Chapter 7 Diagonalization and
Quadratic Forms
• 7.1 Orthogonal Matrices
• 7.2 Orthogonal Diagonalization
• 7.3 Quadratic Forms
1
7.1: Orthogonal Matrices
2
Orthogonal Matrix (I)
 Definition 1 : A square matrix A is an orthogonal
matrix
if A-1 = AT
AAT = ATA = AA-1 = A-1A = I
The matrix Q for QR-Decomposition is an orthogonal matrix!
Note: By convention, it is called an “orthogonal matrix”
3
Orthogonal Matrix (II)
 Example 1
3
 7
AT A =  2
 7
6
 7
3
 7
A =  −6
 7
2
 7
−6
3
2
7
7
7
6 
7
7
3
2 
7
7
6
−3 
7
7 
2
 3
7  7
6   −6
7  7
−3   2
7   7
2
2
3
6
7
7
7

7  1 0 0 
2  = 0 1 0 

7 
−3  0 0 1 
7 
6
Note : the set of all columns {c1, c2, c3} is an orthonormal set!
the set of all rows {r1, r2, r3} is an orthonormal set
4
Orthogonal Matrix (III)
 Example 2 : Rotation Matrix in R2.
 Rotate θ in counterclockwise direction:
cos 
A=
 sin 
− sin  
cos  
 cos 
A A=
 − sin 
T
sin   cos 


cos    sin 
This is A-1 formula and is
used to rotate the old axis
This formula is to
rotate the point
− sin   1 0 
=


cos   0 1 
 is positive for
counterclockwise rotation
5
Properties for Orthogonal Matrix (I)
 Theorem 7.1.1 :
The following statements are equivalent
a)
A is orthogonal.
b) {r1, r2,…, rn} is an orthonormal set in Rn with the
Euclidean inner product.
c) {c1, c2,…, cn} is an orthonormal set in Rn with the
Euclidean inner product.
6
Properties for Orthogonal Matrix (II)
 Proof: (a)  (b)
 −r1 − 
 −r − 
A= 2 




−
r
−
 n 
AT = r1 | r2 |
ri : the i-th row of A
 r1  r1 = r2  r2 =
| rn 
 r1  r1 r1  r2
r  r r  r
AAT =  2 1 2 2


rn  r1 rn  r2
r1  rn 
r2  rn 
=I


rn  rn 
= rn  rn = 1
ri  r j = 0 if i  j  {r1 ,r2 ,...,rn } is an orthonormal set.
7
Properties of Orthogonal Matrix (III)
Theorem 7.1.2 :
a) If A is orthogonal then A-1 is also orthogonal.


AAT = ATA = I ➔ AT = A-1 ,
(A-1)(A-1)T = (A-1)(AT)-1 = (A-1)(A-1)-1 = A-1A = I
b) If AAT = I and BBT = I then AB is orthogonal.
➔Product of two orthogonal matrices is orthogonal.
(AB)(AB)T = (AB)(BTAT) = A (BBT) AT = AIAT = AAT = I
c) If A is orthogonal then det(A) = ±1

Sine : det(A) = det(AT)
1 = det(In) = det(AAT) = det(A)．det(AT) = det(A)2
8
Preserved Properties (I)
Theorem 7.1.3 : The following statements are equivalent.
a)
A is orthogonal.
b)
|| Ax|| = ||x|| ∀ 𝐱 ∈ 𝑹𝑛 (length preservation)
c)
Ax．Ay = x．y ∀ 𝐱, 𝐲 ∈ 𝑹𝑛 (angle preservation)
Note : Rotation, Reflection satisfy length and angle
preservation.
9
Preserved Properties (II)
A linear transformation, T: Rn→ Rn ∀ 𝐱 ∈ 𝑹𝑛

If the standard matrix A is orthogonal , then T is
called an orthogonal operator.

