MATH 415 Study Guide - Final Guide: Augmented Matrix, Euclidean Vector, Ion

• School: University of Illinois
• Department: Mathematics
• Course: MATH 415
• Professor: J Li

Document Summary

: a 2b = c, 4a + 2c = 0 (cid:21) : b 0(cid:27) r2. : a 2b = 1, 4a + 2c = 0 (cid:21) : a and b in r(cid:27) r2 a b c b. Solution. (a) w1 is a subspace of r3, since it is the nullspace of the matrix. 0 (b) this set is not a subspace of r2. Consider the vector (cid:20) 1 is in w4. Since 12 1, we have that (cid:20) 1. 0 (cid:21) is not in w4, since 22 = 4 > 1. A sketch of w4 is the disc of radius 1 in r2 (i. e. , the lled-in circle of radius 1). By plotting our counterexample, we can see visually that w4 is not closed under scalar multiplication. Also note that w4 is not closed under vector addition - can you nd a counterexample? (cid:3) Does not satisfy 0 2(0) = 1. Therefore w5 is not a subspace of r3.