22 Below is a sevenstep proof of part b of Theorem 411 Justi

22. Below is a seven-step proof of part (b) of Theorem 4.1.1. Justify each step either by stating that it is true by hypothesis or by specifying which of the ten vector space axioms applies Hypothesis: Let u be any vector in a vector space V, let 0 be the zero vector in and let k be a scalar. Conclusion: Then ko 0. Proof: (1) ko ku k(0 u) -ku (2) (3) Since ku is in V. -ku is in V (4) Therefore, (ko ku) ku) --ku (-ku) (5) ko (6) ko (7)

Solution

(1) This step is true by axiom number 7 of vector space. Recall that cu+cv = c(u+v). Where u and v are elements of vectors space.

(2) This step is true by axiom number 4 of vector space. Recall that 0 + u = u. Where u is an arbitrary vector of vector space and 0 is zero vector of vector space V.

(3) This step is true by combination of axioms 5 and 6 of vector space. Recall that for every u in V, there exists a corresponding -u in V. Also, if u is in V, then ku is in V. Combining these two, we get that for every ku in V, there exists a -ku in V.

(4) This step is true algebraically because we are just adding an equal amount on both the sides of equation in step 2, therefore, the resultant equation must hold true.

(5) This step is true by axiom 3 of vector space. We are using associativity theorem to rewrite the left hand side of the previous step.

(6) This step is true by axiom 5 of vector space. That is, u-u = 0

(7) This step is true by axiom 4 of vector space. That is, u+0 = u. Any vector added to zero vector remains unchanged.