mat1302-lecture19(2)

mat1302-lecture19(2) - MAT 1302 - Mathematical Methods II...

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MAT 1302 - Mathematical Methods II Alistair Savage Mathematics and Statistics University of Ottawa Winter 2010 – Lecture 19 Alistair Savage (uOttawa) MAT 1302 - Mathematical Methods II Winter 2010 – Lecture 19 1 / 27
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Announcements Second Midterm: Solutions posted on course website. Fourth Assignment: Posted on Virtual Campus. Due at 8:30am, April 6. Covers material up to and including today’s lecture. Teaching evaluations: Beginning of class on Tuesday, March 30. Please come to class and be on time. Last time: Eigenvectors/eigenvalues. Alistair Savage (uOttawa) MAT 1302 - Mathematical Methods II Winter 2010 – Lecture 19 2 / 27
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Eigenvectors, eigenvalues, and eigenspaces Definition (eigenvectors and eigenvalues) Suppose A is a square matrix. If ~ x is a nonzero vector and λ is a scalar such that A ~ x = λ~ x , ~ x 6 = ~ 0 , then λ is an eigenvalue of A , and ~ x is an eigenvector of A (an eigenvector corresponding to the eigenvalue λ ). If λ is an eigenvalue, then the set of solutions to A ~ x = λ~ x (or ( A - λ I ) ~ x = ~ 0) is the eigenspace corresponding to λ . Note: Since the eigenspaces are null spaces of A - λ I , they are subspaces. Alistair Savage (uOttawa) MAT 1302 - Mathematical Methods II Winter 2010 – Lecture 19 3 / 27
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Example 1 A = ± 2 1 4 2 ² , ~ u = ± 1 2 ² , ~ v = ± 0 3 ² Is ~ u or ~ v an eigenvector of A ? Solution: A ~ u = ± 2 1 4 2 ²± 1 2 ² = ± 4 8 ² = 4 ± 1 2 ² = 4 ~ u Therefore ~ u is an eigenvector of A with eigenvalue 4. A ~ v = ± 2 1 4 2 ²± 0 3 ² = ± 3 6 ² Since ± 3 6 ² is not a multiple of ~ v , the vector ~ v is not an eigenvector of A . Alistair Savage (uOttawa) MAT 1302 - Mathematical Methods II Winter 2010 – Lecture 19 4 / 27
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Example 2 Is λ = - 3 an eigenvalue of A = - 4 - 2 - 3 1 - 1 3 - 2 - 4 - 9 ? If so, find a basis for the corresponding eigenspace. Solution: ± A + 3 I ~ 0 ² = - 1 - 2 - 3 0 1 2 3 0 - 2 - 4 - 6 0 row reduce ------→ 1 2 3 0 0 0 0 0 0 0 0 0 Since there are nontrivial solutions, - 3 is an eigenvalue of A . The eigenspace is the solution set. Alistair Savage (uOttawa) MAT 1302 - Mathematical Methods II Winter 2010 – Lecture 19 5 / 27
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± A + 3 I ~ 0 ² = - 1 - 2 - 3 0 1 2 3 0 - 2 - 4 - 6 0 row reduce ------→ 1 2 3 0 0 0 0 0 0 0 0 0 We write the general solution in vector parametric form: x 1 = - 2 x 2 - 3 x 3 x 2 free x 3 free = ~ x = - 2 x 2 - 3 x 3 x 2 x 3 = x 2 - 2 1 0 + x 3 - 3 0 1 with x 2 , x 3 any scalars. Therefore, the eigenspace is Span - 2 1 0 , - 3 0 1 and a basis is - 2 1 0 , - 3 0 1 . Alistair Savage (uOttawa)
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mat1302-lecture19(2) - MAT 1302 - Mathematical Methods II...

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