Proving a subspace

Proving subset of vector space is closed under scalar multiplication. Let V V be the vector space of all continuous functions f f defined on [0, 1] [ 0, 1]. Let S S be a subset of these functions such that ∫1 0 f(x) =∫1 0 xf(x) ∫ 0 1 f ( x) = ∫ 0 1 x f ( x). To prove it is closed under scalar multiplication, I've done the following:

Definition. If V is a vector space over a field K and if W is a subset of V, then W is a linear subspace of V if under the operations of V, W is a vector space over K.Equivalently, a nonempty subset W is a linear subspace of V if, whenever w 1, w 2 are elements of W and α, β are elements of K, it follows that αw 1 + βw 2 is in W.. As a corollary, all vector …8. The number of axioms is subject to taste and debate (for me there is just one: A vector space is an abelian group on which a field acts). You should not want to distinguish by noting that there are different criteria. Actually, there is a reason why a subspace is called a subspace: It is also a vector space and it happens to be (as a set) a ...An invariant subspace of a linear mapping. from some vector space V to itself is a subspace W of V such that T ( W) is contained in W. An invariant subspace of T is also said to be T invariant. [1] If W is T -invariant, we can restrict T to W to arrive at a new linear mapping. provide a useful set of vector properties. Theorem 1.2. If u,v,w ∈ V (a vector space) such that u+w = v +w, then u = v. Corollary 1.1. The zero vector and the additive inverse vector (for each vector) are unique. Theorem 1.3. Let V be a vector space over the field F, u ∈ V, and k ∈ F. Then the following statement are true: (a) 0u = 0 (b ...1 Answer. If we are working with finite dimensional vector spaces (which I assume we are) then there are a few ways to do this. If X ⊆ V X ⊆ V is our vector subspace then we can simply determine what dim X dim X is. If 0 < dim X < dim V 0 < dim X < dim V then we know that X X is a proper subspace. The easiest way to check this is to find a ...book. The idea is that a \generic" line will intersect any subspace in at most one point (geometrically obvious in R3). However, it is a bit tricky to arrange a \generic" line in the present context. Suppose V = V 1 [:::[V n; we choose such a union with nas small as possible. There exists x2V 1 but not in any other V j;j>1, for otherwise we ...DEFINITION A subspace of a vector space is a set of vectors (including 0) that satisfies two requirements: If v and w are vectors in the subspace and c is any scalar, then (i) v Cw is in the subspace and (ii) cv is in the subspace.

If you have to do it otherwise, you can always just check the two conditions for being a subspace, viz closure under addition and scalar multiplication. Share. Cite. Follow answered Apr 22, 2013 at 6:47. Lord_Farin Lord_Farin. 17.6k 9 9 gold badges 49 49 silver badges 126 126 bronze badges1 Answer. To prove a subspace you need to show that the set is non-empty and that it is closed under addition and scalar multiplication, or shortly that aA1 + bA2 ∈ W a A 1 + b A 2 ∈ W for any A1,A2 ∈ W A 1, A 2 ∈ W. The set isn't empty since zero matrix is in the set. W2 = {f ∈ C0[a, b]: f(−x) = f(x) for all x} W 2 = { f ∈ C 0 [ a, b]: f ( − x) = f ( x) for all x }, the set of even continuous functions on [a, b] [ a, b] Okay, I know to show that W W is a subspace of V V: a. W W is non-empty. b. if x1,x2 ∈ W x 1, x 2 ∈ W then x1 +x2 ∈ W x 1 + x 2 ∈ W. c. for k ∈ R, kx1 ∈ W k ∈ R, k x 1 ...As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's …Basis of a Subspace. As we discussed in Section 2.6, a subspace is the same as a span, except we do not have a set of spanning vectors in mind. There are infinitely many choices of spanning sets for a nonzero subspace; to avoid redundancy, usually it is most convenient to choose a spanning set with the minimal number of vectors in it. This is ...Any subspace admits a basis by this theorem in Section 2.6. A nonzero subspace has infinitely many different bases, but they all contain the same number of vectors. We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in Section 3.5.

