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# The Eigenvalue Problem

Module by: Steven J. Cox. E-mail the author

Summary: (Blank Abstract)

## Introduction

Harking back to our previous discussion of The Laplace Transform we labeled the complex number λλ an eigenvalue of BB if λIB λ I B was not invertible. In order to find such λλ one has only to find those ss for which sIB-1 s I B is not defined. To take a concrete example we note that if

B=( 110 010 002 ) B 110 010 002
(1)
then
sIB-1=1s12(s2)( (s1)(s2)s20 0(s1)(s2)0 00s12 ) s I B 1 s 1 2 s 2 s1 s2 s2 0 0 s1 s2 0 0 0 s1 2
(2)
and so λ1=1 λ1 1 and λ2=2 λ2 2 are the two eigenvalues of BB. Now, to say that λj IB λj I B is not invertible is to say that its columns are linearly dependent, or, equivalently, that the null space 𝒩 λj IB 𝒩 λj I B contains more than just the zero vector. We call 𝒩 λj IB 𝒩 λj I B the jjth eigenspace and call each of its nonzero members a jjth eigenvector. The dimension of 𝒩 λj IB 𝒩 λj I B is referred to as the geometric multiplicity of λj λj . With respect to BB above, we compute 𝒩 λ1 IB 𝒩 λ1 I B by solving (IB)x=0 IB x 0 , i.e., ( 0-10 000 001 )( x1 x2 x3 )=( 0 0 0 ) 0 -1 0 000 001 x1 x2 x3 0 0 0 Clearly 𝒩 λ1 IB= c( 100 )T cR 𝒩 λ1 I B c 100 c Arguing along the same lines we also find 𝒩 λ2 IB= c( 001 )T cR 𝒩 λ2 I B c 001 c That BB is 3x3 but possesses only 2 linearly eigenvectors leads us to speak of BB as defective. The cause of its defect is most likely the fact that λ1 λ1 is a double pole of sIB-1 sI B In order to flesh out that remark and uncover the missing eigenvector we must take a much closer look at the transfer function RssIB-1 R s sI B In the mathematical literature this quantity is typically referred to as the Resolvent of BB.

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