Eigenvector question

  • Thread starter makris
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  • #1
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Hi all,

I have the following question.

A = nxn non singular matrix
I = nxn identity matrix
li = eigevalues of A i=1,2...n
ui = eigenvectors corresponding to the previous eigenvalues.

It true that

( A - l1 * I ) * x =0

is satisfied by any vector of the form x = a1 * u1 where a1= arbitrary real number

Lanczos in his book Applied Analysis p. 61 claims that the following quadratic equation in A

( A - l1 * I ) * ( A - l2 * I ) * x = 0

is satisfied by an arbitrary linear combination of the first two eigenvectors

x = a1 * u1 + a2 * u2

It is not very obvious to me why this happens.
(Extending this to include n eigenvectors and eigenvalues will eventually lead to the so called Cayley-Hamilton theorem.)

I was wondering if you could give me a hint starting from first principals.

Thanks
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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( A - l1 * I ) * ( A - l2 * I ) = ( A - l2 * I ) *( A - l1 * I )
so if
T=( A - l1 * I ) * ( A - l2 * I ) = ( A - l2 * I ) *( A - l1 * I )
T(a1 * u1)=a1*T(u1)=0
T(a2 * u2)=a1*T(u2)=0
hence by linearity
T(a1 * u1 + a2 * u2)=0
 

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