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##### (Solution) - In each case either show that the statement is true

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In each case either show that the statement is true or give an example showing that it is false. Throughout, X, Y, Z, X1, X2,..., Xn denote vectors in Rn.
(a) If U is a subspace of Rn and X+Y is in U, then X and Y are both in U.
(b) If U is a subspace of Rn and rX is in U, then X is in U.
(c) If U is a nonempty set and .sX + tY is in U for any s and t whenever X and Y are in U, then U is a subspace.
(d) If U is a subspace of Rn and X is in U, then - X is in U.
(e) If {X, Y} is independent, then {X, Y, X + Y} is independent.
(f) If {X, Y, Z} is independent, then {X, Y) is independent.
(g) If (X, Y) is not independent, then {X, Y, Z) is not independent.
(h) If all of X1, X2,? ? ?,Xn are nonzero, then{X1, X2,..., Xn) is independent.
(i) If one of X1, X2,..., Xn is zero, then {X1 X2,..., Xn} is not independent.
(j) If aX+ bY+ cZ = 0 where a, b, and c are in R, then {X, Y, Z} is independent.
(k) If {X, Y, Z} is independent, then aX + bY+ cZ = 0 for some a, b, and c in R.
(l) If {X1, X2,..., Xn) is not independent, then t1X1 + t2X2 + ? ? ?+ tnXn = 0 for t, in R not all zero.
(m) If {X1, X2,..., Xn) is independent, then t1X1 + t2X2 + ? ? ? + tnXn = 0 for some ti, in R.
(n) Every set of four nonz-ero vectors in R4 is a basis.
(o) No basis of R3 can contain a vector with a component 0.
(p) R3 has a basis of the form {X, X + Y, Y) where X and Y are vectors.
(q) Every basis of R5 contains one column of I5.
(r) Every nonempty subset of a basis of R3 is again a basis of R3.
(s) If {X1 X2, X3, X4} and {Y1,Y2, Y3, Y4} are bases of R4, then {X1 + Y1,X2 + X2,X3 + Y3,X4 + Y4} is also a basis of R4.

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This question was answered on: Jul 11, 2017

Solution~000131230952.zip (18.37 KB)