Other Definition : || T(x) || = || x ||
➔ || A．x || = || x || ∀ 𝐱 ∈ 𝑹𝑛
10
Length and Angle Preservation (I)
(𝑎) ⇒ (𝑏) Assume AAT = I (length preservation)
Note x．y = xTy
|| Ax|| = (Ax．Ax)1/2 = ((Ax) T(Ax))1/2
= (xTATAx)1/2 = (xTx)1/2= (x．x)1/2= ||x||
(𝑎) ⇒ (𝑐) Ax．Ay = x．y (angle preservation)
Ax．Ay = xTATAy = x．ATAy = x．y
11
Length and Angle Preservation (II)
(𝑏) ⇒ (𝑐) Assume || Ax|| = ||x|| ∀ 𝐱 ∈ 𝑹𝑛
Since : u  v = 1
Ax  Ay = 1
=1
4
u+v
Ax + Ay − 1
2
4
2
−1
Ax − Ay = 1
2
4
x + y − 1 x − y = xy
4
4
2
4
2
u−v
2
A(x + y ) − 1
2
4
4
A(x − y )
2
x, y
(𝑐) ⇒ (𝑎) If Ax．Ay = x．Y since Ax．Ay = x．ATAy
➔ AAT = I, A is an orthogonal matrix
Ax．Ay =xTATAy=(xT)(ATAy)=x．ATAy
12
Properties for Orthonormal Basis
Theorem 7.1.4 : If S is an orthonormal basis for an ndimensional inner product space , and if (u)S = (u1, u2,…,
un) and (v)S = (v1, v2,…, vn) then
a)
u = u12 + u22 + ... + un2
b) d(u, v) = (u1 − v1 )2 + (u2 − v2 )2 + ... + (un − vn )2
c)
<u, v> = u1v1 + u2v2 +…+ unvn
➔ Reduce the general norms and inner products to
Euclidean norms and inner products.
13
Change of Orthonormal Basis
Theorem 7.1.5 :If P is the transition matrix for an
orthonormal basis B to another orthonormal
basis B’ then P is orthogonal.
14
Change of Orthonormal Basis
Proof : Two orthonormal basis
B = {u1, u2, …, un }
B’ = {u1’ , u2’ , …, un’ }
The transition matrix from B’ to B
P = PB’→ B = {[u1’]B , [u2’]B ,…, [un’]B }
Rotation between two coordinate systems: orthogonal
Use the length preservation to prove :
For any orthonormal basis of B , we have
||u|| = || [u]B’ || = || [u]B || = || P [u]B’ ||
The norm is the same as the coordinate vector with respect to
Euclidean Inner Product.
Let [u]B’ = x, || u || = || x || = || P x || ➔ P is orthogonal.
15
Application to Rotation in R2 (I)
Rotation in R2
y
u’2
Q
y'
u2
x'
u’1
x
y
y
u1
y
y
u’2
u’1
cos θ
x’
x
θ+ π/2
sin
(θ+ π/2)
cos (θ+ π/2)
x’
x
 Coordinate change from an old basis (standard basis) to a
new basis rotates {u1, u2} old basis to {u1’, u2’ } new basis
for point Q(x, y)
16
Application to Rotation in R2 (II)
 x '  -1  x 
 '  = P  y
 
y 
cos 
P=
 sin 
 cos  
 u  = 

sin



'
1 B
 

cos( + ) 

 − sin  
2
'
 u 2  = 
=

B

cos


 sin( + )  

2 
− sin  
cos  
17
Application to Rotation in R2 (III)
cos  − sin  
 cos  sin  
−1
T
P=
 P = P =  − sin  cos  
sin

cos





(x,y) rotates the original
 x '   cos  sin    x 
axis by -, getting new
 '  =  − sin  cos    y 
y
 
  
coordinates (x‘,y’)
Equivalent to
rotate the original
axis by 
𝜋
𝜋
𝜋
2
 For example if 𝜃 = since cos = sin =
4
4
4
2
θ = (x, y) = (2, -1) in old basis (standard unit)
θ’s coordinate in new basis: (x’, y’)


cos
'
x  
4
 ' = 
 y   − sin 

4
 2
sin 

4  x =  2

   y   2
cos
−
4 
 2

 2 
2



2  2 = 2 
 
2   −1  3 2 

−

2 
 2 
18
7.2: Orthogonal Diagonalization
19
Diagonalization
 The Diagonalization Problem
 Given an n×n matrix A , does there exist an invertible
matrix P s.t. P-1AP is a diagonal matrix.
 d1
D =  0
 0
0
d2
0
0
0 
d3 
 d1k