Ask Question. Asked 9 years, 1 month ago. Modified 8 years, 4 months ago. Viewed 4k times. 0. Let V= P5 P 5 (R) = all the polynomials with real coefficients of degree at most 5. Let U= {rx+rx^4|rϵR} (1) Prove that U is a subspace. (2) Find a subspace W such that V=U⊕W.This result can provide a quick way to conclude that a particular set is not a Euclidean space. If the set does not contain the zero vector, then it cannot be a subspace . For example, the set A in Example 1 above could not be a subspace of R 2 because it does not contain the vector 0 = (0, 0).An invariant subspace of a linear mapping. from some vector space V to itself is a subspace W of V such that T ( W) is contained in W. An invariant subspace of T is also said to be T invariant. [1] If W is T -invariant, we can restrict T to W to arrive at a new linear mapping.7. This is not a subspace. For example, the vector 1 1 is in the set, but the vector 1 1 1 = 1 1 is not. 8. 9. This is not a subspace. For example, the vector 1 1 is in the set, but the vector ˇ 1 1 = ˇ ˇ is not. 10. This is a subspace. It is all of R2. 11. This is a subspace spanned by the vectors 2 4 1 1 4 3 5and 2 4 1 1 1 3 5. 12. This is ...Proving that a Linear Transformation of a Subspace is a Subspace. linear-algebra linear-transformations. 3,673. To show that this is a subspace, we need to show that it is non-empty and closed under scalar multiplication and addition. We know it is non-empty because T(0m) =0n T ( 0 m) = 0 n, so 0n ∈ T(U) 0 n ∈ T ( U).

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If S is a subspace of a vector space V , then 0V ∈ S. Proof. A subspace S will be closed under scalar multiplication by elements of the underlying field F, in.Problem for proving a subspace. Hot Network Questions Has a wand ever been used as a physical weapon? Comparator doesn't compare inputs close to VCC Normal Force Components For Circular Motion Who should I ask for help with nasty financial problems? Does Python's semicolon statement ending feature have any unique use? ...Feb 5, 2016 · Proving Polynomial is a subspace of a vector space. W = {f(x) ∈ P(R): f(x) = 0 or f(x) has degree 5} W = { f ( x) ∈ P ( R): f ( x) = 0 or f ( x) has degree 5 }, V = P(R) V = P ( R) I'm really stuck on proving this question. I know that the first axioms stating that 0 0 must be an element of W W is held, however I'm not sure how to prove ... Proposition 2.4. Let X be a Banach space, and let Z ⊂ X be a linear subspace. The following are equivalent: (i) Z is a Banach space, ehen equipped with the norm from X; (ii) Z is closed in X, in the norm topology. Proof. This is a particular case of a general result from the theory of complete metric spaces. Example 2.3.N ( A) = { x ∈ R n ∣ A x = 0 m }. That is, the null space is the set of solutions to the homogeneous system Ax =0m A x = 0 m. Prove that the null space N(A) N ( A) is a subspace of the vector space Rn R n. (Note that the null space is also called the kernel of A A .) Add to solve later. Sponsored Links.

Since you are working in a subspace of $\mathbb{R}^2$, which you already know is a vector space, you get quite a few of these axioms for free. Namely, commutativity, associativity and distributivity. With the properties that you have shown to be true you can deduce the zero vector since $0 v=0$ and your subspace is closed under scalar ...provide a useful set of vector properties. Theorem 1.2. If u,v,w ∈ V (a vector space) such that u+w = v +w, then u = v. Corollary 1.1. The zero vector and the additive inverse vector (for each vector) are unique. ... subspace of V if and only if W is closed under addition and closed under scalar multiplication. Examples of Subspaces 1. A plane through the origin …Any subspace admits a basis by this theorem in Section 2.6. A nonzero subspace has infinitely many different bases, but they all contain the same number of vectors. We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in Section 3.5.Clearly, in both cases the solutions set is a linear subspace of $\mathbb R^n$ True (and obvious) if $0$ is the only solution. But there are plenty of infinite subsets of $\mathbb R^n$ that are not subspaces.We have proved that W = R(A) is a subset of Rm satisfying the three subspace requirements. Hence R(A) is a subspace of Rm. THE NULL SPACE OFA. The null space of Ais a subspace of Rn. We will denote this subspace by N(A). Here is the definition: N(A) = {X :AX= 0 m} THEOREM. If Ais an m×nmatrix, then N(A) is a subspace of Rn. Proof.In order to prove that the subset U is a subspace of the vector space V, I need to show three things. Show that 0 → ∈ U. Show that if x →, y → ∈ U, then x → + y → ∈ U. Show that if x → ∈ U and a ∈ R, then a x → ∈ U. (1) Since U is given to be non-empty, let x 0 → ∈ U. Since u → + c v → ∈ U, if u → = v → ...Since Y is a Banach space, it is convergent to some element in Y. Call that element Ax, i.e. lim n → ∞Anx = Ax Since x was arbitrary, Ax is defined for any x ∈ X. Thus, A is a map from X to Y defined by x → Ax. We need to show that A is linear, bounded, and Ann → ∞ → A in the operator norm.If X ⊆ V X ⊆ V is our vector subspace then we can simply determine what dim X dim X is. If 0 < dim X < dim V 0 < dim X < dim V then we know that X X is a proper subspace. The easiest way to check this is to find a basis for the subspace and check its length.Exercise. Give an example of a proper subspace of the vector space of polynomials in x x with real coefficients of degree at most 2 2 . Let P 2 P 2 denote the vector space of polynomials in x x with real coefficients of degree at most 2 2 . Consider W = { a x 2: a ∈ R } W = { a x 2: a ∈ R } . Let u = a x 2 u = a x 2 and v = a ′ x 2 v = a ...

