Linear Algebra. Week 8
Dr. Marco A Roque SolEigenvalues and Eigenvectors. The equation
Ax = y
can be viewed as a linear transformation that maps (or transforms) a given vector x into a new vector x.
Given an n × n matrix A we consider the problem of finding a vector x that is transformed into a multiple of itself
Ax = λx but this is equivalent to say that
Ax = λIx ⇒ Ax − λIx = 0 ⇒
The latter equation has nonzero solutions if and only if λ is chosen so that
|A − λI| = 0
This is a polynomial equation of degree n in λ and is called the characteristic equation of the matrix A
Facts
a) It is possible to show that if λ1 and λ2 are two eigenvalues of A
and if λ16= λ2, then their corresponding eigenvectors x(1) and x(2)
are linearly independent.
This result extends to any set λ1, ..., λk of distinct eigenvalues:
their eigenvectors x(1), ..., x(k) are linearly independent. Thus, if each eigenvalue of an n × n matrix is simple, then the n
b) On the other hand, if A has one or more repeated eigenvalues, then there may be fewer than n linearly independent eigenvectors associated with A, since for a repeated eigenvalue with multiplicity m, we may have q < m linearly independent vectors.
c) In the case of an eigenvalue, λi with multiplicity m, if we can
find m eigenvectors x(i1), x(i2), ..., x(im), linearly independent
associated to λi, we say that the matrix is Non-defective.
d) Otherwise, if we are able to find just x(i1), x(i2), ..., x(iq); q < m linearly independent associated to λi, we say that the
Example 8.1
Find the eigenvalues and eigenvectors of the matrix
A = 2 −3 −1 0 −1 0 −1 1 2 Solution
(A − λI) x = 2 − λ −3 −1 0 −1 − λ 0 −1 1 2 − λ x1 x2 x3 = 0 0 0
The eigenvalues are the roots of the equation
− −1 2 − λ 1 0 (2 − λ)2− 1 −1 − λ 0 0 −1 − λ = (1 + λ)(2 − λ)2− 1 = 0
The roots are λ1 = −1, λ2= 1, and λ3= 3 .
3 −3 −1 0 0 0 −1 1 3 x1 x2 x3 = 0 0 0
We can reduce this to the equivalent system
The above system is reduced immediately to the equations
x1+ x2+ 3x3 = 0 8x3= 0
One equation and two unknowns. Hence, one of them is free
variable, let’s say x2 = α. Therefore we have x1= −x2 = −α, and
1 −3 −1 0 −2 0 0 −2 0 = 1 −3 −1 0 −2 0 0 0 0 x1 x2 x3 = 0 0 0
The above system is reduced immediately to the equations
x1− 3x2− x3 = 0 − 2x2= 0
One equation and two unknowns. Hence, one of them is free
variable, let’s say x3 = α. Therefore we have x1= x3 = α, and
x = α 0 α = α 1 0 1 ; α = real number
A particular eigenvector is given by
−1 −3 −1 0 −4 0 −1 1 −1 = −1 −3 −1 0 −4 0 0 4 0 = −1 −3 −1 0 −4 0 0 0 0 x1 x2 x3 = 0 0 0
The above system is reduced immediately to the equations
−x1− 3x2− x3= 0 − 4x2 = 0
One equation and two unknowns. Hence, one of them is a free
variable, let’s say x3 = α. Therefore we have x1= −x3 = −α, and
x = −α 0 α = α −1 0 1 ; α = real number
A particular eigenvector is given by
Thus, the three linearly independent eigenvectors, are x(1) = 1 1 0 x(2) = 1 0 1 x(3)= − 1 0 1
Example 8.2
Find the eigenvalues and eigenvectors of the matrix
A = 0 1 1 1 0 1 1 1 0 Solution
(A − λI) x = −λ 1 1 1 −λ 1 1 1 −λ x1 x2 x3 = 0 0 0
The eigenvalues are the roots of the equation
