MAT-2240 Midterm: MATH 2240 App State Spring2011 Test3 answer key
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If w is a subspace, show it is by running through the subspace test. The equation y = x2 should make us suspicious about w being a subspace. Now (x + u)2 = x2 + 2xu + u2 6= x2 + u2, so closure under vector addition should fail. Thus w is not closed under vector addition so. We could also try to see if scalar multiplication holds or fails. (cx)2 = c2x2 6= cx2 so it seems like closure under scalar multiplication will also fail. Any scalar other than 1 should show this. Thus w is not closed under scalar multiplication so w is not a subspace of r3. Recall: p2 is the space of polynomials of degree 2 or less along with 0. Quickly: w = {ax2 + b(x2 + 3) | a, b r} = span({x2, x2 + 3}) so w is a subspace (since span(s) is always a subspace for any set s).