Dk =  0
0

0
d2k
0
0 

0 
d3k 
 d1−1 0
0 


D −1 =  0 d 2 −1 0 
−1 
 0
0
d
3


If A is diagonalizable, then Ak and A-k can be computed easily.
20
Procedures for Diagonalizing a Matrix
1) Find n linearly independent eigenvectors of A,
say p1, p2, …, pn
2) Form Matrix P having p1, p2, …, pn as its column
vectors
3) The matrix P-1AP=D, Dii= λi , where λi is the
eigenvalue corresponding to pi, for 1 ≦ i≦ n
21
Orthogonal Diagonalization (I)
 1st Question : Which matrices are orthogonally
diagonalizable ?
1)
P-1AP = D
P is an orthogonal matrix , so PT = P-1
2) A = PDP-1 = PDPT
For general case, P needs to
be invertible only
AT = (PDPT)T = PDPT = A
so A must be a symmetric matrix.
22
Orthogonally Diagonalization (II)
Theorem 7.2.1: f A is an n×n matrix, then (a)=(b)=(c)
a) A is orthogonally diagonalizable.
b) A has an orthonormal set of n eigenvectors.
c) A is symmetric.
23
Important Properties for Symmetric Matrix (I)
Theorem 7.2.2 : If A is a symmetric matrix, then
a) The eigenvalues of A are all real numbers.
b) Eigenvectors from different eigenspaces are orthogonal.
(distinct eigenvalues)
24
Important Properties for Symmetric Matrix (II)
Proof (b) : Let v1 and v2 be the eigenvectors
corresponding to distinct λ1 and λ2 , respectively.
show : v1．v2 = 0
since Av1．v2 = (Av1)Tv2 = v1TATv2
= v1．Av2
(AT = A)
since Av1 = λ1v1 Av2 = λ2v2
λ1v1．v2 = v1．λ2v2 = λ2v1．v2
(λ1 – λ2) v1．v2 = 0
➔ v1．v2 = 0
25
Diagonalization of Symmetric Matrices
1. Find a basis for each eigenspace of A.
2. Apply Gram-Schmidt process to each of these
bases to obtain an orthonormal basis for each
eigenspace.
3. Form the matrix P whose columns are the basis
vectors constructed in step 2.
26
Examples for Orthogonally Diagonalization (I)
1 2
𝐴=
A is a symmetric matrix.
2 4
det( I − A) =
 −1
−2
−2
=  2 − 5
 −4
 = 0 and 5
 Step 1 : Find the basis for each eigenspace
a)
λ = 0 (λI – A)x = 0
 −1 −2 

x = 0
−
2
−
4


b)
1 2 
0 0  x = 0


  −2 
Eigenspace for  = 0 : span    
 1  
λ = 5 (λI – A)x = 0
 4 −2

x = 0
−
2
1


  1  
1 − 1 
2

 x = 0 Eigenspace for  = 5 : span   2  
0
0 
  1  
27
Examples for Orthogonally Diagonalization (II)
 Step 2 : Gram-Schmidt
−2
For λ = 0 𝐯1 =
1
Note v1．v2 = 0
v
1 −2
p1 = 1 =
 
v1
51
λ=5
𝐯2 =
1Τ
2
1
v2
2  1 2  1 1 
p2 =
=
 =
 
v2
5  1 
5  2
 Step 3 : Construct P :
so
 Note :
1  −2 1 
P=


5  1 2
0 0 
P AP = 

0 5 
−1
1  2 −1 1  −2 1 

=


5  −1 −2 
5  1 2
and P-1 = PT P is an orthogonal matrix.
P −1 = −
28
Examples for Orthogonally Diagonalization (III)
Example 1:
4 2 2
A =  2 4 2  det( I − A) = ( − 2) 2 ( − 8) = 0
 2 2 4 
1,2 = 2, and 3 = 8
(1) Find the basis for each eigenspace
(a) λ1,2 = 2
 −2 −2 −2 
1 1 1 
-1
-1
 −2 −2 −2  x=0  0 0 0  x = 0 u =  1  u =  0 
1  
2  




 −2 −2 −2 
0 0 0 
 0 
 1 
 -1 
 -1 
 
 2
(2) Apply the Gram-Schmidt to {u1, u2}
 6
 
 1 
 1 
v1 =   v 2 =  
2
 6
 
 2 
 0 
 6
 
 
29
Examples for Orthogonally Diagonalization (IV)
(b) λ3=8
 4 -2 -2 
( I − A) = -2 4 -2  x = 0
-2 -2 4 
Normalization
Note: u1．u3 =0
1
u 3 = 1
1
 1 
 
 3
 1 
v3 =  
 3
 1 
 3
 
u2．u3 =0
 -1

 2
 1
P= 
 2

 0

-1
6
-1
6
2
6
1 

3
1 
3 
1 
3 
30
Spectral Theorem
 If A is a symmetric matrix, then Q-1AQ=D or QTAQ=D=
1


