BY
YAN RU LIN
SCOTT HENDERSON
NIRUPAMA GOPALASWAMI
GROUP 4
11.1 EIGENVALUES &
EIGENVECTORS
Definition
 An eigenvector of a n x n matrix A is a nonzero vector
x such that 𝑨𝒙 = λ𝒙 for some scalar λ.
 A scalar λ is called an eigenvalue of A if there is a
nontrivial solution x of 𝑨𝒙 = λ𝒙; such an x is called an
eigenvector corresponding to λ.
D.C. Lay, "Eigenvectors and Eigenvalues," in Linear Algebra and Its Applications, 3rd ed. Boston, MA: Pearson, 2006, ch. 5, pp. 301-372
Formulation
 𝑨𝒙 = λ𝒙
 This is also equivalent to (𝑨 − λ𝑰)𝒙 = 𝟎
 Now it can be solved easily using a determinant
Essentially, for some value of λ given a transformation matrix A, there
may exist a vector x such that the equation is satisfied. If λ and this
particular vector x do exist, then we call λ the eigenvalue and the x the
corresponding eigenvector.
Example
1. 𝐴𝑠𝑠𝑢𝑚𝑒
5 −1
=𝑨
3 10
5 −1
𝒙 = λ𝒙
3 10
5 −1
3.
𝒙 − λ𝑰𝒙 = 𝟎
3 10
5−λ
−1
4.
𝒙=𝟎
3
10 − λ
5. Solve now using a determinant
2.
Importance
 Eigenvalues and eigenvectors find numerous
applications in these areas:
 Differential Equations
 Dynamical systems
 Engineering design
 Chemistry and physics
 Schrödinger equation (quantum mechanics)
 Vibration analysis
BRAINBITE
A. λ=2 B. λ=0
C. λ=-3 D. λ=1
Remember that…
𝑨𝒙 = λ𝒙
x is given along with A, so λ is solved easily
http://www.maths.usyd.edu.au/u/UG/JM/MATH1014/Quizzes/quiz10.html
Answer: C
11.2 EIGENVALUES SOLUTION
PROCEDURE AND
APPLICATIONS
11.2 Eigenvalues Solution Procedure
and Applications
 Ax = x  (A-I)x = 0
 x=0 is a trivial solution
 Non-trivial solutions exist if and only if:
det( A  I) 
a11  
a21

an1
a12

a22   

an 2
a1n
a2 n


 ann  
0
 Resulting algebraic equation is called the characteristic




equation.
Characteristic polynomial- nth-order polynomial in 
Roots are the eigenvalues {1, 2, …, n}
Solution space is called eigenspace corresponding to {1,
2, …, n}
The solutions obtained are called eigenvectors
Eigenvalue Example
 Characteristic matrix
1 2 
1 0 1  
A  I  
 



3  4
0 1  3
2 
 4   
 Characteristic equation
A  I  (1   )(4   )  (2)(3)  2  3  10  0
 Eigenvalues: 1 = -5, 2 = 2
11.2 Subsection(1) -Quick Tips
 An n x n matrix A means that
are n values to x, and there
will be n eigenvectors and
eigenvalues even if some are
duplicated
 The eigenvalues of a triangular
matrix are the entries on its
main diagonal
 Consider that since λ is scalar,
A must act on eigenvectors
only to “stretch” x and not to
change its direction (see figure)
Unknown. (2011, Oct
27). Eigenvalues and
eigenvectors [Online].
Available: http://en.wikipedia.o
rg/wiki/Eigenvalues_and_eigen
vectors
Example
 Click here to view a demo on eigenvalues and
eigenvectors
 http://web.mit.edu/18.06/www/Demos/eigen-applet-
all/eigen_sound_all.html
11.2 Subsection(2)-Determining Eigenvectors
 First determine eigenvalues: {1, 2, …, n}
 Then determine eigenvector corresponding to each
eigenvalue:
( A  I)x  0  (A  k I)x k  0
 Eigenvectors determined up to scalar multiple
 Distinct eigenvalues
 Produce linearly independent eigenvectors
 Repeated eigenvalues
 Produce linearly dependent eigenvectors
 If n roots are equal then the eigenvalues are said
to of multiplicity n.
Eigenvector Example
 Eigenvalues
1 2  1  5
A
  2
3

