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10 - Eigenvalues and Eigenvector 2

10 - Eigenvalues and Eigenvector 2 - 1 Math 1b Practical...

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1 Math 1b Practical — Eigenvalues and eigenvectors, II February 11, 2011 Example 6. Let P be the matrix of the orthogonal projection onto a subspace U of R n . If u U , then P u = u , and if w U , then P w = 0. That is, elements of U are eigenvec- tors corresponding to eigenvalue 1, and elements of U are eigenvectors corresponding to eigenvalue 0. The (geometric) multiplicity of 1 is dim( U ) and the (geometric) multiplicity of 0 is n dim( U ). Let R be the matrix of the reflection through a subspace U of R n . If u U , then R u = u , and if w U , then R w = w . That is, elements of U are eigenvectors corresponding to eigenvalue +1, and elements of U are eigenvectors corresponding to eigenvalue 1. (I may give a more careful explanation in class.) Example 7. If A and B are square matrices, the eigenvalues of A O O B are those of A AND those of B . More about this in class, or in a later version of this handout. In Example 1 of the part I of this handout, we saw that 5 and 2 were the eigenvalues of A = 1 3 4 2 and that A e 1 = 1 3 4 2 3 4 = 15 20 = 5 e 1 , A e 2 = 1 3 4 2 1 1 = 2 2 = 2 e 2 . As an indication of why this helps us understand A and its powers, first note that A n e 1 = 5 n e 1 and A n e 2 = ( 2) n e 2 . Since e 1 and e 2 form a basis for R 2 , we can understand A n x by writing x as a linear combination of e 1 and e 2 . For example, let x = (9 , 5) . We have (check this) 9 5 = 2 3 4 + 3 1 1
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2 and then A n 9 5 = 2 A n 3 4 +3 A n 1 1 = 2 · 5 n 3 4 +3 · ( 2) n 1 1 = 6 · 5 n + 3 · ( 2) n 8 · 5 n 3 · ( 2) n . That is, we have derived a formula for A n (9 , 5) . One can see from this, for example, that the ratio of its first and second coordinates tends to 3 / 4 as n tends to infinity. Sometimes a matrix has an positive eigenvalue λ so that λ > | μ | for every other eigenvalue μ ; such an eigenvalue is called the dominant eigenvalue . In this example, 5 is the dominant eigenvalue.
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