 0
0
 Q is an orthogonal matrix

n 
 A=QDQ-1=QDQT=
 q1 , q2 ,
1
, qn  
 0
0   q1T 
   =  q qT +  q qT +
2 2 2
  1 1 1
n   qnT 
+ n qn qnT
 Spectral Decomposition
31
Example for Spectral Theorem
4 2 2
-1 1
-1 1 2
A =  2 4 2  1,2 = 2, and 3 = 8 q1 =(
,
,0) q2 =(
,
,
)
2 2
6 6 6
 2 2 4 
1 1
 1 -1 
0
6 6
2 2

 4 2 2



-1
1
1 1


T
T
A =  2 4 2  1,2 = 2, and 3 = 8 q1q1 = 
0  q2 q2 = 
6 6
2 2

 2 2 4 

 0 0 0
 -1 -1




 3 3
1
3
 1 -1 0  
1
A = 2q1q1T +2q2 q2T +8q3q3T = -1 1 0  + 
3
 0 0 0  
 -2
 3
1
3
1
3
-2
3
-2   8
3  3
 
2  8
+
3  3
 
4  8
3   3
8
3
8
3
8
3
q3 =(
-1 -1 -1
,
,
)
3 3 3
-1 
1
3
3


-1 
1
T
q3q3 = 
3
3


2
1
 3
3 
1
3
1
3
1
3
8
3

8
=A
3

8
3 
32
1
3

1
3

1
3 
Example for Spectral Theorem
 Example 2
1 2 
 has eigenvalues λ1 = -3 and λ2 = 2 with
2
−
2


 The matrix A = 
1
2
x1 =   and x2 =   corresponding eigenvectors(verify).
 − 2
1 
Normalizing these basis vectors yields
 1 
 2 
x1  5 
x2  5 
u1 =
=
= 
 and u2 =
x1 − 2 
x2  1 
 5 
 5 
33
Example for Spectral Theorem(pp.401)
 so a spectral decomposition of A is
 1 
 2 
 5  1
 5  2
1 2 
2 
T
T
−
+ (2) 

2 − 2 = 1u1u1 + 2u2u2 = (−3)  2  

1
5
5



−

−
 5
 5 
 5 
2
 1
4 2
−

5  + (2)  5 5 
= (−3)  5
2 1
2 4 
−



 5 5 
5 5
1 
5 
1 2  1 3
Ax = 
= 



2 − 2 1 0
34
Example for Spectral Theorem(pp.401)
2
 1
4
−
 5
 1
5
1 2  1
5
Ax = 
=
(
−
3
)
+
(
2
)
 2 4  
2
 1
2
−
2
1

 
−
 

 5 5 
5
 1
6
−
 5
5
= (−3)   + (2)  
2
3
 
 
 5 
5
 3  12 
 5   5   3
= +  = 
6
6
 −    0 
 5  5 
2
5  1
1  1

5
35
Conclusion (I)
 Let A be an n×n matrix, the followings are equivalent:
a)
A is an orthogonal matrix.
b)
The linear transformation T(x)=Ax presenters length.
||T (x)||=||x|| or ||Ax||=x
c)
Orthogonal Transformation preserves the dot product
T (x)．T (y) = x． y
d)
The columns (rows) of A form an orthonormal basis of Rn
e)
ATA=In
36
Conclusion (II)
 Orthogonal Diagonalization
 Spectral Theorem
 Geometric Meaning of Spectral Theorem
37
7.3: Quadratic Forms
38
Quadratic Forms
 A quadratic form is a function
 f : Rn → R of the form
 f (x) = xTAx; where A is a symmetric matrix.
 If A = [aij] then
 xTAx = a11x12 + a22x22 + … + annxn2 + 2  aij x i x j
i j
39
Quadratic Forms
 A quadratic form is a function
 f : Rn → R of the form
 f (x) = xTAx; where A is a symmetric matrix.
 If A = [aij] then
 xTAx = a11x12 + a22x22 + … + annxn2 + 2
a x x
i j
ij
i
j
 Example :
x 
 2 −3
x =  1 A = 

 −3 5 
 x2 
 2 −3  x1 
2
2
f (x) = xT Ax =  x1 , x2  
=
2
x
+
5
x
− 6 x1 x2
1
2



 −3 5   x2 
40
Example for Quadratic Forms
Example 1:
2 3   x 
 2 x + 6 xy − 7 y =  x, y   3 −7   y 