4


2
 Determine eigenvectors: Ax = x
x1  2 x2  x1
3x1  4 x2  x2
(1   ) x1  2 x2  0

3x1  (4   ) x2  0
 Eigenvector for 1 = -5
 Eigenvector for 1 = 2
6 x1  2 x2  0
3x1  x2  0
1
x1   
 3
 x1  2 x2  0
3x1  6 x2  0
 2
x2   
1 
BRAINBITE
http://www.maths.usyd.edu.au/u/UG/JM/MATH1014/Quizzes/quiz12.html
Answer : c
11.2.2 APPLICATIONS TO ELEMENTARY
SINGULARITIES IN THE PHASE PLANE
 Consider a linear system of ODEs given by
 If the eigenvalues λ is real
Criteria
Type
λ<0
Stable node
λ>0
Unstable node
λ > 0 and λ < 0
Saddle
 If eigenvalues are complex of the form 𝑎 + 𝑖𝑏
Criteria
Type
a<0
Stable focus
a=0
Centre
a>0
Unstable focus
11.2. subsection(3)Special matrices in exercises
(1) Markov Matrix
 Let
A=
 The sum of elements of row or column sum to unity.
 One of the eigenvalue of Markov matrix is 1.
 The rows of [A-I]sum to zero
 [A-I] is singular and columns of A-I are linearly dependent.
M.D. Greenberg, "The Eigenvalue Problem," in Advanced Engineering Mathematics,
2nd ed. Upper Saddle River, New Jersey: Prentice Hall, 1998, ch. 11.3
(2)Tridiagnol matrix
 A Tridiagnol matrix is one in which all element are
zero except the principal diagonals and its two
adjacent diagonals .
 Eigenvalues are given by
(3) Generalized eigenvalue problem
 If B≠1 then Ax=λBx is called generalized eigenvalue
problem.
 Characteristic equation got by det(A - λB)x=0
 Eigenvectors given by (A - λB)x=0
(4) Cayley hamilton theorem
 Theorem- The characteristic equation of any square
matrix A is
λn+ α1 λn-1 +…. αn λ =0
then
An+ α1 An-1+…+ αn -1A+ αn I=0.
i.e A satisfies characteristic equation.
BRAINBITE
http://www.maths.usyd.edu.au/u/UG/JM/MATH1014/Quizzes/quiz12.html
Answer : a
11.3 SYMMETRIC MATRICES
 A square matrix is symmetric if A = AT. This means
that each element aij = aji, as figure[1].
 A symmetric matrix needs not have
real numbers as elements. However,
when it does, it has the remarkable
property of having only real
eigenvalues.
 Proof : recall, a complex number 𝑧 =𝑎 + 𝑖𝑏, then, the
conjugate of z is defined to be 𝑧 = 𝑎 − 𝑖𝑏, and 𝐴𝐵 =
(𝐴)(𝐵), 𝑧𝑧 = 𝑧 2 …[2]
[1] http://www.aiaccess.net/English/Glossaries/GlosMod/e_gm_symmetric_matrix.htm
[2]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering
Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
Let A be an n x n matrix. Let 𝐸 be the eigenvector
corresponding to its eigenvalue 𝜆, and get
𝐸=
𝑒1
𝑒2
⋮
𝑒𝑛
, 𝐸 𝑇 = 𝑒1 , 𝑒2 ⋯ 𝑒𝑛 ,
𝐸 𝑇 𝐴𝐸 = 𝐸 𝑇 𝜆𝐸 = 𝜆𝐸 𝑇 𝐸 = 𝜆 𝑒1 , 𝑒2 ⋯ 𝑒𝑛
= 𝜆( 𝑒1
2
+ 𝑒2
2
𝑒1
𝑒2
⋮
𝑒𝑛
+ ⋯ + 𝑒𝑛 2 )
If A is a real and symmetric matrix, then 𝑨 = 𝐀 and
𝑨𝑇 = 𝑨, now compute[1]…
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
𝐸 𝑇 𝑨𝐸 = 𝐸 𝑇 𝑨𝐸 = 𝐸 𝑇 𝑨𝐸 is a 1x1 matrix (a number), and so
is the same as its transpose
⇒ 𝐸 𝑇 𝑨𝐸 = (𝐸 𝑇 𝑨𝐸)𝑇 = 𝐸 𝑇 𝑨(𝐸 𝑇 )𝑇 = 𝐸 𝑇 𝑨𝐸
⇒ 𝐸 𝑇 𝑨𝐸 = 𝐸 𝑇 𝑨𝐸 = 𝜆( 𝑒1 2 + 𝑒2 2 + ⋯ + 𝑒𝑛 2 )
Therefore, the number 𝐸 𝑇 𝑨𝐸 , being equal to its conjugate,
is a real number. And ( 𝑒1 2 + 𝑒2 2 + ⋯ + 𝑒𝑛 2 ) is
certainly real. Therefore 𝜆 is real[1].
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
 Let A be a real symmetric matrix. Then eigenvectors
associated with distinct eigenvalues are orthogonal.
 Let A be a real symmetric matrix. Then there is a real,
orthogonal matrix that diagonalizes A[1].
 Let A be a real, n x n symmetric matrix. Then its
eigenvector provide an orthogonal basis for n-space.
Therefore, if an eigenvalue is repeated by k times.
Then the eigenspace is of dimension k, and we can
find another set of orthogonal vector by linear
combination[2].
[1] Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering
Mathematics 5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
[2] M.D. Greenberg, "The Eigenvalue Problem," in Advanced Engineering Mathematics, 2nd ed. Upper
Saddle River, New Jersey: Prentice Hall, 1998, ch. 11.3, pp. 554-569.
Symmetric Matrix Examples
1  2
 3 0 2 
let A   0 2 0  and get 2  1
 2 0 0 
3  4


