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University of Illinois at Urbana-Champaign. April 4 2018 ... known from the study of differential equations). Can we search for such solutions directly?
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Chapter 5: Diagonalization and Eigenvalues and Eigenvectors Section 1: Eigenvalues and eigenvectors and computational examples Ivan Contreras, Sergey Dyachenko and Bob Muncaster University of Illinois at Urbana-Champaign April 4 2018 Ivan Contreras, Sergey Dyachenko and Bob MuncasterUniversity of Illinois at Urbana-ChampaignApplied Linear Algebra——– April 4 2018 1 / 8
We confront now an unusual but significant problem in linear algebra. It is called the eigenvalue problem. It is best motivated by the study of linear differential systems.
Here is a differential equation from calculus, and its solution: du dt
= 4 u ⇒
∫ (^) du
u
∫ 4 dt ︸ ︷︷ ︸ separation of variables
⇒ ln |u| = 4 t + C
⇒ u(t) = ke^4 t^ , C and k arbitrary constants
This example suggests that exponential solutions are important (a fact well known from the study of differential equations). Can we search for such solutions directly? Just substitute u = ke λ t^ with k and λ as unknowns: du dt
= λ ke λ t^ = 4 u = 4 ke λ t
Cancel k and e λ t^ to get λ = 4. And now k is arbitrary.
Let A be an n × n matrix (such as the 2 × 2 above). Then write the eigenvalue problem Ax = λ x in the alternate form Ax − λ Ix = 0 or (A − λ I )x = 0. Set B = A − λ I. Then the eigenvalue problem is
Bx = 0 i.e. find the null space of B and hope it is non-trivial
While this shows that the problem is one we can handle, it is slightly deceptive. Since B = A − λ I we should really write B( λ ) to emphasize the dependence on λ , after all, λ is an unknown at the moment. Thus we have
B( λ )x = 0 i.e. find λ so that the null space of B( λ ) IS non-trivial
Fact: As we will soon see, there are at most n values of λ that work (i.e. for which N(B( λ )) is non-trivial), and some may be complex numbers rather than real numbers! The goal is to find each λ and then find a basis of N(B( λ )) for each one.
Conclusion: When λ is such that N(B( λ )) is non-trivial, the solutions of Ax = λ x forms a nontrivial subspace of R n.
Definition: If A is a square matrix, then λ is an eigenvalue of A if there is a vector x 6 = 0 such that Ax = λ x. We then say that x is an eigenvector of A corresponding to λ. The set of all x’s corresponding to λ is call the eigenspace corresponding to λ.
So how do we find these special λ ’s? The null space of a square matrix is non-trivial if and only if the matrix is singular, i.e. its determinant is zero.
Ex: For our 2 × 2 example above we have
0 = det
− λ
= det
4 − λ − 5 2 − 3 − λ
= ( 4 − λ )(− 3 − λ ) − ( 2 )(− 5 ) = λ^2 − λ − 2 = ( λ + 1 )( λ − 2 ) λ = − 1 or λ = 2
Ex: We have our eigenvalues for the 2 × 2 problem. Let’s find an eigenvector for each (or a basis for the eigenspace) Case 1: λ = − 1
(A − λ I )x =
k m
⇒ k = m
⇒ N(A − λ I ) = Span{
Case 2: λ = 2
(A − λ I )x =
k m
⇒ 2 k = 5 m
⇒ N(A − λ I ) = Span{
Let us not forget where the eigenvalue problem came from: a system of differential equations. Let us write that system in matrix form by setting
, u =
v w
du dt
dv /dt dw /dt
du dt
= Au
We have looked for solutions of the form u(t) = xe λ t^ and discovered that λ is an eigenvalue of A with corresponding eigenvector x. Therefore we have found two solutions:
u 1 (t) =
v 1 (t) w 1 (t)
= e−t
e−t e−t
u 2 (t) =
v 2 (t) w 2 (t)
= e^2 t
5 e^2 t 2 e^2 t
Now we need some results from the theory of linear differential equations.
Fact 1: EVERY solution of du/dt = Au has the form u = c 1 u 1 + c 2 u 2 provided u 1 and u 2 are linearly independent solutions (assuming A is 2 × 2 and u is in R^2 ).
Fact 2: The ci are determined by specifying an “initial condition” for u (i.e, for v and w ):
u( 0 ) =
v ( 0 ) w ( 0 )
(say)
For example: [ 8 5
= c 1 u 1 ( 0 ) + c 2 u 2 ( 0 ) = c 1
c 1 c 2
By G-J or using inverses we find [ c 1 c 2
Thus
u(t) = 3 e−t
v (t) = 3 e−t^ + 5 e^2 t w (t) = 3 e−t^ + 2 e^2 t Some observations about eigenvalue problems: The characteristic polynomial p( λ ) can always be written as a product of factors
det(A − λ I ) = p( λ ) = (− 1 )n( λ − λ 1 )( λ − λ 2 ) · · · ( λ − λ n)
in terms of its n roots (some of which may be complex numbers). If we set λ = 0, we see that
det A = λ 1 λ 2 · · · λ n
