERENCE SYSTEM
SATISFY THE SIMPLIFIED “DIAGONALIZED” FORMULA (5.10). THE
AXES OF SUCH A COORDINATE SYSTEM ARE CALLED THE “PRINCIPAL
AXES” OF S.
THE ORIGINAL EQUATION IN TERMS OF x1 , x2 , x3 , IS LESS
TRANSPARENT. IT HAS MIXED TERMS. THE ORIGINAL I,J,K
SYSTEM MAY NOT BE TOO GOOD TO DESCRIBE S. IT MAY NOT BE
THE MOST NATURAL GIVEN THE POSITION OF THE QUADRIC S.
WE SEE THAT THE APPEARANCE OF MIXED TERMS IN THE EQUATION
FOR S COMES EXACTLY FROM THE ROTATED POSITION OF S. A
ROTATION OF THE COORDINATE AXES IS THUS CAPABLE OF GETTING
RID OF THEM.
--------
7.
Using MatLab to plot S and superimposing its principal
axes.
The strategy.
a) Plot S using meshgrid and isosurface. Play with the
scale until you get a nice informative figure. Don’t use
too many points in meshgrid if you don’t want the computer
to be excruciatingly slow. (100 points in each variable
will produce 10^6 points in the 3-d array. That’s still
ok.) Use an m-file for flexibility and for saving your
work. For an example see(10) below.
b)Put the equation of S in the form
Rewrite it as
x' Ax  1
(or
q( x )  1 (or q( x )  0 ).
x' Ax  0 ).
The familiar command
(7.1)
[V,R]= eig(A)
provides the eigenvectors v1=V(:,1), v2=V(:,2),v3=V(:,3).
What was stated in section 4 is perfectly fine also in the
case when some eigenvalue of A is repeated, i.e., when
p( )  det( I  A) , a third order polynomial, has a double or
triple root. But we want to examine what’s peculiar in that
case, both mathematically and in MatLab, compared to the
most frequent situation in which all three eigenvalues are
different.
-------8. When all three eigenvalues of A are different. There is
no choice for v1 (except, obviously, for multiples), nor
for v2, nor for v3. This is an easy exercise in linear
algebra that students may safely attempt. In producing V,
MatLab normalizes v1, v2, and v3 to length =1. It could
choose –v1 instead of v1, or –v2 instead of v2, etc. But,
other than similar obvious options, it has no choice. Life
without choices is supposed to be easy. In this case it
is. The three axes along v1, v2, v3 are the principal axes
of S.
Use hold on to preserve your previous figure of S.
The axis corresponding to v1 goes through the origin and is
directed by v1. Hence its parametric equation is
(8.1)
x=t v1(1),
y=t v1(2), z=t v1(3)
To prevent overwriting the variables x,y,z --which you
might need later on-- use, e.g.,
t=linspace (-20,20,100)
(the 20 is guessed, e.g., from
playing around with scales in (a))
xx1= t*v1(1);
yy1=t*v1(2); zz1=t*v1(3); this gives you
axis1. Similarly you get axis2, etc. Finally,
(8.2) plot3(xx1,yy1,zz1,‘r’,xx2,yy2,zz2,’b’,
xx3,yy3,zz3,’g’)
This plots the three axes in red, blue and green to help
you track them down. You may also use dotted lines along
with color, or other variations you like.
-------9. Example. Here is an m-file
%quadric 2. x^2 +2*y^2 +3 z^2 +xy +2 xz +6 yz=1;
%hyperboloid of one sheet
%matrix of quadric A=[1 .5 1;.5 2 3;1 3 3];
x=linspace(-20,20,100);
y=linspace(-20,20,100);
z=linspace(-20,20,100);
[X,Y,Z]=meshgrid(x,y,z);
Q=X.^2 +2*(Y.^2) +3*(Z.^2) +X.*Y +2*(X.*Z) +6*(Y.*Z);
isosurface(X,Y,Z,Q,1);
axis equal
xlabel('x')
ylabel('y')
zlabel('z')
title ('hyperboloid of one sheet')
hold on
%Plot the principal axes.
