Engineering Mathematics for GATE ME/CE/EC papers. Section 1 : Linear Algebra

1. Section 1 : Linear
Algebra
Topic 3 : Eigen Values and Eigen Vectors

2. Definition
 Let A be an n × n matrix. If there exist a real value λ and a non-zero n × 1 vector x satisfying
Ax = λx
then we refer to λ as an eigenvalue of A, and x as an eigenvector of A corresponding to λ.
 e.g. A =
1 −1
2 4
x =
1
−2
 Ax = 3
1
−2
 Hence, 3 is an eigenvalue of A. Vector
1
−2
is an eigenvector of A corresponding to 3.

3. Finding Eigenvalues
Where, λI = λ
1 0
0 1
 Identity Matrix
 If det (A – λI) ≠ 0 , then above equation has a unique solution x = 0.
 However, the goal here is to find λ, so det (A – λI) = 0 must be chosen.
 Considering A from previous example,
 det (A – λI) =
1 − λ −1
2 4 − λ
= (1 − λ)(4 − λ) + 2 = λ2 − 5 λ + 6
 λ2 − 5 λ + 6 = 0
 λ1 = 3 λ2 = 2
 These are all the eigenvalues of A.
Ax = λx
(A – λI)x = 0

4. Finding Eigenvectors
 Considering these two eigenvalues λ1 = 3 λ2 = 2
 Let us first look for the eigenvectors of A for λ1 = 3. Namely, we want to find x =
𝑥1
𝑥2
to
satisfy:
(A − λ1I)x = 0
1 − 3 −1
2 4 − 3
𝑥1
𝑥2
=
0
0
−2 −1
2 1
𝑥1
𝑥2
=
0
0
 2 x1 + x2 = 0 
𝑥1
𝑥2
=
1
−2
Hence, any
𝑥1
𝑥2
satisfying 2 x1 + x2 = 0 is a solution to the above
system. The set of such vectors can be represented in a parametric form: x1 = t and x2 = −2t
for any t ∈ R. or t
1
−2

5. Finding Eigenvectors
 Similarly for, λ2 = 2
(A − λ2I)x = 0
1 − 2 −1
2 4 − 2
𝑥1
𝑥2
=
0
0
−1 −1
2 2
𝑥1
𝑥2
=
0
0
 x1 + x2 = 0 
𝑥1
𝑥2
=
1
−1
Hence, any
𝑥1
𝑥2
satisfying x1 + x2 = 0 is a solution to the above system.
The set of such vectors can be represented in a parametric form: x1 = s and x2 = −s for any s
∈ R. or s
1
−1
in vector form.
 Solution : t
1
−2
+ s
1
−1
t, s ∈ R