 
2
2
4 0   x 
 4 x − 5 y =  x, y  0 −5  y 

 
2
2
0
 xy =  x, y   1
 2
1 
2  x 
 y

0  

 1 2 −1  x1 
 2 7 3  x 
2
2
2
x
+
7
x
−
3
x
+
4
x
x
−
2
x
x
+
6
x
x
=
x
,
x
,
x


1
2
3
1
2
1
3
2
3
1
2
3


 2
 −1 3 −3  x3 
 Only consider A is a symmetric matrix
2 1   x 
2 x 2 + 6 xy − 7 y 2 =  x, y  
 
 5 −7   y 
41
The Principal Axes Theorem
Making a substitution for 𝐱 = 𝑃𝐲 if P is invertible and
orthogonal
𝐱 𝑇 𝐴𝐱 = 𝑃𝐲 𝑇 𝐴 𝑃𝐲 = 𝐲 𝑇 𝑃𝑇 𝐴𝑃𝐲 = 𝐲 𝑇 𝑃𝑇 𝐴𝑃 𝐲
𝐱 𝑇 𝐴𝐱 = 𝐲 𝑇 𝐷𝐲 = 𝑦1
𝑦2
… 𝑦𝑛
= 𝜆1 𝑦12 + 𝜆2 𝑦22 + ⋯ + 𝜆𝑛 𝑦𝑛2
𝜆1
0
⋮
0
0
𝜆2
⋮
0
… 0
… 0
⋱ ⋮
… 𝜆𝑛
𝑦1
𝑦2
⋮
𝑦𝑛
42
The Principal Axes Theorem
Theorem 7.3.1:
If A is a symmetric n × n matrix, then there is an orthogonal
change of variable that transforms form 𝐱 𝑇 𝐴𝐱 into 𝐲 𝑇 𝐷𝐲.
Specifically, if P orthogonally diagonalizes A, then making
the change of variable 𝐱 = 𝑃𝐲 in the quadratic form 𝐱 𝑇 𝐴𝐱
yields the quadratic form
𝐱 𝑇 𝐴𝐱 = 𝐲 𝑇 𝐷𝐲 = 𝜆1 𝑦12 + 𝜆2 𝑦22 + ⋯ + 𝜆𝑛 𝑦𝑛2
In which 𝜆1 , 𝜆2 , … , 𝜆𝑛 are the eigenvalues of A corresponding
to the eigenvectors that form the successive columns of P.
43
The Principal Axes Theorem
Example 2:
Find an orthogonal change of variable that
eliminates the cross product terms in the quadratic
form 𝑄 = 𝑥12 − 𝑥32 + 4𝑥1 𝑥2 + 4𝑥2 𝑥3 , and express 𝑄 in
terms of the new variables.
44
The Principal Axes Theorem
𝑄 = 𝑥 𝑇 𝐴𝑥 = 𝑥1
𝜆−1
2
0
𝑥2
𝑥3
1
−2
0
−2
0
2
0 𝑥1
2 𝑥2
−1 𝑥3
2
𝜆
−2
0
−2 = 𝜆3 − 9𝜆 = 𝜆 𝜆 + 3 𝜆 − 3 = 0
𝜆+1
2
1
2
−
−
3
3
3
1
2
2
𝜆 = 0:
, 𝜆 = −3: − , 𝜆 = 3:
3
3
3
2
2
1
3
3
3
2
1
2
−
−
3
3
3 𝑦
𝑥1
1
1
2 2
𝑥2 =
𝑦
2
−
3
3 3 𝑦3
𝑥3
2 2
1
3 3
3
0 0 0 𝑦1
𝑄 = 𝐲 𝑇 𝑃𝑇 𝐴𝑃 𝐲 = 𝑦1 𝑦2 𝑦3 0 −3 0 𝑦2 = −3𝑦22 + 3𝑦32
0 0 3 𝑦3
45
Quadratic Forms in Geometry
a) 𝑎𝑥 2 + 2𝑏𝑥𝑦 + 𝑐𝑦 2 + 𝑑𝑥 + 𝑒𝑦 + 𝑓 = 0
46
Rotated to the Standard Position
y
𝑥
𝑦
𝑥
𝑦
𝑎 𝑑 𝑒 𝑥
𝑧 𝑑 𝑏 𝑓 𝑦 = 𝑘 and
𝑒 𝑓 𝑐 𝑧
𝑎 0 0 𝑥
𝑧 0 𝑏 0 𝑦 =𝑘
0 0 𝑐 𝑧
x
47
Rotated to the Standard Position
Example 3:
(a) Identify the conic whose equation is 5𝑥 2 − 4𝑥𝑦 +
8𝑦 2 − 36 = 0 by rotating the xy-axes to put the
conic in standard position.
(b) Find the angle θ through which you rotated the
xy-axes in part(a).
48
Rotated to the Standard Position
 Sol(a):
5 −2
𝐱 𝑇 𝐴𝐱 = 36 where 𝐴 =
. Characteristic polynomial of A is
−2 8
𝜆−5
2
= (𝜆 − 4)(𝜆 − 9)
2
𝜆−8
𝜆 = 4:
−1
2