We will prove the main theorem by using invariant subspaces and showing that if Wis T-invariant, then the characteristic polynomial of T Wdivides the characteristic polynomial of T. So, let us recall the de nition of a T-invariant space: De nition 2. Given a linear transformation T: V !V, a subspace WˆV is called T-invariant if for all x 2W, T ...

Problem for proving a subspace. Hot Network Questions Has a wand ever been used as a physical weapon? Comparator doesn't compare inputs close to VCC Normal Force Components For Circular Motion Who should I ask for help with nasty financial problems? Does Python's semicolon statement ending feature have any unique use? ...2. To check that W W is a vector subspace you need to check the 3 following conditions: i) W W is non empty (clear if V V is non empty), ii)if x ∈ W x ∈ W and y ∈ W y ∈ W, then x +y ∈ W x + y ∈ W. iii)If α ∈ K α ∈ K, and x ∈ W x ∈ W, then αx ∈ W α x ∈ W. For your second question, you need to check these three ...The origin of V V is contained in A A. aka a subspace is a subset with the inherited vector space structure. Now, we just have to check 1, 2 and 3 for the set F F of constant functions. Let f(x) = a f ( x) = a, g(x) = b g ( x) = b be constant functions. (f ⊕ g)(x) = f(x) + g(x) = a + b ( f ⊕ g) ( x) = f ( x) + g ( x) = a + b = a constant (f ...Mar 19, 2007 · The "steps" can be combined, since one can easily prove (you could try that, too) that the following two conditions for "being a subspace" are equivalent (if V is a vector space over a field F, and M a non-empty candidate for a subspace of V): (1) for every x, y in M, x + y is in M & for every x in M and A in F, Ax is in M (2) for every x, y in ... We would like to show you a description here but the site won't allow us.Since you are working in a subspace of $\mathbb{R}^2$, which you already know is a vector space, you get quite a few of these axioms for free. Namely, commutativity, associativity and distributivity. With the properties that you have shown to be true you can deduce the zero vector since $0 v=0$ and your subspace is closed under scalar ...Since \(\text{Span}\{v_1,v_2,\ldots,v_p\}\) satisfies the three defining properties of a subspace, it is a subspace. Now let \(V\) be a subspace of \(\mathbb{R}^n\). If \(V\) is the zero subspace, then it is the span of the empty set, so we may assume \(V\) is nonzero. Choose a nonzero vector \(v_1\) in \(V\).I'm learning about proving whether a subset of a vector space is a subspace. It is my understanding that to be a subspace this subset must: Have the $0$ vector. Be closed under addition (add two elements and you get another element in the subset).Jan 14, 2018 · 1 Answer. If we are working with finite dimensional vector spaces (which I assume we are) then there are a few ways to do this. If X ⊆ V X ⊆ V is our vector subspace then we can simply determine what dim X dim X is. If 0 < dim X < dim V 0 < dim X < dim V then we know that X X is a proper subspace. The easiest way to check this is to find a ...