The roots are λ1 = 2, λ2 = −1, and λ3= −1 . 1) For λ1= 2 −λ1 1 1 1 −λ1 1 1 1 −λ1 x1 x2 x3 = −2 1 1 1 −2 1 1 1 −2 x1 x2 x3 = 0 0 0
The above system is reduced immediately to the equations 2x1− x2− x3 = 0 x2− x3 = 0
Two equations and three unknowns. Hence, one of them is a free
variable, let’s say x3 = α. Therefore x2= x3 = α and
The above system is reduced immediately to the single equation x1+ x2+ x3 = 0
One equation and three unknowns. Hence, two of them are free
veriables, let’s say x1 = α, x2 = β, and x3 = −α − β . Thus we
In this way two linearly independent eigenvectors associated to λ2 = −1 are ( α = 1β = 0 ; α = 0β = 1) x(2) = 1 0 − 1 x(3)= 0 1 − 1
Thus, the three linearly independent eigenvectors, are
x(1)= 1 1 1 x(2)= 1 0 − 1 x(3) = 0 1 − 1
Example 8.3
Find the eigenvalues and eigenvectors of the matrix
A = 4 6 6 1 3 2 1 −5 −2 Solution
(A − λI) x = 4 − λ 6 6 1 3 − λ 2 −1 −5 −2 − λ x1 x2 x3 = 0 0 0
The eigenvalues are the roots of the equation
The roots are λ1 = 1, λ2 = 2, and λ3 = 2 . 1) For λ1= 1 (A − λ1I) x = 4 − λ 6 6 1 3 − λ 2 −1 −5 −2 − λ x1 x2 x3 = 3 6 6 1 2 2 −1 −5 −3 x1 x2 x3 = 0 0 0
We can reduce this to the equivalent system
The above system is reduced immediately to the equations
x1+ x2+ 2x3 = 0 4x2+ 2x3 = 0
Two equations and three unknowns. Hence, one of them, is a free variable, let’s say x3 = α, x2 = 12α, and x3 = −2x3− x2 = −3α .
In this way, there is one linearly independent eigenvector associated to λ2,3 = 2 , namely, x(2)= 3 1 − 2
Therefore, there are just two linearly independent eigenvectors
x(1) = 1 1 1 x(2) = 4 1 − 3
Example 8.4
Find the eigenvalues and eigenvectors of the matrix
A = 1 0 0 2 1 −2 3 2 1 Solution
(A − λI) x = 1 − λ 0 0 2 1 − λ −2 3 2 1 − λ x1 x2 x3 = 0 0 0
The eigenvalues are the roots of the equation
The roots are λ1 = 1, λ2 = 1 + 2 i , and λ3 = 1 − 2 i . 1) For λ1= 1 (A − λ1I) x = 1 − λ 0 0 2 1 − λ −2 3 2 1 − λ = 0 0 0 2 0 −2 3 2 0 x1 x2 x3 = 0 0 0
We can reduce this to the equivalent system
The above system is reduced immediately to the equations
x1= 0 − 2 ix2− 2x3= 0, 2x2− 2 ix3 = 0
Thus, we have one equation and two unknowns. Hence, one of
them 1 is a free veriable, let’s say x2 = α, x3= −i α . Hence
The above system is reduced immediately to the equations
x1 = 0; 2i x2− 2x3 = 0; 2x2+ 2i x3 = 0
Thus, we have one equation and two unknowns. Hence, one of
them 1, is a free variable, let’s say x2 = α, x3= i α . Thus we
In this way, there is two real linearly independent eigenvectors associated to λ2= 1 + 2 i , namely, x(2) = 0 1 0 x(3)= 0 0 1
Hence, we have three linearly independent eigenvectors, namely
x(1)= 1 − 3/2 1 x(2) = 0 1 0 x(3) = 0 0 1
OBS
Let A be a real-valued n × n matrix. If x(1) = x(R)± i x(I ) are
complex eigenvectors of the matrix A with complex eigenvalues
λ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for A
with eigenvalues λ = u ± i v
Finally, let’s introduce another concept
The Dot Product in Rn
For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dot
x · y =< x, y >= x1 x2 . . . xn y1 y2 .. . yn = x1y1+ x2y2+ ... + xnyn OBS
Theorem
Let A be an n × n matrix. If A is symetric, ( A = AT ) then
1) All eigenvalues are real. 2) A is always Nondefective.
