Show that every proper subspace u of r 2 is a line through the origin. [hint: if d is a nonzero vector in u, let l = rd = {rd | r in r} denote the line with direction vector d. if u is in u but not in l, argue geometrically that every vector v in r 2 is a linear combination of u and d.]

Explanation:

Given any pair of elements from U, then the line formed by those 2 elements must be on U because U is a subspace of R². Lets suppose we have two elements v = (v1,v2) and w = (w1,w2), v and 2 not aligned. We call Lv the line with direction v that contains v (and the origin vector). This line contain all vector multiples of v.

If v1 ≠ 0, then we should have a vector of the form (w1 , k) on Lv, obtained by multypling v by w1/v1. Note that k must be different from w2, because v and w are not aligned. (w1, k) ∈ U, and the difference (w1, k) - (w1,w2) = (0,k-w2) is non zero vector multiple of (0,1). Therefore, (0,1) ∈ U.

By a similar argument, if v2 ≠ 0, then (1,0) ∈ U. Any linear combination between (1,0) and (0,1) is on U, as a result. U must be R².

If v1 = 0, then both v2 and w1 must be different from 0, otherwise v and w would be aligned. Since (0,v1) ∈ U, (0,1) should be on U, because it is a multiple. And the element (w1,0) = w- w2*(0,1) is also an element of U. Thus, (1,0) is on U and therefore U is R². A similar argument works if v2 = 0.

answer: are you

step-by-step explanation:

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

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