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.

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b