 We can see that a real, symmetric matrix provides a set
of real eigenvalues. And the corresponding eigenvectors
are
 0  1  2 
1, 0, 0 
     
 0   2   1
     
 These form an orthogonal set of vectors[1].
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
 The orthonormal form is divided by its length and they
can be used as columns of an orthogonal matrix.

1
2 
0 5 5 
Q  1 0 0 


 0 25  15 




 We can find Q-1 = QT, and A can be diagonalized by Q[1].
 2 0 0
Q 1 AQ   0 1 0 
 0 0 4


[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
Useful properties[1]
 If A, B are symmetric n × n matrices, then A+B is
symmetric.
𝐴 + 𝐵 𝑇 = 𝐴𝑇 + 𝐵𝑇 = 𝐴 + 𝐵
 If A, B are symmetric n × n matrices, then AB is not
symmetric.
𝐴𝐵 𝑇 = 𝐵𝑇 𝐴𝑇 = 𝐵𝐴 ≠ 𝐴𝐵
 If C is any n × n matrix. Then 𝐵 = 𝐶 𝑇 𝐶 is a symmetric
matrix.
(𝐶 𝑇 𝐶)𝑇 = 𝐶 𝑇 (𝐶 𝑇 )𝑇 = 𝐶 𝑇 𝐶
 If D is a diagonal matrix, then D is symmetric.
[1]http://www.math.panam.edu/
BRAINBITE
Find which of followings is not the eigenvalue of the 4x4
matrix and its corresponding orthogonal eigenvector.
a. 𝜆 = 0, 𝑣 = [1 0 0 0]𝑇
0 0
0 0
0 1 −2 0
b. 𝜆 = 0, 𝑣 = [0 0 0 1]𝑇
0 −2 1 0
0 0
0 0
c. 𝜆 = −1, 𝑣 = [0 1 1 0]𝑇
d. 𝜆 = 1, 𝑣 = [0 − 1 − 1 0]𝑇
e. 𝜆 = 3, 𝑣 = [0 − 1 1 0]𝑇
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.3, pp. 354-362.
answer : d
11.4 DIAGONALIZATION
Background
 Diagonal matrices have good properties for simplifying
calculations
 Exploit these properties by diagonalizing the matrix
A in 𝑨𝒙 = λ𝒙 or 𝑨𝒙 = 𝒙′ for DE
 Essentially, a change of base is required (𝐱 = 𝐐𝒙) so
that…
 𝑨𝑸𝒙 = 𝑸𝒙′
 𝒙′ = 𝑸−1 𝑨𝑸 𝒙 = 𝑫𝒙
Properties and Restrictions
 A is diagonalizable if and only if it has n LI
eigenvectors
 If the above condition is met, 𝑸 = {𝑒1 , 𝑒2 , … , 𝑒𝑛 } (i.e.
the eigenvectors of A form a matrix Q)
 Symmetric matrices are always diagonalizable.
 𝑸𝑇 = 𝑸−1 due to property of LI eigenvectors
Example




