HIT121 Engineering Maths 2
K. Mutangi
29 February 2016
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
1 Eigenvalues and eigenvectors
De nitions
Characteristic polynomial: If A is an n n matrix, the
polynomial Pn () of degree n in the scalar de ned as
Pn () = det (A I ) is called the characteristic polynomial of
A.
Eigenvalues: The roots of the equation Pn () = 0 are called
eigenvalues of A.
Eigenvectors: The column vectors X1 , X2 , . . . , Xn satisfying
the equation (A i I )Xi = 0 are the eigenvectors of A.
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
1 Eigenvalues and eigenvectors
De nitions
In general, we seek solutions to the homogeneous system of
equations AX = X or (A I )X = 0. It should be noted
that a matrix with complex coe cients will have complex
eigenvalues. It is however possible for a matrix with real
entries to have complex eigenvalues.
If an eigenvalue is repeated r times corresponding to the
presence of a factor ( )r in Pn (), the number r is called
the algebraic multiplicity of .
Spectrum: is the set of all eigenvalues 1 , 2 , . . . , n .
Spectral radius of A: R = max(|1 |, |2 |, . . . , |n |)
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
1 Eigenvalues and eigenvectors
Theorem
Linear independence of eigenvectors: the eigenvectors
X1 , X2 , . . . , Xm corresponding to m distinct eigenvalues of an n n
matrix A are linearly indepndent.
De nitions
When an eigenvalue with algebraic multiplicity r > 1 has s
dierent eigenvectors associated with it, where s < r then s is
called the geometric multiplicity of the eigenvalue.
The set of all eigenvectors associated with an eigenvalue with
geometric multiplicity s together with the null vector 0 forms
the eigenspace associated with the eigenvalue.
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Example
Find the characteristic polynomial,
the eigenvalues and the
2 1 1
eigenvectors of the matrix A = 3 2 3
3 1 2
Solution
2 1 1
3 2 3
P3 () = |A I | =
3 1 2
2 3 1 3 3
= (2 )
1 2 3 2
3 2
1
3 1
3 2
= + 2 + 2
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Solution
The characteristic equation is P3 () = 0, hence
3 + 22 + 2 = 0 with roots = 2, 1 and -1 as eigenvalues.
we can now nd eigenvectors for each eigenvalue as follows:
Case 1 1 = 2
22 1 1 0
x1
(A I )X = 3 22 3 x2 = 0
3 1 2 2 x3 0
0 1 1 0
x1
3 0 3 x2 = 0
3 1 4 x3 0
x2 x3 = 0, 3x1 3x3 = 0, 3x1 + x2 4x3 = 0
x2 = x3 and x1 = x3 x1 = x2 = x3
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
thus the rst eigenvector is given by:
1 1
k1
X1 = k1 = k1 1 = 1
k1 1 1
This is possible after setting x3 = k1 and then choosing k1 = 1.
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Case 2: 2 = 1
21 1 1 0
x1
3 21 3 x2 = 0
3 1 2 1 x3 0
x1 + x2 x3 = 0 (1)
3x1 + x2 3x3 = 0 (2)
3x1 + x2 3x3 = 0 (3)
From equation (1) x2 = x3 x1 and from (2) and (3),
x2 = 3x3 3x1 , implying that x1 = x3 and x2 = 0. The second
eigenvector is thus found as follows:
1 1
k2
X2 = 0 = k2 0 = 0 , if we let x3 = k2 and k2 = 1
k2 1 1
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Case 3: 3 = 1
3 1 1 0
x1
3 3 3 x2 = 0
3 1 1 x3 0
3x1 + x2 x3 = 0 (4)
3x1 + 3x2 3x3 = 0 (5)
3x1 + x2 x3 = 0 (6)
From (4), x2 = x3 3x1 and substitute in (5) to obtain the
following: 3x1 + 3[x3 3x1 ] 3x3 = 3x1 9x1 + 3x3 3x3 = 0
implying that x1 = 0 and x2 = x3 = k3 , thus:
0 0
X3 = k3 = 1 if we let k3 = 1.
k3 1
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Example
1 2 1
Let A = 1 0 1 . Find the characteristic polynomial,
4 4 5
eigenvalues and eigenvectors of A.
Solution: Class exercise, students should be able to nd
P3 ()= 3 62 + 11 6 and the eigenvalues are
1 = 1, 2 = 2 and 3 = 3. The associated eigenvectors are as
follows
1 2 1
X1 = 1 , X2 = 1 , X3 = 1
2 4 4
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Theorem
Eigenvalues and eigenvectors of a symmetric matrix:
Let A be an n n real symmetric matrix. Then
(i) The eigenvalues of A are all real;
(ii) The eigenvectors of A corresponding to distinct eigenvalues are
mutually orthogonal.
Theorem
Similar matrices have the same characteristic polynomial.
Proof.
(Proof: Kolman B, page 351). Note that for A similar to B, then
B = P 1 AP for some nonsingular matrix P.
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Properties of Eigenvalues and Eigenvectors
Theorem
Property 1
Let A be an n n matrix and 1 , 2 , . . . , n be the eigenvalues of
A. Then:
(i) AT has the same eigenvalues 1 , 2 , . . . , n .
1 1
(ii) A1 (if it exists) has eigenvalues 1
1 , 2 , . . . , n .
(iii) The matrix A I has eigenvalues 1 , 2 , . . . , n .
(iv) For any non-negative integer k , the matrix Ak has eigenvalues
k1 , k2 , . . . , kn
K. Mutangi HIT121 Engineering Maths 2
Linear Algebra
Properties of Eigenvalues and Eigenvectors
Theorem
Property 2: For any square matrix A, the sum of eigenvalues is
equal to the sum of diagonal elements of A(trace(A)).
Property 3: For any square matrix A, the product of
eigenvalues is equal to the determinant of A.
Property 4: The eigenvectors of a square matrix A
corresponding to distinct eigenvalues are linearly
independent,i.e One cannot be written as a linear combination
of the other eigenvectors.
K. Mutangi HIT121 Engineering Maths 2
