C__DOCUME~1_MAXWID~1_LOCALS~1_Temp_plugtmp-27_notes5-2 - ,...

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SECTION 5.2 HOW TO FIND EIGENVALUES For λ to be an eigenvalue of the (square) matrix A , the vector equation A v = λ v must have a nontrivial solution for v . This equation is called the characteristic equation of the matrix A . A scalar λ is an eigenvalue of A exactly when λ is a root or solution of the characteristic equation. EXAMPLE. Find the eigenvalues of the matrix 5 - 2 3 0 1 0 6 7 - 2 .
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Remember that λ = r is a root of the polynomial p ( λ ) exactly when λ - r is a factor of the polynomial. If p ( λ ) = λ 3 ( λ - 3) 2 ( λ +5), then we say that 0 is a root of multiplicity 3, 3 is a root of multiplicity 2, and - 5 is a root of multiplicity 1. The multiplicities are important in eigenvector/eigenvalue considerations, so we often list the roots in this case as 0
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Unformatted text preview: , , , 3 , 3 ,-5. When we use multiplicities and include complex roots, if any, then a polynomial of degree n has exactly n roots. It turns out that matrices A and B related by B = P-1 AP for some invertible matrix P share many important properties. Such matrices are said to be similar . FACT. Similar matrices have the same characteristic polynomial and so have the same eigenvalues with the same multiplicities. EXAMPLE, IF TIME. Analyze the long-term behavior of the system dened by x k +1 = A x k when A = "-1 / 2 1 3 / 2 # and x = " 1 10 # . HOMEWORK: SECTION 5.2...
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C__DOCUME~1_MAXWID~1_LOCALS~1_Temp_plugtmp-27_notes5-2 - ,...

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