MATH136 Lecture Notes - Lecture 34: Invertible Matrix, Coefficient Matrix, Tx1
14 May 2018
School
Department
Course
Professor
Wednesday, July 19
−
Lecture 34 : Eigenvectors. (Refers to 6.2)
Concepts:
1. Define the family of all eigenvectors associated to an eigenvalue.
2. Find all eigenvectors associated to an eigenvalue.
3. Define an eigenspace.
34.1 Theorem
−
Let λ1 be a number. The number λ1 is an eigenvalue of A if and only if
the matrix equation Ax = λ1x with variable x has a non-trivial solution.
Proof :
λ1 is an eigenvalue of A ⇔ λ1 is a solution to det(A
−
λI) = 0
⇔ det(A
−
λ1I) = 0
⇔ A − λ1I is not invertible
⇔ (A − λ1I)x = 0 has a non-trivial solution
⇔ Ax − λ1Ix = 0 has a non-trivial solution
⇔ Ax = λ1Ix has a non-trivial solution
⇔ Ax = λ1x has a non-trivial solution.
Different ways of viewing an eigenvalue. The following are equivalent:
1. λ1 is an eigenvalue of A
2. det(A − λ1I) = 0
3. (A − λ1I)x = 0 has a non-trivial solution
4. Ax = λ1x has a non-trivial solutions
5. Null(A − λ1I) ≠ {0}
6. A − λ1I is non-invertible.
34.1.1 Example
−
Show that 1 is an eigenvalue of the following matrix A
Solution: By the above theorem it suffices to verify that the following homogeneous
system (A – (1)I)x = 0 has a non-trivial solution. That is,
has a non-trivial solution. This coefficient matrix easily row-reduces to
Since (A – (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A.
Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have
non-trivial solutions and so 2 and 3 are eigenvalues of A.
34.2. Definition − Suppose λ1 is an eigenvalue of n × n matrix A. Then any non-zero
vector x1 in Null(A
−
λ1I) is called an eigenvector of A corresponding to
λ
1. That is, x1 is
an eigenvector associated to λ1 if and only if it is a non-zero solution of the vector
equation Ax = λ1x.
34.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always
exists in relation to some eigenvalue. So to say that x1 is an eigenvector of the matrix
A invites the following question: “x1 is associated to which of the eigenvalues of A?”.
An eigenvector of A can only correspond to a single eigenvalue. If x1 is an
eigenvector of A associated to the eigenvalues λ1 and λ2 of A then λ1 = λ2. This is
because x1 cannot be the zero vector. To see this consider:
The expression “λ1x1 = λ2x1 implies λ1 = λ2” since an eigenvector cannot be the zero
vector.
34.2.2 Remark – Extending the notion of “eigenvalue” so that it applies to linear
mappings. The notions of “eigenvalue” and “eigenvector” of a matrix A can be
extended to a linear map (transformation) T : ℝn → ℝn.
Definition –We say that λ1 is an eigenvalue of the linear map T : ℝn → ℝn if
T(x) = λ1x
has a non-trivial solution for x. We say that x1 is an eigenvector of the linear
transformation T corresponding to the eigenvalue λ1 if x1 is a non-trivial solution to
T(x) = λ1x.
- If λ1 is an eigenvalue of T : ℝn → ℝn and x1 is an eigenvector corresponding to
λ1 then, “T maps x1 to a scalar multiple λ1x1 of x1”.
34.2.2.1 Example – Suppose Rπ : ℝ2 → ℝ2 is the linear transformation which
rotates points in ℝ2 counterclockwise about the origin by an angle of π radians.
Suppose we seek an eigenvalue and its associated eigenvectors for this linear map.
See that if x is any ordered pair in ℝ2, Rπ(x) = −x. Then λ1 = −1 is an eigenvalue of
Rπ. The eigenvectors associated to λ1 = −1 are all non-zero vectors in ℝ2. Note that
−1 is the only eigenvalue of Rπ (why?) and so must be of algebraic multiplicity 2.
The set of all eigenvectors corresponding to an eigenvalue – If λ1 is an eigenvalue
of a matrix A, all the eigenvectors associated to λ1 form a set or “package”. This set is
the nullspace, Null(A
−
λ1I ), of the matrix A
−
λ1I, minus the 0-vector. Remember
that Null(A
−
λ1I ) is a subspace (i.e., all solutions of the system (A
−
λ1I )x = 0; so it
is closed under linear combinations.
34.2.3 Definition – We will refer to the set
Null(A
−
λ1I ) = Eλ1
as the eigenspace associated to λ1. Every vector in Eλ1 is an eigenvector associated to
λ1 except for the vector 0. To each eigenvalue λI of A we can associate a particular
eigenspace Eλi.
The following diagram illustrates how we compute, given a matrix A, its eigenvalues
and determine their respective eigenspaces.
An eigenspace Eλ1 of an n by n matrix A is by definition Null(A
−
λ1I ) and so is a
subspace of ℝn. By solving for Null(A
−
λ1I ) by row-reduction will always produce
its basis.
34
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Document Summary
Wednesday, july 19 lecture 34 : eigenvectors. (refers to 6. 2) Concepts: define the family of all eigenvectors associated to an eigenvalue, find all eigenvectors associated to an eigenvalue, define an eigenspace. 34. 1 theorem let 1 be a number. The number 1 is an eigenvalue of a if and only if the matrix equation ax = 1x with variable x has a non-trivial solution. 1 is an eigenvalue of a 1 is a solution to det(a i) = 0. (a 1i)x = 0 has a non-trivial solution. Ax 1ix = 0 has a non-trivial solution. Ax = 1ix has a non-trivial solution. Ax = 1x has a non-trivial solution. 34. 1. 1 example show that 1 is an eigenvalue of the following matrix a. Solution: by the above theorem it suffices to verify that the following homogeneous system (a (1)i)x = 0 has a non-trivial solution. Since (a (1)i)x = 0 has a non-trivial solution then 1 is an eigenvalue of a.