A=[1 .5 1;.5 2 3;1 3 3];
[U,R]=eig(A);
%cols of U hold evecs; diag of R holds evals;
%AU=UR; A=U*R*inv(U);
v1=U(:,1); r1=R(1,1); %1st evec and its eval
v2=U(:,2); r2=R(2,2); %etc.
v3=U(:,3); r3=R(3,3);
t=(-30:.1:30);
xx1=v1(1).*t; yy1=v1(2).*t; zz1=v1(3).*t; %parametric eq of axis1 (for v1), i.e.,
line through origin directed by v1
xx2=v2(1).*t; yy2=v2(2).*t; zz2=v2(3).*t; %etc
xx3=v3(1).*t; yy3=v3(2).*t; zz3=v3(3).*t; %used xx (rather than x) to prevent
overwriting some other x that might be around,etc.
plot3(x1,y1,z1,'b',x2,y2,z2,'r',x3,y3,z3,'g');
grid
box
% R=
% -0.5915
0
0
%
0 0.8032
0
%
0
0 5.7884
--------
10.
When A has repeated eigenvalues.
The goal is to obtain three orthonormal eigenvectors
v1,v2,v3 of A and use them as columns of V. They exist by
the theorem in Section (4). Once V is constructed, we
proceed exactly as in (8). This is almost the end of the
story. (But read (11)!)
Example:
A=
1
1
1
1
1
1
1
1
1
>> [V,R]=eig(A)
V=
0.4082 0.7071 0.5774
0.4082 -0.7071 0.5774
-0.8165
0
0.5774
R=
-0.0000
0
0
0
0
0
0
0
3.0000
Here   0 is a repeated eigenvalue with multiplicity=2.
MatLab came up with eigenvectors v1,v2 v3, the columns of V.
Are they orthonormal? Section 5 explains how to check for this:
you compute V’*V . If you get the identity I, they are. If you
don’t, they aren’t. In this example,
V'*V
ans =
1.0000
0.0000
-0.0000
0.0000
1.0000
0.0000
-0.0000
0.0000
1.0000
Hence it works. The principal axes of the quadric
q(x,y,z)=x^2+y^2+z^2+2xy+2xz+2yz=1 associated with A are along
v1=V(:1),v2=V(:,2), v3=V(:,3). You plot the axes exactly as you
did in (8).
The novelty is mathematical (and not a MatLab quirk):
Namely, two eigenvectors v1 and v2 correspond to   0 . They are
not multiples of each other (look at the 0 in v2: obviously v2 is
not a multiple of v1 or viceversa.) Hence all linear
combinations v  c1v1  c2 v2 will also be eigenvectors with
eigenvalue 0 (do you see why?) Hence the eigenvectors
corresponding to   0 fill up a whole plane . Within that
plane, MatLab could have chosen a different pair of orthonormal
vectors. Why it chose these ones and not others is a bit of a
mystery. But who cares? Having the orthogonal matrix V is all
you need to proceed as in (8). Follow the steps and see what you
get for S. This example makes very clear the role of repeated
eigenvalues. Instead of the three unique principal axes you got
in (8), you have infinitely many ones. One is a line along v3,
the eigenvector for the non-repeated eigenvalue   3 . Here there
is no choice. But any two lines through the origin that are
orthogonal to each other and to v3 will do as principal axes. S
is perfectly rotationally symmetric around the axis of v3.
-------11. When A has repeated eigenvalues and MatLab has a
stability quirk.
Occasionally, when there are repeated eigenvalues,(7.1)
furnishes a V where the columns are NOT orthogonal. This is rare,
especially if you use decimals. It is an unlikely event for A to
have repeated eigenvalues.
Example of what can happen and what to do about it.
exactly as follows. DON’T USE DECIMALS!
>> B= [2,0,0;0,2,0;0,0,3];
%B is symmetric
>>
>>
>>
>>
>>
% T’= inv(T).
Construct
w1= (1/sqrt(3))*[1;1;1];
w2=(1/sqrt(2))*[1;-1;0];
w3=(1/sqrt(6))*[1;1;-2];
T=[w1,w2,w3];
A=T*B*T’;
>> A
A =
Check T*T’=I
%A defined. It’s symmetric.
