3. Eigenvalues and Eigenvectors
• If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial
solution x ∈ ℜⁿ, then λ is an eigenvalue of A and x is a corresponding
eigenvector of A.
– Ax=λx=λIx
– (A-λI)x=0
• The matrix (A-λI ) is called the characteristic matrix of a where I is the
Unit matrix.
–
• The equation det (A-λI )= 0 is called characteristic equation of A and the
roots of this equation are called the eigenvalues of the matrix A.The set
of all eigenvectors is called the eigenspace of A corresponding to λ.The
set of all eigenvalues of a is called spectrum of A.
4. Characteristic Equation
▪ If A is any square matrix of order n, we can form the matrix , where
is the nth order unit matrix.
▪ The determinant of this matrix equated to zero,
▪ i.e.,
is called the characteristic equation of A.
0
λ
a
...
a
a
...
...
...
...
a
...
λ
a
a
a
...
a
λ
a
λ
A
nn
n2
n1
2n
22
21
1n
12
11
I
5. • On expanding the determinant, we get
• where k’s are expressible in terms of the elements a
•The roots of this equation are called Characteristic roots
or latent roots or eigen values of the matrix A.
•X = is called an eigen vector or latent vector
0
k
...
λ
k
λ
k
λ
1)
( n
2
n
2
1
n
1
n
n
4
2
1
...
x
x
x
6. Properties of Eigen Values:-
1. The sum of the eigen values of a matrix is the sum of the
elements of the principal diagonal.
2.The product of the eigen values of a matrix A is equal to its
determinant.
3. If is an eigen value of a matrix A, then 1/ is the eigen value
of A-1 .
4.If is an eigen value of an orthogonal matrix, then 1/ is also its
eigen value.
7. PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of A, then
i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA,
where k is a non – zero scalar.
ii. are the eigen values of the inverse
matrix A-1.
iii. are the eigen values of Ap, where p is any positive
integer.
8. Algebraic & Geometric Multiplicity
▪ If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times
then n is called the algebraic multiplicity of λ.The number of linearly
independent eigenvectors is the difference between the number of
unknowns and the rank of the corresponding matrix A- λI and is
known as geometric multiplicity of eigenvalue λ.
9. Cayley-Hamilton Theorem:-
• Every square matrix satisfies its own characteristic equation.
• Let A = [aij]n×n be a square matrix then,
n
n
nn
2
n
1
n
n
2
22
21
n
1
12
11
a
...
a
a
....
....
....
....
a
...
a
a
a
...
a
a
A
10. Let the characteristic polynomial of A be
(λ)
Then,
The characteristic equation is
11 12 1n
21 22 2n
n1 n2 nn
φ(λ) = A - λI
a - λ a ... a
a a - λ ... a
=
... ... ... ...
a a ... a - λ
| A - λI|=0
11. n n-1 n-2
0 1 2 n
n n-1 n-2
0 1 2 n
We are to prove that
p λ +p λ +p λ +...+p = 0
p A +p A +p A +...+p I= 0 ...(1)
Note 1:- Premultiplying equation (1) by A-1 , we
have
I
n-1 n-2 n-3 -1
0 1 2 n-1 n
-1 n-1 n-2 n-3
0 1 2 n-1
n
0 =p A +p A +p A +...+p +p A
1
A =- [p A +p A +p A +...+p I]
p
12. This result gives the inverse of A in terms of (n-1) powers of A
and is considered as a practical method for the computation of the
inverse of the large matrices.
Note 2:- If m is a positive integer such that m > n then any positive
integral power Am of A is linearly expressible in terms of those of
lower degree.
13. Example 1:-
Verify Cayley – Hamilton theorem for the matrix
A = . Hence compute A-1 .
Solution:-The characteristic equation of A is
2
1
1
1
2
1
1
1
2
tion)
simplifica
(on
0
4
9λ
6λ
λ
or
0
λ
2
1
1
1
λ
2
1
1
1
λ
2
i.e.,
0
λI
A
2
3
21. Similarity of Matrix
▪ IfA & B are two square matrices of order n then B is said to be similar
to A, if there exists a non-singular matrix P such that,
B= P-1AP
1. Similarity matrices is an equivalence relation.
2. Similarity matrices have the same determinant.
3. Similar matrices have the same characteristic polynomial and
hence the same eigenvalues. If x is an eigenvector corresponding to
the eigenvalue λ, then P-1x is an eigenvector of B corresponding to
the eigenvalue λ where B= P-1AP.
