{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution08_121

# Solution08_121 - Solutions to PS 8(Math 121 J Scholtz...

This preview shows pages 1–2. Sign up to view the full content.

Solutions to PS 8 (Math 121) J. Scholtz January 6, 2007 Question 1: 7.1/2(a) We are to find the basis of the generalazied eigenvectors and the corresponding Jordan form of matrix A : A = 1 1 - 1 3 This matrix gives us a double eigenvalue of λ = 2. Hence we need to find all the vectors such that: A = - 1 1 - 1 1 a b = 0 This is satisfied by a (1 , 1). Therefore we need to search for another generalized eigenvector. So we need to find a vector such that ( A - 2 I ) 2 v = 0 but ( A - 2 I ) v = 0. The solution to the first equation is any vector since ( A - 2 I ) 2 = 0, therefore we just need to find a vector such that ( A - 2 I ) = 0, hence anything linearly independent from a (1 , 1), pick v = ( - 1 , 0), so that ( A - 2 I ) v = (1 , 1). Then the Jordan form is: J = 2 1 0 2 Question 2: 7.1/3(a) First, by action on the standard basis we can establish that: A = 2 - 1 0 0 2 - 2 0 0 2 The only (triple) eigenvalue is again λ = 2. We can see that the rank( A - 2 I ) = 2 and so there must be a three-cycle of generalized eigenvectors, such that ( A - 2 I ) 3 v = 0 but ( A - 2 I ) 2 v = 0. But we know that ( A - 2 I ) 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}