3) The eigenvectors corresponding to different eigenvalues are orthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn form
The general theory of a system of n first order linear equations x10 = p11x1+ p12x2+ . . . + p1nxn+ g1(t) x20 = p21x1+ p22x2+ . . . + p2nxn+ g2(t) .. . ... xn0 = pn1x1+ pn2x2+ . . . + pnnxn+ gn(t) or X0 = P(t)X + g(t)
we assume that P and g are continuous on some interval α < t < β; that is, each of the scalar functions
p11, ..., pnn, g1, ..., gn is continuous there.
Theorem
If the vector functions x(1) and x(2) are solutions of the homogeneus system ( g(t) = 0 ) then the linear combination c1x(1)+ c2x(2) is also a solution for any constants c1 and c2.
By repeated application of Theorem, we can conclude that if x(1), ..., x(k) are solutions of the homogeneous system, then
c1x(1)+ ... + ckx(k)
is also a solution for any constants c1, ..., ck.
Theorem
If the vector functions x(1), ..., x(n) are linearly independent
solutions of the homogeneous system for each point in the interval α < t < β, then each solution x = φ(t) of the homogeneous system can be expressed as a linear combination of x(1), ..., x(n) in exactly one way.
If the constants c1, ..., cn are thought of as arbitrary, then the
above equation includes all solutions of the system, and it is customary to call it the general solution.
Any set of solutions x(1), ..., x(n) of the homogeneus system that is linearly independent at each point in the interval α < t < β is said to be a fundamental set of solutions for that interval.
Theorem
W [x(1), x(2). . . x(n)] = x(1)1 x(2)1 . . . x(n)1 x(1)2 x(2)2 . . . x(n)2 .. . ... . . . ... x(1)n x(2)n . . . x(n)n
either is identically zero or else never vanishes. To prove this theorem is necessary to establish that
dW
dt = [p11+ p22+ ... + pnn]W Hence
Theorem
Let x(1), ..., x(n)be the solutions of the homogeneus system that satisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),
..., x(n)(t
0) = e(n), respectively, where t0 is any point in α < t < β
Then, x(1), ..., x(n) form a fundamental set of solutions of the homogeneous system.
Finally in the case that the solution is complex-valued, we have the following result.
Theorem
Consider the homogeneous system
X0 = P(t)X
We will concentrate most of our attention on systems of
homogeneous linear differential equations with constant coefficients
x0 = Ax
where A is a constant n × n matrix. Unless stated otherwise, we will assume further that all the elements of A are real (rather than complex) numbers.
The case n = 2 is particularly important and lends itself to
visualization in the x1x2− plane, called the phase plane. By
Now, for the system
x0 = Ax
we look for solutions of the form
x = veλt
where the exponent λ and the vector v are to be determined. Substituting x in the system gives
λveλt = Aveλt
Thus, to solve the system of differential equations, we must solve the above system of algebraic equations. That is, we need to find the eigenvalues and eigenvectors of the matrix A.
If we assume that A is a real-valued matrix, then we must consider the following possibilities for the eigenvalues of A:
Example 8.5 Consider the system
x0 = Ax =1 1
4 1
x
Solution
λ1= 3, λ2 = −1
If λ1 = 3, then the system reduces to the single equation
−2v1+ v2= 0, =⇒ v2= 2v1
and a corresponding eigenvector is
v(1)=1
Similarly, corresponding to λ2 = −1, we find that a corresponding eigenvector is v(2)= 1 − 2
The Wronskian of these solutions is W [x(1), x(2)](t) = e3t e−t 2e3t −2e−t = −4e2t 6= 0
Hence, the solutions x(1) and x(2) form a fundamental set, and the
general solution of the system is
