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Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #10 (due Thursday, November 17) All problems are from Leon’s book (8th edition). Section 5.5: 6c, 29a, 29b, 30a, 30b Section 5.6: 1a, 3, 4, 7, 8 MATH 304 Linear Algebra Lecture 22: Eigenvalues and eigenvectors (continued). Characteristic polynomial. Eigenvalues and eigenvectors of a matrix Definition. Let A be an n × n matrix. A number λ ∈ R is called an eigenvalue of the matrix A if A v = λ v for a nonzero column vector v ∈ R n . The vector v is called an eigenvector of A belonging to (or associated with) the eigenvalue λ . Remarks. • Alternative notation: eigenvalue = characteristic value , eigenvector = characteristic vector . • The zero vector is never considered an eigenvector. Diagonal matrices Let A be an n × n matrix. Then A is diagonal if and only if vectors e 1 , e 2 , . . . , e n of the standard basis for R n are eigenvectors of A . If this is the case, then the diagonal entries of the matrix A are the corresponding eigenvalues: A = λ 1 O λ 2 . . . O λ n ⇐⇒ A e i = λ i e i Eigenspaces Let A be an n × n matrix. Let v be an eigenvector of A belonging to an eigenvalue λ . Then A v = λ v = ⇒ A v = ( λ I ) v = ⇒ ( A − λ I ) v = . Hence v ∈ N ( A − λ I ), the nullspace of the matrix A − λ I . Conversely, if x ∈ N ( A − λ I ) then A x = λ x . Thus the eigenvectors of A belonging to the eigenvalue λ are nonzero vectors from N ( A − λ I ). Definition. If N ( A − λ I ) negationslash = { } then it is called the eigenspace of the matrix A corresponding to the eigenvalue λ . How to find eigenvalues and eigenvectors? Theorem Given a square matrix A and a scalar λ , the following statements are equivalent: • λ is an eigenvalue of A , • N ( A − λ I ) negationslash = { } , • the matrix A − λ I is singular, • det( A − λ I ) = 0. Definition. det( A − λ I ) = 0 is called the characteristic equation of the matrix A . Eigenvalues λ of A are roots of the characteristic equation. Associated eigenvectors of A are nonzero solutions of the equation ( A − λ I ) x = . Example. A = parenleftbigg a b c d parenrightbigg . det( A − λ I ) = vextendsingle vextendsingle vextendsingle vextendsingle a − λ b c d − λ vextendsingle vextendsingle vextendsingle vextendsingle = ( a − λ )( d − λ ) − bc = λ 2 − ( a + d ) λ + ( ad − bc ). Example. A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ....
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 Spring '07
 Parzen
 Linear Algebra, Det

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