129a_ans_5.3 - 1 2 1 1 3 3 1 3 4 1 0 0 0 2 0 0 0 3 1 2 1 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Spring 2010 Answer of Section 5.3 5.3 #1 " 408 - 952 168 - 392 # 5.3 #4 " 4 + 3 · 2 k 12(2 k - 1) 1 - 2 k 4 · 2 k - 3 # 5.3 #5 eigenvalues are 5 and 1 a basis for eigenspace E (5) is 1 1 1 a basis for eigenspace E (1) is 1 0 - 1 , 2 - 1 0 5.3 #11 - 1 4 - 2 - 3 4 0 - 3 1 3 =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 2 1 1 3 3 1 3 4 1 0 0 0 2 0 0 0 3 1 2 1 1 3 3 1 3 4 -1 5.3 #26 YES, it is possible that the matrix is NOT diagonalizable because the third eigenspace may be one-dimensional....
View Full Document

This note was uploaded on 09/08/2010 for the course MATH 129A at San Jose State University .

Ask a homework question - tutors are online