Engineering Calculus Notes 389

# Engineering Calculus Notes 389 - 3.9 THE PRINCIPAL AXIS...

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Unformatted text preview: 3.9. THE PRINCIPAL AXIS THEOREM 377 Finding Eigenvectors How do we ﬁnd the eigenvectors of a matrix A? If we knew the eigenvalues, then for each eigenvalue λ we would need to simply solve the system of linear equations → → A− = λ− u u → for − . To ﬁnd the eigenvalues, we use our observations about singularity u and determinants in Appendix E. We rewrite the eigenvector equation in the form → → →− A− − λ− = 0 u u or, using the distributive law for matrix multiplication, → →− (A − λI )− = 0 . u (3.33) You should verify that the matrix A − λI is obtained from A by → subtracting λ from each diagonal entry, and leaving the rest alone. Now − u is by assumption a nonzero vector, and this means that the matrix A − λI is singular (since it sends a nonzero vector to the zero vector). From Appendix E, we see that this forces det(A − λI ) = 0. (3.34) Now given a 3 × 3 matrix A, Equation (3.34) is an equation in the unknown λ; you should verify (Exercise 4) that the left side of this equation is a polynomial of degree three in λ; it is called the characteristic polynomial of A, and Equation (3.34) is called the characteristic equation of A. As a corollary of Proposition 3.9.2 we have Remark 3.9.5. Every eigenvalue of A is a zero of the characteristic polynomial p(λ) = det (A − λI ). If A is symmetric and 3 × 3, then p(λ) is a cubic polynomial with three real zeroes, and these are the eigenvalues of A. Thus, we ﬁnd the eigenvalues of A ﬁrst, by ﬁnding the zeroes of the characteristic polynomial (i.e., solving the characteristic equation (3.34)); then for each eigenvalue λ we solve the system of equations (3.33) to ﬁnd the corresponding eigenvectors. Let us apply this to two examples. ...
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