QuestionAugust 24, 2025

Use the vector u=(u_(1),ldots ,u_(n)) to verify the following algebraic properties of R^n a u+(-u)=(-u)+u=0 b. c(du)=(cd)u for all scalars c and d

Use the vector u=(u_(1),ldots ,u_(n)) to verify the following algebraic properties of R^n a u+(-u)=(-u)+u=0 b. c(du)=(cd)u for all scalars c and d
Use the vector u=(u_(1),ldots ,u_(n)) to verify the following algebraic properties of R^n a u+(-u)=(-u)+u=0 b. c(du)=(cd)u for all scalars c and d

Solution
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Answer

a. u + (-u) = (-u) + u = 0; b. c(du) = (cd)u for all scalars c and d. Explanation 1. Verify u + (-u) = 0 For vector u = (u_1, \ldots, u_n), -u = (-u_1, \ldots, -u_n). Adding these gives u + (-u) = (u_1 - u_1, \ldots, u_n - u_n) = (0, \ldots, 0) = 0. 2. Verify (-u) + u = 0 Similarly, (-u) + u = (-u_1 + u_1, \ldots, -u_n + u_n) = (0, \ldots, 0) = 0. 3. Verify c(du) = (cd)u For scalars c and d, du = (du_1, \ldots, du_n). Then c(du) = c(du_1, \ldots, du_n) = (cdu_1, \ldots, cdu_n). Also, (cd)u = (cdu_1, \ldots, cdu_n). Thus, c(du) = (cd)u.

Explanation

1. Verify $u + (-u) = 0$<br /> For vector $u = (u_1, \ldots, u_n)$, $-u = (-u_1, \ldots, -u_n)$. Adding these gives $u + (-u) = (u_1 - u_1, \ldots, u_n - u_n) = (0, \ldots, 0) = 0$.<br /><br />2. Verify $(-u) + u = 0$<br /> Similarly, $(-u) + u = (-u_1 + u_1, \ldots, -u_n + u_n) = (0, \ldots, 0) = 0$.<br /><br />3. Verify $c(du) = (cd)u$<br /> For scalars $c$ and $d$, $du = (du_1, \ldots, du_n)$. Then $c(du) = c(du_1, \ldots, du_n) = (cdu_1, \ldots, cdu_n)$. Also, $(cd)u = (cdu_1, \ldots, cdu_n)$. Thus, $c(du) = (cd)u$.
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