MATH 4130 Midterm: Exam 2 Solution 1

• School: UCCS
• Department: Mathematics
• Course: MATH 4130
• Professor: All

Document Summary

Exam ii linear algebra i: problem 1. Solution: if is an eigenvalue and t v = v for some v 6= 0, then v = t v = t 2v = t (t v) = t ( v) = This implies = 0 or = 1 are the only possible eigenvalues of t . If = 0 is an eigenvalue, the corresponding eigenspace is e0 = n (t 0i) = n (t ). If = 1 is an eigenvalue, then e1 = {v v |t v = v} = r(t ) (last equality needs a bit of proof ) (2) show that t is diagonalizable. [hint: use that fact that the relation t 2 = t implies n (t ) r(t ) = v and t is the projection on r(t ) along n (t ). You do not need to prove this here!]