for part (d) I understand why it is not closed underaddition I do not understand

Question

for part (d) I understand why it is not closed underaddition I do not understand why it is closed under multiplicationbecasue you end up with (ka1, ka1, ka1 +kc1 + k , kc1) How does that fulfill the fact that the middle term must bea1 + c1 + 1... in this example, the middleexample is ka1 + kc1 + k ...k instead of 1 for part (d) I understand why it is not closed underaddition I do not understand why it is closed under multiplicationbecasue you end up with (ka1, ka1, ka1 +kc1 + k , kc1) How does that fulfill the fact that the middle term must bea1 + c1 + 1... in this example, the middleexample is ka1 + kc1 + k ...k instead of 1

Explanation / Answer

(ka1, ka1, ka1 + kc1 +k , kc1) This can not satisfy closed under scalar multiplication. for instance if you had the a vector in R2 say [x , y + 1] x,y are elements of Reals, you would have a line going throughpoint (0,1). Then you multiply it by scalar c, which does notequal 1, then if it is a vector space the line must have the sameorientation and position therefore it must go through (0, 1) and c[x, y +1] = [cx, cy +c] now when x and y are = 0 the line is at point (0, c) and is off theline and is therefore not closed! I agree with you, and am confused about why your text saysthis apparent fallacy

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