DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

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90 II. DETERMINANTS AND EIGENVALUES4. Eigenvalues and Eigenvectors for n × n MatricesOne way in which to understand a matrix A is to examine its effects on thegeometry of vectors in R n . For example, we saw that det A measures the relativechange in area or volume for figures generated by vectors in two or three dimensions.Also, as we have seen in an exercise in Chapter I, Section 2, the multiplesAe 1 ,Ae 2 ,...,Ae n of the standard basis vectors are just the columns of the matrixA. More generally, it is often useful to look at multiples Av for other vectors v.Example 1. LetA =[ ]2 1.1 2Here are some examples of products Av.vAv[ ] [ [ [ [ ]1 1 1 0.5 00 0.5]1]1]1[ ] [ ] [ ] [ [ ]2 2.5 3 2 11 2 3 2.5]2These illustrate a trend for vectors in the first quadrant.Vectors vTransformed vectors AvVectors pointing near one or the other of the two axes are directed closer to thediagonal line. A diagonal vector is transformed into another diagonal vector.Let A be any n × n matrix. In general, if v is a vector in R n , the transformedvector Av will differ from v in both magnitude and direction. However, somevectors v will have the property that Av ends up being parallel to v; i.e., it pointsin the same direction or the opposite direction. These vectors will specify ‘natural’axes for any problem involving the matrix A.
4. EIGENVALUES AND EIGENVECTORS FOR n × n MATRICES 91vvAVPositive eigenvalueAvNegative eigenvalueVectors are parallel when one is a scalar multiple of the other, so we make thefollowing formal definition. A non-zero vector v is called an eigenvector for thesquare matrix A if(1) Av = λvfor an appropriate scalar λ. λ is called the eigenvalue associated with the eigenvectorv.In words, this says that v ≠ 0 is an eigenvector for A if multiplying it by A hasthe same effect as multiplying it by an appropriate scalar. Thus, we may thinkof eigenvectors as being vectors for which matrix multiplication by A takes on aparticularly simple form.It is important to note that while the eigenvector v must be non-zero, the correspondingeigenvalue λ is allowed to be zero.Example 2. LetA =[ ]2 1.1 2We want to see if the system[ ][ ] [ ]2 1 v1 v1=λ1 2 v 2 v 2has non-trivial solutions v 1 ,v 2 . This of course depends on λ. If we write this systemout, it becomesor, collecting terms,In matrix form, this becomes(2)2v 1 + v 2 = λv 1v 1 +2v 2 =λv 2(2 − λ)v 1 + v 2 =0v 1 +(2−λ)v 2 =0.[ ]2 − λ 1v = 0.1 2−λ
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