22. Diagonalization
▪ A matrix A is said to be diagonalizable if it is similar to diagonal
matrix.
▪ A matrix A is diagonalizable if there exists an invertible matrix P such
that P-1AP=D where D is a diagonal matrix, also known as spectral
matrix.The matrix P is then said to diagonalize A of transform A to
diagonal form and is known as modal matrix.
23. Reduction of a matrix to Diagonal Form
▪ If a square matrix A of order n has n linearly independent eigen
vectors then a matrix B can be found such that B-1AB is a diagonal
matrix.
▪ Note:-The matrix B which diagonalises A is called the modal matrix
of A and is obtained by grouping the eigen vectors ofA into a square
matrix.
24. Example:-
Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3.
Solution:- Characteristic equation is
=> λ = 1, 2, 3
Hence eigenvalues of A are 1, 2, 3.
3
0
0
1
2
0
2
1
1
0
λ
-
3
0
0
1
λ
-
2
0
2
1
λ
1-
25. Corresponding to λ = 1, let X1 = be the eigen
vector then
3
2
1
x
x
x
0
0
1
k
X
x
0
x
,
k
x
0
2x
0
x
x
0
2x
x
0
0
0
x
x
x
2
0
0
1
1
0
2
1
0
0
X
)
I
(A
1
1
3
2
1
1
3
3
2
3
2
3
2
1
1
26. Corresponding to λ = 2, let X2 = be the eigen
vector then,
3
2
1
x
x
x
0
1
-
1
k
X
x
-k
x
,
k
x
0
x
0
x
0
2x
x
x
0
0
0
x
x
x
1
0
0
1
0
0
2
1
1
-
0
X
)
(A
2
2
3
2
2
2
1
3
3
3
2
1
3
2
1
2
0
,
I
2
27. Corresponding to λ = 3, let X3 = be the eigen
vector then,
3
2
1
x
x
x
2
2
-
3
k
X
x
k
-
x
,
k
x
0
x
0
2x
x
x
0
0
0
x
x
x
0
0
0
1
1
-
0
2
1
2
-
0
X
)
(A
3
3
1
3
3
3
2
3
3
2
1
3
2
1
3
3
2
2
3
,
2
I
3
k
x
31. Orthogonally Similar Matrices
▪ If A & B are two square matrices of order n then B is said to be orthogonally
similar to A, if there exists orthogonal matrix P such that
B= P-1AP
Since P is orthogonal,
P-1=PT
B= P-1AP=PTAP
1. A real symmetric of order n has n mutually orthogonal real eigenvectors.
2. Any two eigenvectors corresponding to two distinct eigenvalues of a real
symmetric matrix are orthogonal.
32. Diagonalises the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
32
Example :-
2
0
4
0
6
0
4
0
2
6
6,
2,
λ
0
λ)
16(6
λ)
λ)(2
λ)(6
(2
0
λ
2
0
4
0
λ
6
0
4
0
λ
2
33. I
1
1 2
3
1
1
2
3
1 3
2
1 3
1 1 2 3 1
1 1
x
when λ = -2,let X = x be the eigen vector
x
then (A + 2 )X = 0
4 0 4 x 0
0 8 0 x = 0
4 0 4 x 0
4x + 4x = 0 ...(1)
8x = 0 ...(2)
4x + 4x = 0 ...(3)
x = k , x = 0, x = -k
1
X = k 0
-1
34. 2
2
I
0
1
2
3
1
2
3
1 3
1 3
1 3 2
2 2 3
x
whenλ = 6,let X = x betheeigenvector
x
then (A - 6 )X = 0
-4 0 4 x 0
0 0 x = 0
4 0 -4 x 0
4x + 4x = 0
4x - 4x = 0
x = x and x isarbitrary
x must be so chosen that X and X are orthogonal among th
.
1
emselves
and also each is orthogonal with X
35.
2 3
3 1
3 2
3
1 α
Let X = 0 and let X = β
1 γ
Since X is orthogonal to X
α - γ = 0 ...(4)
X is orthogonal to X
α + γ = 0 ...(5)
Solving (4)and(5), we get α = γ = 0 and β is arbitrary.