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(4) Axler, Chapter 1 problem 8: Prove that the intersection of any collection of subspaces of V is itself a subspace of V . Proof: Note - in class I said it ...DEFINITION A subspace of a vector space is a set of vectors (including 0) that satisfies two requirements: If v and w are vectors in the subspace and c is any scalar, then (i) v Cw is in the subspace and (ii) cv is in the subspace.Learn to determine whether or not a subset is a subspace. Learn the most important examples of subspaces. Learn to write a given subspace as a column space or null space. Recipe: compute a spanning set for a null space. Picture: whether a subset of R 2 or R 3 is a subspace or not. Vocabulary words: subspace, column space, null space. We will prove the main theorem by using invariant subspaces and showing that if Wis T-invariant, then the characteristic polynomial of T Wdivides the characteristic polynomial of T. So, let us recall the de nition of a T-invariant space: De nition 2. Given a linear transformation T: V !V, a subspace WˆV is called T-invariant if for all x 2W, T ...Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products.Share. Watch on. A subspace (or linear subspace) of R^2 is a set of two-dimensional vectors within R^2, where the set meets three specific conditions: 1) The set includes the zero vector, 2) The set is closed under scalar multiplication, and 3) The set is closed under addition.Proposition 1.6. For any v2V, the linear orbit [v] of vis an invariant subspace of V. Moreover it is the minimal invariant subspace containing v: if WˆV is an invariant subspace and v2W, then [v] ˆW. Exercise 1.2. Prove Proposition 1.6. Exercise 1.3. Let SˆV be any subset. De ne the orbit of T on Sas the union of the orbits of T on sfor all s2S.To show that H is a subspace of a vector space, use Theorem 1. 2. To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. EXAMPLE: Is V a 2b,2a 3b : a and b are real a subspace of R2? Why or why not?Proving subset of vector space is closed under scalar multiplication. Let V V be the vector space of all continuous functions f f defined on [0, 1] [ 0, 1]. Let S S be a subset of these functions such that ∫1 0 f(x) =∫1 0 xf(x) ∫ 0 1 f ( x) = ∫ 0 1 x f ( x). To prove it is closed under scalar multiplication, I've done the following: ….

I'm learning about proving whether a subset of a vector space is a subspace. It is my understanding that to be a subspace this subset must: Have the $0$ vector. Be closed under addition (add two elements and you get another element in the subset). Predictions about the future lives of humanity are everywhere, from movies to news to novels. Some of them prove remarkably insightful, while others, less so. Luckily, historical records allow the people of the present to peer into the past...Subspace topology. In topology and related areas of mathematics, a subspace of a topological space X is a subset S of X which is equipped with a topology induced from that of X called the subspace topology (or the relative topology, or the induced topology, or the trace topology[citation needed] ).Problem Statement: Let T T be a linear operator on a vector space V V, and let λ λ be a scalar. The eigenspace V(λ) V ( λ) is the set of eigenvectors of T T with eigenvalue λ λ, together with 0 0. Prove that V(λ) V ( λ) is a T T -invariant subspace. So I need to show that T(V(λ)) ⊆V(λ) T ( V ( λ)) ⊆ V ( λ).A basis is a set of linearly independent vectors that span a vector space. In this video, we are given a set of vectors and prove that it 1) spans the vector...This is a subspace if the following are true-- and this is all a review-- that the 0 vector-- I'll just do it like that-- the 0 vector, is a member of s. So it contains the 0 vector. Then if v1 and v2 are both members of my subspace, then v1 plus v2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. Mar 25, 2021 · Prove that a subspace contains the span. Let vectors v, w ∈ Fn v, w ∈ F n. If U U is a subspace in Fn F n and contains v, w v, w, then U U contains Span{v, w}. Span { v, w }. My attempt: if U U contains vectors v, w v, w. Then v + w ∈ U v + w ∈ U and av ∈ U a v ∈ U, bw ∈ U b w ∈ U for some a, b ∈F a, b ∈ F. Oct 8, 2019 · In the end, every subspace can be recognized to be a nullspace of something (or the column space/span of something). Geometrically, subspaces of $\mathbb{R}^3$ can be organized by dimension: Dimension 0: The only 0-dimensional subspace is $\{(0,0,0)\}$ Dimension 1: The 1-dimensional subspaces are lines through the origin. Proving a subspace, To prove that a subspace W is non empty