� σ A A � I is singular det A � I 0 713 x x N A � I is the set of all

? σ a a ? i is singular det a ? i 0 713 x x n a ? i

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λ σ ( A ) ⇐⇒ A λ I is singular ⇐⇒ det( A λ I ) = 0 . (7.1.3) x = 0 x N ( A λ I ) is the set of all eigenvectors associated with λ. From now on, N ( A λ I ) is called an eigenspace for A . Nonzero row vectors y such that y ( A λ I ) = 0 are called left- hand eigenvectors for A (see Exercise 7.1.18 on p. 503). Geometrically, Ax = λ x says that under transformation by A , eigenvec- tors experience only changes in magnitude or sign—the orientation of Ax in n is the same as that of x . The eigenvalue λ is simply the amount of “stretch” or “shrink” to which the eigenvector x is subjected when transformed by A . Figure 7.1.1 depicts the situation in 2 . Ax = λ x x Figure 7.1.1 66 The words eigenvalue and eigenvector are derived from the German word eigen , which means owned by or peculiar to. Eigenvalues and eigenvectors are sometimes called characteristic values and characteristic vectors, proper values and proper vectors, or latent values and latent vectors.
COPYRIGHTED It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] Buy online from SIAM Buy from AMAZON.com Copyright c 2000 SIAM 7.1 Elementary Properties of Eigensystems 491 Let’s now face the problem of finding the eigenvalues and eigenvectors of the matrix A = 7 4 5 2 appearing in (7.1.1). As noted in (7.1.3), the eigen- values are the scalars λ for which det( A λ I ) = 0 . Expansion of det( A λ I ) produces the second-degree polynomial p ( λ ) = det( A λ I ) = 7 λ 4 5 2 λ = λ 2 5 λ + 6 = ( λ 2)( λ 3) , which is called the characteristic polynomial for A . Consequently, the eigen- values for A are the solutions of the characteristic equation p ( λ ) = 0 (i.e., the roots of the characteristic polynomial), and they are λ = 2 and λ = 3 . The eigenvectors associated with λ = 2 and λ = 3 are simply the nonzero vectors in the eigenspaces N ( A 2 I ) and N ( A 3 I ) , respectively. But deter- mining these eigenspaces amounts to nothing more than solving the two homo- geneous systems, ( A 2 I ) x = 0 and ( A 3 I ) x = 0 . For λ = 2 , A 2 I = 5 4 5 4 −→ 1 4 / 5 0 0 = x 1 = (4 / 5) x 2 x 2 is free = N ( A 2 I ) = x x = α 4 / 5 1 . For λ = 3 , A 3 I = 4 4 5 5 −→ 1 1 0 0 = x 1 = x 2 x 2 is free = N ( A 3 I ) = x x = β 1 1 . In other words, the eigenvectors of A associated with λ = 2 are all nonzero multiples of x = (4 / 5 1) T , and the eigenvectors associated with λ = 3 are all nonzero multiples of y = (1 1) T . Although there are an infinite number of eigenvectors associated with each eigenvalue, each eigenspace is one dimensional, so, for this example, there is only one independent eigenvector associated with each eigenvalue.

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