𝑥(𝑡)
3 5
𝒙 + 𝑨𝒙 = 𝟎 where 𝒙 =
and 𝑨 =
𝑦(𝑡)
1 2
Make substitution of 𝒙 = 𝑸𝒙
𝒙′′ + 𝑸−1 𝑨𝑸𝒙 = 𝟎 where 𝑸−1 𝑨𝑸𝒙 = 𝑫𝒙
Solve for eigenvalues and eigenvectors to find D and Q
′′
Example
 Eigenvalues are of A are λ1 = 4.79129 and λ2 =0.208712
2.79129
−1.79129
 Eigenvectors are
and
1
1
4.79129
0
 Therefore, 𝑫 =
when 𝐐 =
0
0.208712
2.79129 −1.79129
1
1
Example
 Using D, two uncoupled differential equations arise
(instead of coupled like before)
𝒙(𝒕)′′ + 𝟑𝒙 𝒕 + 𝟓𝑦(𝑡) = 𝟎
𝒚(𝒕)′′ + 𝟏𝒙 𝒕 + 𝟐𝒚(𝒕) = 𝟎 Diagonalization
𝒙′′ + 𝟒. 𝟕𝟗𝟏𝟐𝟗𝒙 = 𝟎
𝒚′′ + 𝟎. 𝟐𝟎𝟖𝟕𝟏𝟐𝒚 = 𝟎
 These equations are simple ODE and are solved using
the solution 𝐴 sin(𝜔𝑡 + 𝜑).
 ω can be solved for easily whereas A and φ are
constants of integration
Example
 Since we assumed 𝒙 = 𝑸𝒙, now we can solve for the
real x(t) and y(t)
𝑥(𝑡)
𝐴1 sin(𝜔1 𝑡 + 𝜑1 )