we usually prove that the zero vector exists in the subspace. But then is it necessary to prove the existence of zero vector. Can't we prove the existence of any vector instead? Can someone please explain with an example where we can prove that W is a subspace by taking the existence of any …, Solution 1. To show a subset is a subspace, you need to show three things: Show it is closed under addition. Show it is closed under scalar multiplication. Show that the vector 0 0 is in the subset. To show 1, as you said, let w1 = (a1,b1,c1) w 1 = ( a 1, b 1, c 1) and w2 = (a2,b2,c2) w 2 = ( a 2, b 2, c 2)., Orthogonal Complements. Definition of the Orthogonal Complement. Geometrically, we can understand that two lines can be perpendicular in R 2 and that a line and a plane can be perpendicular to each other in R 3.We now generalize this concept and ask given a vector subspace, what is the set of vectors that are orthogonal to all vectors in the subspace., Let (X, d) ( X, d) be a metric space and Y ⊂ X. Y ⊂ X. Let T T be the subspace topology on Y Y as a subspace of X X and let T′ T ′ be the topology on Y Y induced by the metric d. d. Let C C be the set of all open d d -balls of X. X. Let B = {Y ∩ c: c ∈ C}. B = { Y ∩ c: c ∈ C }. Now C C is a base for the topology on X X so B B is ..., And so now that we know that any basis for a vector space-- Let me just go back to our set A. A is equal to a1 a2, all the way to an. We can now say that any basis for some vector, for some subspace V, they all have the same number of elements. And so we can define a new term called the dimension of V., if the image of T is an n-dimensional subspace of the (n-dimensional) vector space W. But the only full-dimensional subspace of a nite-dimensional vector space is itself, so this happens if and only if the image is all of W, namely, if T is surjective. In particular, we will say that a linear transformation between vector spaces V and, Strictly speaking, A Subspace is a Vector Space included in another larger Vector Space. Therefore, all properties of a Vector Space, such as being closed under addition and …, 1 Answer. To prove a subspace you need to show that the set is non-empty and that it is closed under addition and scalar multiplication, or shortly that aA1 + bA2 ∈ W a A 1 + b A 2 ∈ W for any A1,A2 ∈ W A 1, A 2 ∈ W. The set isn't empty since zero matrix is in the set. , a subspace Uof V such that U\nullT= f0gand rangeT= fTuju2Ug. Proof. Proposition 2.34 says that if V is nite dimensional and Wis a subspace of V then we can nd a subspace Uof V for which V = W U. Proposition 3.14 says that nullT is a subspace of V. Setting W= nullT, we can apply Prop 2.34 to get a subspace Uof V for which V = nullT U, We will prove the main theorem by using invariant subspaces and showing that if Wis T-invariant, then the characteristic polynomial of T Wdivides the characteristic polynomial of T. So, let us recall the de nition of a T-invariant space: De nition 2. Given a linear transformation T: V !V, a subspace WˆV is called T-invariant if for all x 2W, T ... , , Proving polynomial to be subspace Ask Question Asked 9 years, 1 month ago Modified 8 years, 4 months ago Viewed 4k times 0 Let V= P5 P 5 (R) = all the …, Proving polynomial to be subspace. Let V= P5 P 5 (R) = all the polynomials with real coefficients of degree at most 5. Let U= {rx+rx^4|rϵR} (1) Prove that U is a subspace. (2) Find a subspace W such that V=U⊕W. For the first proof, I know that I have to show how this polynomial satisfies the 3 conditions in order to be a subspace but I …, Using a counterexample, we demonstrate that a set is not a vector subspace. This is Chapter 6 Problem 10 from the MATH1231/1241 Algebra notes. Presented by D..., Sep 26 at 22:25. Add a comment. 41. Compact sets need not be closed in a general topological space. For example, consider the set with the topology (this is known as the Sierpinski Two-Point Space ). The set is compact since it is finite. It is not closed, however, since it is not the complement of an open set., Proving subset of vector space is closed under scalar multiplication. Let V V be the vector space of all continuous functions f f defined on [0, 1] [ 0, 1]. Let S S be a subset of these functions such that ∫1 0 f(x) =∫1 0 xf(x) ∫ 0 1 f ( x) = ∫ 0 1 x f ( x). To prove it is closed under scalar multiplication, I've done the following:, The union of two subspaces is a subspace if and only if one of the subspaces is contained in the other. The "if" part should be clear: if one of the subspaces is contained in the other, then their union is just the one doing the containing, so it's