A
2.1667
0.1667
-0.3333
0.1667
2.1667
-0.3333
-0.3333
-0.3333
2.6667
>> [V,R]=eig(A)
V =
-0.9129
0.1826
-0.3651
0.4082
0.4082
-0.8165
0.5435
0.6100
0.5767
R =
%R has repeated eval 2
2.0000
0
0
0
3.0000
0
0
0
2.0000
-0.0000
1.0000
-0.0000
-0.5953
-0.0000
1.0000
>> V'*V
ans =
1.0000
-0.0000
-0.5953
%
Hence v1 is not  v3. Notice that v1, v3 correspond to the
same eigenvalue   2 . On the other hand, v1  v2 and v3  v2. They
have no choice because they correspond to the different
eigenvalues 2 and 3.
This V is useless to carry out the program in (8).
What to do.
Change the given matrix A in an imperceptible way
keeping it symmetric. Let’s add, e.g., 10^(-10) to A(1,1)
Anew=A;
Anew(1,1)=A(1,1)+10^(-10);
Here’s what MatLab returns:
Anew =
2.1667
0.1667
-0.3333
0.1667
2.1667
-0.3333
-0.3333
-0.3333
2.6667
>> [Vnew,Rnew]=eig(Anew)
Vnew =
-0.9129
0.1826
-0.3651
0.4082
0.4082
-0.8165
0.0000
0.8944
0.4472
0
3.0000
0
0
0
2.0000
-0.0000
1.0000
-0.0000
-0.0000
-0.0000
1.0000
Rnew =
2.0000
0
0
>> Vnew'*Vnew
ans =
1.0000
-0.0000
-0.0000
% The vectors v1new,v2new, v3new are orthonormal and all is now
fine. We proceed as in (8). Notice: Anew looks like A, Rnew
looks like R, but Vnew looks very different from V in its third
colummn. Adding the little 10(^-10)has changed A into a symmetric
matrix Anew that is A for our intents and purposes. But Anew has
three different eigenvalues. Hence the corresponding eigenvectors
are orthogonal. Two eigenvalues are very, very close to each
other. Check it using
format long
. This method invites a lot
of thinking about instabilities and other matters. Some reflexion
shows that it is quite legitimate. For those who do not like
what we did, a totally different method follows.
-------12.
The
orth
command.
12.1. Orthonormalizing two vectors in R^3. Start with any two
vectors w1 and w2 in R^3. Think them as column vectors. Let L be
the subspace of R^3 of all linear combinations of w1 and w2.
L can have dimension 2 (plane), 1 (line), or 0 (L is just the
origin).
Form the matrix W=[w1,w2].
It is 3 by 2.
Then
V=orth(W)
produces a matrix V whose columns are an orthonormal basis of L.
Example 1
w1=[1;2;3],
w2=[4;5;6],
W=[w1,w2],
W =
1
2
3
4
5
6
>> V=orth(W)
V =
-0.4287
-0.5663
-0.7039
0.8060
0.1124
-0.5812
MatLab returned a V with two columns v1=V(:,1), v2=V(:,2) showing
that L has dimension 2 and providing an orthonormal basis v1, v2
of L. The command has “orthonormalized the vectors w1, w2”.
Checking that v1,v2 are orthonormal:
V'*V
ans =
1.0000
0
Example 2
0
1.0000
w1=[1;2;3], w2=[-2;-4;-6],
W=[w1,w2],
W =
1
2
3
-2
-4
-6
>> V=orth(W)
V =
-0.2673
-0.5345
-0.8018
V has only one column. Hence L has dimension 1. It’s a line .
the column matrix V provides an orthonormal basis of L. This
only says that V has length=1.
Check it: V'*V
ans =
1
12.2. Orthonormalizing
The command orth
Example 3
k vectors
in R^n
works in R^n in full generality
.