=𝑸
𝑦(𝑡)
𝐴2 sin(𝜔2 𝑡 + 𝜑2 )
𝑥(𝑡)
2.79129

=
𝑦(𝑡)
1
−1.79129 𝐴1 sin(𝜔1 𝑡 + 𝜑1 )
𝐴2 sin(𝜔2 𝑡 + 𝜑2 )
1
 x(t) and y(t) have been solved completely and easily
compared to not using properties of diagonal matrices
11.5 APPLICATIONS TO FIRST
ORDER SYSTEMS WITH
CONSTANT COEFFICIENTS
11.5 APPLICATIONS TO FIRST ORDER
SYSTEMS WITH CONSTANT COEFFICIENTS
 Consider an initial value problem
 In matrix form
 The solution to the differential equation is given by
 Where
A= coefficients of variables
Q= modal matrix =[e1,e2….en]
D= Diagonal matrix where jth diagonal elements
are jth eigenvalue of A.
 The solution can also be expressed of the form
Where
M.D. Greenberg, "The Eigenvalue Problem," in Advanced Engineering
Mathematics, 2nd ed. Upper Saddle River, New Jersey: Prentice Hall, 1998, ch. 11.3
Example
 Consider the equations
 Solution :
 Replacing the values of A,D,Q and Q-1 in the following
equation
 we get
11.6 QUADRATIC FORMS
 A (complex) quadratic form is an expansion
𝑛
𝑗=1
𝑛
𝑘=1 𝑎𝑗𝑘 𝑧𝑗 𝑧𝑘 ,
in which each 𝑎𝑗𝑘 and 𝑧𝑗 is a
complex number[1]. For n=2, this is
𝑎11 𝑧1 𝑧1 + 𝑎12 𝑧1 𝑧2 + 𝑎21 𝑧2 𝑧1 + 𝑎22 𝑧2 𝑧2
 The quadratic form is real if each 𝑎𝑗𝑘 and 𝑧𝑗 is real, and
we usually write 𝑧𝑗 as 𝑥𝑗 , and the form is
𝑛
𝑛
𝑗=1 𝑘=1 𝑎𝑗𝑘 𝑥𝑗 𝑥𝑘 .
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
 It is often convenient to write a quadratic form in a
matrix form. If A = [𝑎𝑖𝑗 ], 𝑧 =
𝑎11
then 𝑧 𝑇 𝐴𝑍 = (𝑧1 , 𝑧2 ⋯ 𝑧𝑛 ) ⋮
𝑎𝑛1
=
𝑛
𝑗=1
𝑧1
𝑧2
⋮
𝑧𝑛
⋯ 𝑎1𝑛
⋱
⋮
⋯ 𝑎𝑛𝑛
𝑧1
𝑧2
⋮
𝑧𝑛
𝑛
𝑘=1 𝑎𝑗𝑘 𝑧𝑗 𝑧𝑘 [1]
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
Quadratic forms example
 Let 𝐴 =
𝑥1 , 𝑥2
1 4
, then
3 2
1 4
3 2
𝑥1
𝑥2
= 𝑥1 2 + 7𝑥1 𝑥2 + 2𝑥2 2
But we can also rewrite the quadratic form as 𝑥1 2 +
7
1
7
7
𝑥1
2
𝑥1 𝑥2 + 𝑥2 𝑥1 + 2𝑥2 2 = 𝑥1 , 𝑥2 7
2
2
2 𝑥2
2
The advantage of latter form is that A is a symmetric
matrix[1].
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
Classification of The Quadratic Form[1]
Q = 𝑥 𝑇 𝐴𝑥: A quadratic form is said to be:
 Negative definite: Q < 0 when x ≠ 0
 Negative semidefinite: Q ≤ 0 for all x and Q = 0 for
some x ≠ 0
 Positive definite: Q > 0 when x ≠ 0
 Positive semidefinite: Q ≥ 0 for all x and Q = 0 for
some x ≠ 0
 Indefinite: Q > 0 for some x and Q < 0 for some other
x
[1]http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&ved=0CGQQFjAE&url=http%3A
%2F%2Fwww.econ.iastate.edu%2Fclasses%2Fecon501%2FHallam%2Fdocuments%2FQuad_Forms_000.pdf
&ei=wp26TvrgMsiJsAKNzqXOCA&usg=AFQjCNGQ_OibQn6rhf0wrBTNSVMVOltoaQ&sig2=Y061Hf2_fqb
XqjUYyADczQ
Classification example[1]
1 0 0
 For 𝐴 = 0 2 0 , then
0 0 4
1 0 0 𝑥1
𝑄 = 𝑥 𝑇 𝐴𝑥 = (𝑥1 , 𝑥2 , 𝑥3 ) 0 2 0 𝑥2
0 0 4 𝑥3
= 𝑥12 + 2𝑥22 + 4𝑥32
For any real vector 𝑥 ≠ 0, that 𝑄 will be positive (so
called positive definite).
[1]http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&ved=0CGQQFjAE&url=http%3A
