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Vector subspace

Kapitoly: Vector spaces, Examples of vector spaces, Vector subspace, Linear combinations of vectors, Linear wrapper, Bases of vector space, Dimensions of vector space, Transition matrix

A vector subspace is a subset of some vector space that is still closed to addition and multiplication by a scalar.

Definition

Consider some vector space V. Then the vector subspace W would be some subset of the space V, with the set W also being a vector space. Thus a subset W ⊆ V must satisfy the following two conditions for it to be a vector subspace of the space V: for all x, y ∈ W and for each a ∈ ℝ:

  • x+y ∈ W,
  • a · x ∈ W.

Thus, we must choose a subset of V such that these vectors are closed under addition and multiplication.

The subspace of the space R3

Let's try to take the space ℝ3 and find some subspace.

  1. For example, what about all triples of the form [a, a, a], where a∈ ℝ. That is, triples like $\left[1, 1, 1\right], \left[\frac12, \frac12, \frac12\right]$ or [−π, −π, −π]. Such a subspace, let's denote it W1, would be a subset of the space ℝ3, i.e. W1⊆ ℝ3, because ℝ3 simply contains all triples.

Such a space would be closed with respect to the vector addition operation because

$$\left[a,a,a\right]+\left[b,b,b\right]=\left[a+b, a+b,a+b\right].$$

After adding the two vectors, we would obtain a new vector which would again have the same three members, and such a vector is contained in the set W1. The set W1 thus satisfies the first condition of a vector subspace. But does it satisfy the second condition?

Yes, it does, because the k-multiple again changes all three component vectors, but it changes them exactly the same:

$$k\cdot\left[a,a,a\right]=\left[k\cdot a,k\cdot a,k\cdot a\right].$$

Again, we get a vector that has all three components the same, and such a vector is in the set W1. The set W1 is thus a vector subspace of ℝ3.

  • We can try to take all triples of [a, b, c] such that a, b, c ∈ <−1, 1>. We denote this set by W2 and the elements are, for example, triples of $\left[\frac12, -\frac27, \frac89\right]$ or $\left[0, \frac{\pi}{4}, 1\right]$. Are the elements of W2 closed to addition? Surely they are not, because $\left[1, 1, 1\right]+\left[\frac12, \frac13, \frac14\right]$ is equal to the triple $\left[\frac32, \frac43, \frac54\right]$, which does not belong to W2. Thus the set W2 does not form a subspace of the space V, and the set W2 does not form a space at all.

  • Now take all triples of the form [a, 0, b], where a, b∈ ℝ. These are all triples that have the second component zero. Let us denote this set by W3. Is this set closed to addition? Yes, it is, because

$$\left[a, 0, b\right]+\left[c, 0, d\right]=\left[a+c, 0, b+d\right]$$

The result is the triple [a + c, 0, b + d], which has the second component zero and the other two have real numbers on them. Surely this triple is in W3. Is W3 closed to multiplication? Yes, it is, because

$$k\cdot\left[a, 0, b\right]=\left[k\cdot a, 0, k\cdot b\right]$$

Again we get the triple [k · a, 0, k · b], which always has the second component zero and the other components are real numbers. Such a triple is an element of W3. The set W3 thus forms a subspace of the space V. Note that if the triples were of the form [a, b, 0] or [a, 0, 0], they would also form a vector subspace.

Subspace of polynomials

In the previous article, we showed a vector space composed of polynomials. To recap: a polynomial is an expression of the form

$$p(x)=a_0 + a_1x + a_2x^2 + \ldots + a_n x^n.$$

We assume that an≠0. The degree of such a polynomial is then just the number n. The real numbers a0, …, an are called the coefficients of the polynomial. An example of a polynomial is the expression 4 + 3x − 7x2. Let's try to find some vector subspace. So let's have a vector space P(x) of all polynomials.

  1. First try the set of all polynomials of degree n, where n≥0 denote this set by W1. We have already stated in previous articles that this is not a vector space at all, so it is hardly a vector subspace of the space P(x). Let us repeat so that if, for example, we multiply the polynomial in W1 by zero, we get the polynomial p(x) = 0, which by definition has degree −1. Thus this element is certainly not in any W1.

  2. Let's try to take all polynomials of degree n and smaller as W2. So for n = 2 all polynomials of degree 2, 1, 0, −1 will be in W2. Is this set closed to addition? Yes it is, because the sum of two polynomials of degree n will never be greater than n, it can only be less. Similarly, this set is closed to multiplication, because any k-fold polynomial of degree p(x) can result in either a polynomial of the same degree or a zero polynomial. Both possibilities are covered by the set W2, so W2 forms a subspace of the space P(X).

Basic properties of subspaces

Consider the vector space V and its two subspaces W1 and W2. Then

  1. The intersection W1 ∩ W2 is the vector subspace of V.

    Let us have a vector space ℝ3 and subspaces W1 and W2 such that W1 consists of triples of the form [a, b, 0] and W2 consists of triples of the form [a, 0, b], where a, b ∈ ℝ. Thus the intersection W1 ∩ W2 will be the set containing triples of the form [a, 0, 0] - such a set is obviously a subspace of the space ℝ3.

    But of course this is not a proof. It would look like this: take any vector space V and its two subspaces W1 and W2. We now want to prove that for all x, y ∈ W1∩ W2 it holds that x+y∈ W1∩ W2.

    Since the vectors x and y are contained in the intersection of W1∩ W2, then surely it must hold that x, y∈ W1 and also x, y ∈ W2 (this is true by definition of the intersection of sets). Since W1 and W2 are also vector spaces, it must also hold that x+y∈ W1 and at the same time x+y∈ W2 (this holds by definition of vector spaces). But now we have that x+y∈ W1 and at the same time x+y∈ W2, which implies that x+y∈ W1∩ W2 (again by the definition of intersection of sets - if an element lies in the set W1 and also lies in the set W2, it must lie in their intersection).

    Next, we need to check the closure of the multiplication. We need to prove that for all k∈ ℝ and x∈ W1∩ W2, k · x∈ W1∩ W2 holds. The procedure will be the same. Since x∈ W1∩ W2, so are x∈ W1 and x∈ W2. Since W1 and W2 are spaces, so are k · x∈ W1 and k · x ∈ W2 and therefore k · x ∈ W1 ∩ W2. $\Box$

  2. The union of W1 ∪ W2 is not necessarily a subspace of V.

    To prove the second property, we just need to find a suitable counterexample. We make do with the previous spaces W1 and W2, which have triples of the form [a, b, 0] and [a, 0, b], respectively. If we do their union, we get all triples that must have a zero in the second or third place. So the vectors [1,0,1] and [1, 1, 0] are certainly in the union W1 ∪ W2. But in doing so, their sum is equal to [2, 1, 1], which is a vector that has no zero anywhere. This vector is neither in W1, nor in W2, so it can't be in W1 ∪ W2. So we have found two vectors whose sum is not in the same set, so it can't be a vector space.

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