0
Taking β = 1, X = 1
0
1 1 0
Modal matrix is M = 0 0 1
-1 1
0
37. Quadratic Forms
DEFINITION:-
A homogeneous polynomial of second degree in any number of
variables is called a quadratic form.
For example,
ax2 + 2hxy +by2
ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and
ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw
are quadratic forms in two, three and four variables
38. In n – variables x1,x2,…,xn, the general quadratic form
is
In the expansion, the co-efficient of xixj = (bij + bji).
38
n
1
j
n
1
i
ji
ij
j
i
ij b
b
where
,
x
x
b
).
b
(b
2
1
a
where
x
x
a
x
x
b
b
a
and
a
a
where
b
b
2a
Suppose
ji
ij
ij
j
i
n
1
j
n
1
i
ij
j
i
n
1
j
n
1
i
ij
ii
ii
ji
ij
ij
ij
ij
39. Hence every quadratic form can be written as
get
we
form,
matrix
in
forms
quadratic
of
examples
said
above
the
writing
Now
.
x
,...,
x
,
x
X
and
a
A
where
symmetric,
always
is
A
matrix
the
that
so
AX,
X'
x
x
a
n
2
1
ij
j
i
n
1
j
n
1
i
ij
y
x
b
h
h
a
y]
[x
by
2hxy
ax
(i) 2
2
41. Two Theorems On Quadratic Form
Theorem(1): A quadratic form can always be expressed with respect to
a given coordinate system as
where A is a unique symmetric matrix.
Theorem2: Two symmetric matrices A and B represent the same
quadratic form if and only if
B=PTAP
where P is a non-singular matrix.
Ax
x
Y T
42. Nature of Quadratic Form
A real quadratic form X’AX in n variables is said to be
i. Positive definite if all the eigen values ofA > 0.
ii. Negative definite if all the eigen values of A < 0.
iii. Positive semi definite if all the eigen values ofA 0 and at least one
eigen value = 0.
iv. Negative semi definite if all the eigen values of
A 0 and at least one eigen value = 0.
v. Indefinite if some of the eigen values ofA are + ve and others – ve.
43. Find the nature of the following quadratic forms
i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx
ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy
Solution:-
i. The matrix of the quadratic form is
43
Example :-
1
1
3
1
5
1
3
1
1
A
44. The eigen values of A are -2, 3, 6.
Two of these eigen values being positive and one being
negative, the given quadratric form is indefinite.
ii. The matrix of the quadratic form is
The eigen values of A are 2, 3, 6. All these eigen values being
positive, the given quadratic form is positive definite.
3
1
1
1
5
1
1
1
3
A
45. Linear Transformation of a Quadratic
Form
▪ Let X’AX be a quadratic form in n- variables and let X = PY ….. (1)
where P is a non – singular matrix, be the non – singular
transformation.
▪ From (1), X’ = (PY)’ =Y’P’ and hence
X’AX =Y’P’APY =Y’(P’AP)Y
=Y’BY …. (2)
where B = P’AP.
46. Therefore,Y’BY is also a quadratic form in n- variables. Hence it
is a linear transformation of the quadratic form X’AX under the
linear transformation X = PY and B = P’AP.
Note. (i) Here B = (P’AP)’ = P’AP = B
(ii) ρ(B) = ρ(A)
Therefore, A and B are congruent matrices.
47. Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form.
Or
Diagonalises the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by
linear transformations and write the linear transformation.
Or
Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the
sum of squares.
47
Example:-
48. Solution:- The given quadratic form can be written as X’AX where
X = [x, y, z]’ and the symmetric matrix
A =
Let us reduce A into diagonal matrix. We know tat A = I3AI3.
3
4
4
4
0
2
4
2
3
1
0
0
0
1
0
0
0
1
3
4
4
4
0
2
4
2
3
1
0
0
0
1
0
0
0
1
3
4
4
4
0
2
4
2
3
51. The canonical form of the given quadratic form is
Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1.
Note:- In this problem the non-singular transformation which
reduces the given quadratic form into the canonical form is X = PY.
i.e.,
2
3
2
2
2
1
3
2
1
3
2
1
y
y
3
4
3y
y
y
y
1
0
0
0
3
4
0
0
0
3
y
y
y
AP)Y
(P'
Y'
3
2
1
1
1
2
0
1
3
2
0
0
1
y
y
y
z
y
x
