Chpater3&4 MA351Spring2011

   This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.

3.1 11 Ax = 0 to find the span,very basic but very important

3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already.

3.2 4 Fact 3.2.2.

3.2 12 Linearly dependent

3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.

3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3


4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.

4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.

By Bingrou Zhou

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Correspondence Chess Grandmaster and Purdue Alumni

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