a subspace. Now suppose neither subspace is contained in the other subspace. , Let S be a subspace of the inner product space V. The the orthogonal complement of S is the set S⊥ = {v ∈ V | hv,si = 0 for all s ∈ S}. Theorem 3.0.3. (1) If U and V are subspaces of a vector space W with U ∩V = {0}, then U ⊕V is also a subspace of W. (2) If S is a subspace of the inner product space V, then S⊥ is also a subspace of V., The subspaces of \(\mathbb{R}^3\) are {0}, all lines through the origin, all planes through the origin, and \(\mathbb{R}^3\). In fact, these exhaust all subspaces of \(\mathbb{R}^2\) and \(\mathbb{R}^3\) , respectively. To prove this, we will need further tools such as the notion of bases and dimensions to be discussed soon., An invariant subspace of a linear mapping. from some vector space V to itself is a subspace W of V such that T ( W) is contained in W. An invariant subspace of T is also said to be T invariant. [1] If W is T -invariant, we can restrict T to W to arrive at a new linear mapping., Homework Statement Let U and W be subspaces of a vector space V Show that the set U + W = {v ∈ V : v = u + w, where u ∈ U and w ∈ W} is a subspace of V Homework Equations The Attempt at a Solution I understand from this that u and w are both vectors in a vector space V and that u+w..., 1 Answer. To show that this is a subspace, we need to show that it is non-empty and closed under scalar multiplication and addition. We know it is non-empty because T(0m) =0n T ( 0 m) = 0 n, so 0n ∈ T(U) 0 n ∈ T ( U). Now, suppose c ∈ R c …, When you want a salad or just a little green in your sandwich, opt for spinach over traditional lettuce. These vibrant, green leaves pack even more health benefits than many other types of greens, making them a worthy addition to any diet. ..., If you’re a taxpayer in India, you need to have a Personal Account Number (PAN) card. It’s crucial for proving your identify and proving that you paid your taxes that year. Here are the steps you can take to apply online., Clearly, in both cases the solutions set is a linear subspace of $\mathbb R^n$ True (and obvious) if $0$ is the only solution. But there are plenty of infinite subsets of $\mathbb R^n$ that are not subspaces., Definition 2. A subset U ⊂ V of a vector space V over F is a subspace of V if U itself is a vector space over F. To check that a subset U ⊂ V is a subspace, it suffices to check only a couple of the conditions of a vector space. Lemma 6. Let U ⊂ V be a subset of a vector space V over F. Then U is a subspace of V if and only if, Just to be pedantic, you are trying to show that S S is a linear subspace (a.k.a. vector subspace) of R3 R 3. The context is important here because, for example, any subset of R3 R 3 is a topological subspace. There are two conditions to be satisfied in order to be a vector subspace: (1) ( 1) we need v + w ∈ S v + w ∈ S for all v, w ∈ S v ..., Proving isomorphism between between a subspace and a quotient space. Ask Question Asked 9 years, 2 months ago. Modified 6 years, 2 months ago. Viewed 5k times 2 $\begingroup$ I've been thinking about ..., Currently I'm reading linear algebra books by Leon and Friedberg. In Friedberg's book, to be a subspace, a subset of a vector space should (1). contain zero vector, (2). be closed under scalar multiplication and (3). be closed under vector addition. But condition (1) …, 2. To check that W W is a vector subspace you need to check the 3 following conditions: i) W W is non empty (clear if V V is non empty), ii)if x ∈ W x ∈ W and y ∈ W y ∈ W, then x +y ∈ W x + y ∈ W. iii)If α ∈ K α ∈ K, and x ∈ W x ∈ W, then αx ∈ W α x ∈ W. For your second question, you need to check these three ..., Except for the typo I pointed out in my comment, your proof that the kernel is a subspace is perfectly fine. Note that it is not necessary to separately show that $0$ is contained in the set, since this is a consequence of closure under scalar multiplication., In each case, either prove that S S forms a subspace of R3 R 3 or give a counter example to show that it does not. Case: z = 2x, y = 0 z = 2 x, y = 0. Okay, there are 3 conditions that need to be satisfied for this to work. Zero vector has to be a possibility: Okay, we can find out that this is true. [0, 0, 0] [ 0, 0, 0] E S., technically referring to the subset as a topological space with its subspace topology. However in such situations we will talk about covering the subset with open sets from the larger space, so as not to have to intersect everything with the subspace at every stage of a proof. The following is a related de nition of a similar form. De nition 2.4.