5
5
,
𝜆
=
9:
1
2 , thus 𝑃 =
5
5
2
5
1
5
−1
5
2
5
The conic in the x’y’-coordinate system is
write as 4𝑥
′2
2
𝑥′
′2
+ 9𝑦 = 36 or
9
𝑥′
𝑦′
4
0
0 𝑥′
= 36,
9 𝑦′
𝑦 ′2
+
=1
4
49
Rotated to the Standard Position
 Sol(b):
2
5
1
5
−1
2
1
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃
5
𝑃=
, that 𝑐𝑜𝑠𝜃 = , 𝑠𝑖𝑛𝜃 = ,
2 = 𝑠𝑖𝑛𝜃
5
5
𝑐𝑜𝑠𝜃
5
sin𝜃
1
1
tan𝜃 =
= . Thus 𝜃 = tan−1 ≈ 26.6°
cos𝜃
2
2
50
Positive Definite Definition
 Definition 1 : A quadratic form xTAx is said to be
𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒆 if xTAx > 0 for 𝐱 ≠ 0
𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒆 if xTAx < 0 for 𝐱 ≠ 0
𝒊𝒏𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒆 if xTAx has both positive and negative
values
51
Positive Semidefinite Matrix
Example 4:
3 1 1
𝐴 = 1 0 2 are nonnegative, but matrix is indefinite
1 2 0
since its eigenvalues are 𝜆 = 1,4, −2.
3 1 1 𝑥1
𝐱 𝑇 𝐴𝐱 = 𝑥1 𝑥2 𝑥3 1 0 2 𝑥2 = 3𝑥12 + 2𝑥1 𝑥2 +
1 2 0 𝑥3
2𝑥1 𝑥3 + 4𝑥2 𝑥3
𝐱 𝑇 𝐴𝐱 = 4 for 𝑥1 = 0, 𝑥2 = 1, 𝑥3 = 1 and
𝐱 𝑇 𝐴𝐱 = −4 for 𝑥1 = 0, 𝑥2 = 1, 𝑥3 = −1
52
Positive Semidefinite Matrix
Theorem 7.3.3:
If A is a symmetric 2×2 matrix, then:
𝐱 𝑇 𝐴𝐱 = 1 represents an ellipse if A is positive definite.
(b) 𝐱 𝑇 𝐴𝐱 = 1 has no graph if A is negative definite.
(c) 𝐱 𝑇 𝐴𝐱 = 1 represents a hyperbola if A is indefinite.
5𝑥 2 − 4𝑥𝑦 + 8𝑦 2 − 36 = 0 is an ellipse with a major axis of length 6 and
5
1
2
a minor axis of length 4. rewriting is 𝑥 2 −
+ 𝑦 2 = 1, 𝐴 =
36
9𝑥𝑦
9
5
1
−
1
1
36
18
has
eignevalue
𝜆
=
,
𝜆
=
. Eigenvalues is positive so A is
1
1
2
9 2
4
−
18
9
positive definite and the equation represents an ellipse. Moreover, the
2
2
axes of the ellipse have lengths
= 6 and
= 4.
(a)
𝜆1
𝜆2
53
Principal Submatrix
Theorem7.3.4
A symmetric matrix A is positive definite if and only if the
determinants of every principal matrix is positive
- Principal Matrix: Up-left square submatrix
 a11
a
 21
 a31
a12
a22
a32
 a11
A2 = 
 a21
a13 
a23  → principal matrix : A1 =  a11 
a33 
 a11
a12 
a
,
A
=
a22  3  21
 a31
a12
a22
a32
a13 
a23 
a33 
54
Example for Positive Definite Matrices
 Example 5 :
 2 −1 0 
A =  −1 2 −1
 0 −1 2 
2 =2
2 −1
=3
−1 2
2 −1 0
−1 2 −1 = 4
0 −1 2
 Eigenvalues: 2 − 2, 2, 2 + 2  0
55