Whats the answer for these subspaces of vector spaces questions :

Question

Q1:verify that W is subspace of V.In each caes,assume that V has the standard operations.

1.W={(x₁,x₂,x₃,0): x₁,x₂, and x₃ are real numbers}

V = R⁴

2.W={(x,y,2x-3y): x and y are real numbers}

V = R³

3.W is the set of all 2 x 2 matrices of the form:⌈

⌈0 a⌉

⌊b 0⌋

V = M_2.2

Q2:determine whether the set W is a subspace of R³ with the standard operations.justify your answer.

4.W={(x₁,0,x₃): x₁ and x₃ are real numbers}

5.W={(x₁,x₂,4): x₁ and x₂ are real numbers}

6.W={(s,s-t,t): s and t are real numbers}

Answer 1

a subspace W of a vector space V is a subset of V closed under linear combination:

if w and w' are in W and s is a scalar, sw + w' is in W.

w = (x, y, z, 0) and w' = (x', y', z', 0)

sw + w' = s(x,y,z,0) + (x',y',z',0) = (sx + x', sy+y', sz + z', 0).

letting sx + x' = a, sy + y' = b, sz + z' = c, this is (a, b, c, 0), which is in W.

w = (x,y,2x-3y), w' = (x',y',2x'-3y')

sw + w' = s(x,y,2x-3y) + (x',y',2x'-3y') = (sx + x', sy + y', s(2x+3y) + (2x' + 3y'))

.............. = (sx + x', sy + y', 2sx + 3sy + 2x' + 3y') = (sx + x', sy + y', 2(sx + x') + 3(sy + y'))

letting sx + x' = a and sy + y' = b, then this is (a, b, 2a + 3b), which is in W.

w =

[0 a]

[b 0]

w' =

[0 a']

[b' 0]

sw + w' = s[0 a] + [0 a'] = [0 sa + a']

................. [b 0]....[b' 0]....[sb + b' 0]

letting sa + a' = x and sb + b' = y, this is [0 x], which is in W.

....................................................................[y 0]

yes. s(x, 0, y) + (x', 0, y') = (sx + x', 0, sy +y'),

no. (2, 2, 4) is in W but 1/2(2, 2, 4) = (1, 1, 2) is not. So it isn't closed under linear combination

yes. s(x, x-y, y) + (x', x'-y', y') = (sx + x', (sx+x') - (sy+y'), sy + y')