See Attached For this question, no proof is required, but you may receive partia

Question

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For this question, no proof is required, but you may receive partial credit for work that is shown. Give an example of a nonempty subset U of R2 such that U is closed under vector addition and closed under taking additive inverses, but U is not a subspace of R2. Give an example of a nonempty subset U of R2 such that U is closed under scalar multiplication, but U is not a subspace of R2. Let U be the subspace of R4 defined by U = {(xl, x2, x3, x4) : x1 + x2 = x3 and 2x1 + x2 = x4} Find a basis for U. Hint: dim U = 2.

Explanation / Answer

(a)U = {x, 1/x} x belongs to R-{0}

obviously it is closed under addition, its inverse is (-x, -1/x), but it is not a subspace since (x,1/x) +(y,1/y) does not belong to U.

(b)the same example as above can be used here.
U = {x, 1/x} x belongs to R-{0}

it is closed under multiplication but not a subspace

(c)basis = {(1,0,1,2), (0,1,1,1)}

(x1, x2, x3, x4) = x1.(1,0,1,2) + x2.(0,1,1,1)

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