Question

Determine whether the set w is a subspace of R3 with the standard operations. If not, state why. (Select all that apply.) W = {(x1, 1/X1, X3): X1 and xy are real numbers, X1 + 0) W is a subspace of R W is not a subspace of R because it is not closed under addition. Wis not a subspace of R because it is not closed under scalar multiplication. X

Answer 1

real wa ) (x, 1 1/2, 383): X, and Xz are numbers x, TO u and V are in W, then utv is in de 1) if 2) If u fs in w and c is any scalar, scalar, then cu is in W. New consider the given subset is of R3 as represented 123): X, and az are real below w={ca, it nunturs ܐܥ take s let u= (xq ,X4) EW xa VE ( X5, kus X7) E W Now verity closure condition I for the points u auud V EN ut v = (X21 ' xy) +(85) as : X7) azt is 1/2 + 1 25 X4 + xy) EW Utv tu W iso Hence closed under addition closure Now verify condition 2 for the point. ㅕ and scalar CER as represented below. UE W cu = c(X2Xy) (cxr, a cх4) EW .

Since the product two real numbers and division real number. Munibers real also Q of two Scalar multiplication. Hence W is closed under Since both the closure Conditions are satisfied the given set W is a subspace of