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Mathematics

MATH 136

Robert Sproule

Winter

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Friday, March 28 − Lecture 33 : Eigenvectors associated to an eigenvalue.
Concepts:
1. Define the family of all eigenvectors associated to an eigenvalue.
2. Find all eigenvectors associated to an eigenvalue.
3. Define an eigenspace.
33.1 Theorem − Let λ be a number. The number λ is an eigenvalue ofA if and only if
1 1
the matrix equation Ax = λ x w1th variable x has a non-trivial solution.
Proof :
λ 1s an eigenvalue ofA ⇔ λ is a 1olution to det(A − λI) = 0
⇔ det(A − λ I)1= 0
⇔ A − λ I 1s not invertible
⇔ (A − λ I)1 = 0 has a non-trivial solution
⇔ Ax − λ Ix1= 0 has a non-trivial solution
⇔ Ax = λ Ix 1as a non-trivial solution
⇔ Ax = λ x h1s a non-trivial solution.
Different ways of viewing an eigenvalue. The following are equivalent:
1. λ i1 an eigenvalue ofA
2. (A − λ I1x = 0 has a non-trivial solution
3. Ax = λ x 1as a non-trivial solution
4. Null(A − λ I)1≠ {0} has a non-trivial solution
33.1.1 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. 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.
33.1.2 Different ways of viewing an eigenvalue. We have the following different
ways of viewing an eigenvalue of A:
λ1= eigenvalue of A ⇔ det(A − λ I) = 0 1
⇔ A − λ I 1s not invertible,
⇔ (A − λ I)1 = 0 has a non-trivial solution.
⇔ Null(A − λ I) 1ontains infinitely many vectors
⇔ Ax = λ x for infinitely many vectors x
1
33.2. Definition − Suppose λ is a1 eigenvalue of n × n matrix A. Then any non-zero
vector x in Null(A − λ I) is called an eigenvectorof A corresponding to λ .
1 1 1
33.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always
exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix
1
A invites the following question: “x is 1ssociated to which of the eigenvalues of A?”.
An eigenvector of A can only correspond to a single eigenvalue. If x is an 1
eigenvector of A associated to the eigenvalues λ and λ1of A th2n λ = λ . Thi1 is 2
because x c1nnot be the zero vector. To see this consider:
The expression “λ 1 1= λ 2 1mplies λ = λ1” si2ce an eigenvector cannot be the zero
vector.
33.2.2 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 : ℝ → ℝ .n n
Definition –We say that λ is 1n eigenvalue of the linear map T: ℝ → ℝ if n n
T(x) = λ x1
has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear
transformation T corresponding to the eigenvalue λ 1 if x1is a non-trivial solution to
T(x) = λ x1 n n
- If λ i1 an eigenvalue of T: ℝ → ℝ and x is an eigenvec1or corresponding to
λ 1hen, “T maps x to a1scalar multiple λ x of x ”1 1 1
2 2
33.2.2.1 Example – Suppose R : ℝ → ℝ πs the linear transformation which
rotates points in ℝ 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 ℝ , R (x) = −x. Then λ = −1 is an eigenvalue of
π 1 2
R π The eigenvectors associated to λ = −1 ar1 all non-zero vectors in ℝ .
The set of all eigenvectors corresponding to an eigenvalue – If λ is an eigen1alue
of a matrix A, all the eigenvectors associated to λ form a1set or “package”. This set is
the nullspace, Null(A − λ I ), of the mat

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