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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 Spring '08
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 Calculus, Linear Algebra, Linear Equations, Eigenvectors, Equations, Vectors, Quadratic equation, Elementary algebra, Complex number

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