>> w1=[1;2;3;4;5];
>> W
w2=w1.^2;
w3=w1.^3;
W=[w1,w2,w3];
W =
1
2
3
4
5
1
4
9
16
25
1
8
27
64
125
>> V=orth(W)
V =
-0.0084
-0.0596
-0.1936
-0.4503
-0.8696
0.2035
0.4878
0.6313
0.4127
-0.3897
0.7219
0.4539
-0.1524
-0.4454
0.2265
Since V has three columns, the dimension of L is 3. Recall that L
is the subspace in R^5 of all linear combinations of w1, w2, w3.
Furthemore, the columns of V provide an orthonormal basis of L.
Checking the orthonormality:
V'*V
ans =
1.0000
-0.0000
0.0000
-0.0000
1.0000
0.0000
0.0000
0.0000
1.0000
12.3. Rephrasing what orth does in terms of matrices.
For any
arbitrary n by k matrix W, V=orth(W) is an n by r matrix. Its
columns are an orthonormal basis of the range of W and r is the
dimension of that range.
Reason: the range of W is precisely the subspace L in R^n
generated by all the linear combinations of the columns of W.
-------13.
Using
orth
for quadrics with repeated eigenvalues.
Back to (11).
[V,R]=eig(A) produced a useless V. Its
v1, v3 are not orthogonal. But v1 and
correponding to the eigenvalue   2 .
the subspace of R^3 generated by v1 and
  2.
first and third columns
v3 are eigenvectors of A
Hence all vectors in L,
v3, are eigenvectors for
Now
> VV=orth([v1,v3])
VV =
-0.8153
-0.2393
-0.5273
0.4106
-0.8810
-0.2352
shows that L has dimension 2 (it’s a plane) and provides an
orthonormal basis vv1=VV(:,1), vv3=VV(:,2) of L; vv1 and vv3 are
still eigenvectors of A with eigenvalue   2 because they lie in
L. Both are orthogonal to v2 (which has a different eigenvalue
  3 ).
Conclusion vv1,v2,vv3 are orthonormal eigenvectors of A with
eigenvalues 2,3,2 respectively. You can use them to build a new
matrix Vnew that will diagonalize the quadric exactly as V did in
(8).
-------14.
Quadrics with additional linear terms.
No paraboloids or saddles appeared in what we did so far.
That’s because they have equations like
x2  y2  z  0
x2  y2  z  0
These equations are not just quadratic.
linear term (power=1).
They have an additional
Let S be described by the equation
(14.1)
f(x)=
x’*A*x
+ b*x=c
where c is a given number and b=[b1,b2,b3] is a given row vector.
Here is what you do to analyze and understand S. We have already
done most of the work. An example that may help our discussion
is
f ( x )  x'*A * x  x1  2 x2  3x3  0 with
0
0 
1

A  0  1.5 1.5 


0 1.5  1.5
So, here b=[1,2,3] and c=0.
a) first focus on x’Ax. Find V and R and write x’Ax=y’Ry, with
V’x=y, x=Vy, exactly as before. Hence
f ( x )  y '*R * y  b * V * y
Calling d  b *V we rewrite
f ( x )  y '*R * y  d * y  c
The term y '*R * y is already diagonalized and we get for the
equation of S
(14.2)
1 y12  2 y2 2  3 y32
 d1 y1  d 2 y2  d 3 y3  c
b) Now comes the only novelty: Complete the squares whenever
you can. If, e.g., 1  0 you can associate the 1st and 4th terms.
Doing the high school algebra,
(14.3)
1 y12  d1 y1  1 ( y1  p1 ) 2  1 p12 with
p1 
 d1
21
But if 1  0 you cannot complete that square
In our example, it turns out that 1  3, 2  0, 3  1 and you can
hence do (14.3) for y1 and y 3
but not for y 2 .
Changing variables once more, call Y1  y1  p1 ,
nothing to y 2 .
(14.3)
Y3  y3  p3 and do
Now (14.2) looks like
1Y12  3Y32  d 2 y2  c  1 p12  3 p32
Students should compute all these numbers,including
d 2 , p 2 , p3 ;
d 2  3.5355 is thus different from 0. and S is, …is what? Please
complete Exercise 3 in hwk 10. It clinches quite a bit of the
material we saw over the last few weeks.
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