%2F%2Fwww.econ.iastate.edu%2Fclasses%2Fecon501%2FHallam%2Fdocuments%2FQuad_Forms_000.pdf
&ei=wp26TvrgMsiJsAKNzqXOCA&usg=AFQjCNGQ_OibQn6rhf0wrBTNSVMVOltoaQ&sig2=Y061Hf2_fqb
XqjUYyADczQ
Graphical Analysis[1]
 Consider the indefinite matrix A is given by
−2 2
2 2
 The quadratic form then is given by
−2 2 𝑥1
𝑇
𝑄 = 𝑥 𝐴𝑥 = (𝑥1 , 𝑥2 )
2 2 𝑥2
= −2𝑥12 + 4𝑥1 𝑥2 + 2𝑥22
𝐴=
= 4𝑥22 − ( 2𝑥1 − 2𝑥2 )2
 Then Q > 0 for some x and Q < 0 for some other x, so
called indefinite form.
[1]http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&ved=0CGQQFjAE&url=http%3A%2F%
2Fwww.econ.iastate.edu%2Fclasses%2Fecon501%2FHallam%2Fdocuments%2FQuad_Forms_000.pdf&ei=wp26Tv
rgMsiJsAKNzqXOCA&usg=AFQjCNGQ_OibQn6rhf0wrBTNSVMVOltoaQ&sig2=Y061Hf2_fqbXqjUYyADczQ
Graphical Analysis[1]
 The graphic in 3-demension governed by Q = −2𝑥12 +
4𝑥1 𝑥2 + 2𝑥22 as follows
 Where it is clear that Q takes both positive and
negative values.
[1]http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&ved=0CGQQFjAE&url=http%3A%2F%
2Fwww.econ.iastate.edu%2Fclasses%2Fecon501%2FHallam%2Fdocuments%2FQuad_Forms_000.pdf&ei=wp26Tv
rgMsiJsAKNzqXOCA&usg=AFQjCNGQ_OibQn6rhf0wrBTNSVMVOltoaQ&sig2=Y061Hf2_fqbXqjUYyADczQ
 In some problems involving quadratic forms,
calculations are simplified if we transform from
𝑥1 , 𝑥2 ⋯ 𝑥𝑛 coordinate system to a 𝑦1 , 𝑦2 ⋯ 𝑦𝑛 system
in which there are no mixed product terms. That is,
we want to choose so that 𝑛𝑗=1 𝑛𝑘=1 𝑎𝑗𝑘 𝑥𝑗 𝑥𝑘 =
𝑛
2
𝑏
𝑦
𝑗
𝑗
𝑗=1
 The 𝑦1 , 𝑦2 ⋯ 𝑦𝑛 coordinates are called principal axes
for the quadratic form, where the rotation of axes is
used to eliminate mixed product terms in the
equation of conic[1].
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
 Let A be a real symmetric matrix with eigenvalues 𝜆1 ⋯ 𝜆𝑛 ,
and Q is an orthogonal matrix formed by their
corresponding eigenvectors that diagonalizes A. Then the
change of variables X=QY transforms 𝑛𝑗=1 𝑛𝑘=1 𝑎𝑗𝑘 𝑥𝑗 𝑥𝑘
to 𝑛𝑗=1 𝜆𝑗 𝑦𝑗 2 [1].
𝑛
𝑗=1
 Proof:
𝑛
𝑘=1 𝑎𝑗𝑘 𝑥𝑗
𝑥𝑘 = 𝑋 𝑇 𝐴𝑋 =
𝑄𝑌)𝑇 𝐴 𝑄𝑌 = (𝑌 𝑇 𝑄𝑇 𝐴𝑄𝑌 = 𝑌 𝑇 𝑄𝑇 𝐴𝑄 𝑌
= 𝑦1 ⋯ 𝑦𝑛
𝜆1
0
0
0
⋱
0
0
0
𝜆𝑛
𝑦1
2 + ⋯+ 𝜆 𝑦 2
=
𝜆
𝑦
1 1
𝑛 𝑛
⋮
𝑦𝑛
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
Principal Axis Example
 Analyze the conic 4𝑥1 2 − 3𝑥1 𝑥2 + 2𝑥2 2 = 8
First write the quadratic form as 𝑿𝑇 𝑨𝑿 = 8, where 𝐴 =
−3
4
2
, then the eigenvalues of A are (6 ± √13)/2.
−3
2
2
By the principal axis theorem, there is an orthogonal
matrix Q that transforms the equation of the conic to
standard form (canonical form):
6+√13
𝑦1 2
2
+
6−√13
𝑦2 2
2
=8
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.
 This is an ellipse in the 𝑦1 , 𝑦2 plane. The figure[1]
shows a graph of this ellipse.
𝑋 = 𝑄𝑌 ⇒
𝑌 = 𝑄 −1 𝑋 = 𝑄𝑇 𝑋
[1]Peter V. O’Neil, “Eigenvalues,Diagnolization and Special Matrices” in Advanced Engineering Mathematics
5th edition. Birmingham, AL: B. Stenquist, 2003, ch. 8.4, pp. 363-367.