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Week11-12

# Week11-12 - Chapter 4 Eigenvalues and Eigenvectors Week...

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Chapter 4: Eigenvalues and Eigenvectors April 17, 2009 Week 11-12 1 Eigenvalues and eigenvectors If a linear transformation T : R n R n has the form T x 1 x 2 . . . x n = λ 1 0 ··· 0 0 λ 2 ··· 0 . . . . . . . . . . . . 0 0 ··· λ n x 1 x 2 . . . x n , then we can easily see that T ( e i ) = λ i e i ( i = 1 , 2 ,...,n ), i.e., each e i is dilated λ i times by T . Example 1.1. The linear transformation T : R 2 R 2 , deﬁned by T ( x ) = 2 0 0 3 ‚• x 1 x 2 = 2 x 1 3 x 2 , is to dilate the ﬁrst coordinate two times and the second coordinate three times. Example 1.2. Let T : R 2 R 2 be a linear transformation deﬁned by T ( x ) = 4 - 2 3 - 3 ‚• x 1 x 2 . What can we say geometrically about T ? Consider the basis B = { u 1 , u 2 } of R 2 , where u 1 = 2 1 , u 2 = 1 3 . Then T ( u 1 ) = 6 3 = 3 2 1 = 3 u 1 , T ( u 2 ) = - 2 - 6 = - 2 1 3 = - 2 u 2 . For any vector v = c 1 u 1 + c 2 u 2 , we have [ v ] B = c 1 c 2 , and T ( v ) = c 1 T ( u 1 ) + c 2 T ( u 2 ) = 3 c 1 u 1 - 2 c 2 u 2 , 1

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Thus [ T ( v )] B = 3 c 1 - 2 c 2 = 3 0 0 - 2 ‚• c 1 c 2 . If one uses the basis B to describe vector v with coordinate vector [ v ] B , then the coordinate vector of T ( v ) under the basis B simply described as £ T ( v ) / B = 3 0 0 - 2 £ v / B . This means that the matrix of T relative to the basis B is as simple as a diagonal matrix. The above discussion demonstrates that for a linear transformation T : V V , the nonzero vectors v satisfying the condition T ( v ) = λ v (1.1) for some scalar λ is important in simplifying a linear transformation T . Deﬁnition 1.1. Given a linear transformation T : R n R n with T ( x ) = A x . A nonzero vector v in R n is called an eigenvector of T (the matrix A ) if there exists a scalar λ such that T ( v ) = A v = λ v . (1.2) The scalar λ is called an eigenvalue of T (the matrix A ) and the nonzero vector v is called an eigenvector of T (the matrix A ) corresponding to the eigenvalue λ . Example 1.3. Let A = 1 6 5 2 . Then u = 6 - 5 is an eigenvector of A . However, but v = 3 - 2 is not an eigenvector of A . Proposition 1.2. For any n × n matrix A , the value 0 is an eigenvalue of A ⇐⇒ det A = 0 . Proof. Note that det A = 0 ⇐⇒ A is not invertible ⇐⇒ Nul A 6 = { 0 } . The set of eigenvectors of A corresponding to the zero eigenvalue is the set Nul A - { 0 } . Theorem 1.3. Let v 1 ,..., v p be eigenvectors of a matrix A corresponding to distinct eigenvalues λ 1 ,...,λ p , respectively. Then v 1 ,..., v p are linearly independent. Proof.
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Week11-12 - Chapter 4 Eigenvalues and Eigenvectors Week...

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