9. Let S = {[ x y]; in R2 : xy ≥ 0} . Determine whether...

9. Let S = {[ x y]; in R2 : xy ≥ 0} . Determine whether S is a subspace of R2.

(A) S is a subspace of R2.

(B) S is not a subspace of R2 because it does not contain the zero vector.

(D) S is not a subspace of R2 because it is not closed under scalar multiplication.

Solutions

Expert Solution

Colby Messinger answered 1 year ago

Next > < Previous

Related Solutions

6.14 Let f = {(x, y) ∈ R2 : y = x5 + 4x3 + x...

6.14 Let f = {(x, y) ∈ R2 : y = x5 + 4x3 + x + 1}. Prove that (a) f is onto. (b) f is 1-1. Prove that g = {(x, y) ∈ R2 : x = y5 + 4y3 + y + 1} is a function. (You will need to use calculus to prove part (1).)

Pick an arbitrary consumption bundle (x, y) ∈ R2+ (meaning x ≥ 0, y ≥ 0)....

Pick an arbitrary consumption bundle (x, y) ∈ R2+ (meaning x ≥ 0, y ≥ 0). Draw or describe the better than set (upper contour set), the worse than set (lower contour set) and the indifference set in a graph for the following situations: • I like consuming x, and the more the better. I am completely indifferent to the amount of y that I consume. • As long as my consumption of x is less than x∗, the more...

Given ( x + 2 )y" + xy' + y = 0 ; x0 = -1

Q1, what is negations of(∃x ∈ Z)x ≤ 0 ⇒ (∀y ∈ Z). xy > 0...

Q1, what is negations of(∃x ∈ Z)x ≤ 0 ⇒ (∀y ∈ Z). xy > 0 Q2, Prove by induction that 14n + 12n − 5 n is divisible by 7 for all n > 0

x^2 y ′′ + xy′ + λy = 0 with y(1) = y(2) and y ′...

x^2 y ′′ + xy′ + λy = 0 with y(1) = y(2) and y ′ (1) = y ′ (2) Please find the eigenvalues and eigenfunctions and eigenfunction expansion of f(x) = 6.

Determine whether each statement is true or false, and prove or disprove, as appropriate. (a) (∀x∈R)(∃y∈R)[xy=1].(∀x∈R)(∃y∈R)[xy=1]....

Determine whether each statement is true or false, and prove or disprove, as appropriate. (a) (∀x∈R)(∃y∈R)[xy=1].(∀x∈R)(∃y∈R)[xy=1]. (b) (∃x∈R)(∀y∈R)[xy=1].(∃x∈R)(∀y∈R)[xy=1]. (c) (∃x∈R)(∀y∈R)[xy>0].(∃x∈R)(∀y∈R)[xy>0]. (d) (∀x∈R)(∃y∈R)[xy>0].(∀x∈R)(∃y∈R)[xy>0]. (e) (∀x∈R)(∃y∈R)(∀z∈R)[xy=xz].(∀x∈R)(∃y∈R)(∀z∈R)[xy=xz]. (f) (∃y∈R)(∀x∈R)(∃z∈R)[xy=xz].(∃y∈R)(∀x∈R)(∃z∈R)[xy=xz]. (g) (∀x∈Q)(∃y∈Z)[xy∈Z].(∀x∈Q)(∃y∈Z)[xy∈Z]. (h) (∃x∈Z+)(∀y∈Z+)[y≤x].(∃x∈Z+)(∀y∈Z+)[y≤x]. (i) (∀y∈Z+)(∃x∈Z+)[y≤x].(∀y∈Z+)(∃x∈Z+)[y≤x]. (j) (∀x,y∈Z)[x<y⇒(∃z∈Z)[x<z<y]].(∀x,y∈Z)[x<y⇒(∃z∈Z)[x<z<y]]. (k) (∀x,y∈Q)[x<y⇒(∃z∈Q)[x<z<y]].(∀x,y∈Q)[x<y⇒(∃z∈Q)[x<z<y]].

Question