Please show how I should support my answer
U = {[x y x] : x ≠ 0} I said no because there is no zero vector...am I right?
U = {X ∈ ℝ^3 : AX=3X} I said yes, it is closed under scalar multiplication, is that right? What else should i say?
U = {[a b c] : a+2b+c=39} I have no idea how to prove this
10 points! Thanks!
U = {[x y x] : x ≠ 0} I said no because there is no zero vector...am I right?
U = {X ∈ ℝ^3 : AX=3X} I said yes, it is closed under scalar multiplication, is that right? What else should i say?
U = {[a b c] : a+2b+c=39} I have no idea how to prove this
10 points! Thanks!
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For a set to be a subspace, you need to show that
0 is in the set and that multiplications and addition are closed
1. Correct, 0 is not an element.
2. Yes, (x+y) ∈ U since A(x+y) = A(x)+A(y) = 3x+3y = 3(x+y)
A(cx) = cA(x) = c3x = 3(cx), and A0 = 3(0)
3. (a, b, c) = (39-2b-c, b, c) = b(-2, 1, 0) + c(-1, 0, 1) + (39, 0, 0)
and 0 is not an element of the set
0 is in the set and that multiplications and addition are closed
1. Correct, 0 is not an element.
2. Yes, (x+y) ∈ U since A(x+y) = A(x)+A(y) = 3x+3y = 3(x+y)
A(cx) = cA(x) = c3x = 3(cx), and A0 = 3(0)
3. (a, b, c) = (39-2b-c, b, c) = b(-2, 1, 0) + c(-1, 0, 1) + (39, 0, 0)
and 0 is not an element of the set