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Linear wrapper

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

If we have a set of vectors and we count all their linear combinations, then we get the linear envelope of this set of vectors.

Definition of linear envelope

Consider the vectors x1, …, xn. We already know that we can assemble a new vector from these vectors using a linear combination. We choose the real numbers a1, …, an and get the new vector y by multiplying and adding

$$ \mathbf{y}=a_1 \cdot \mathbf{x}_1 + \ldots + a_n \cdot \mathbf{x}_n $$

In this way, if at least one of the vectors xi is different from the null vector, we can make infinitely many "other" vectors from the set of vectors x1, …, xn - we just need to somehow choose different coefficients a1, …, an appropriately.

If we keep generating vectors over and over, we will eventually get all the vectors that can be obtained by linearly combining the vectors x1, …, xn. We then call such a set the linear envelope of the vectors x1, …, xn. We denote the linear envelope of the set of vectors X by pointed brackets: <X>. Formally, we could write the linear envelope as follows

$$ \left<\mathbf{x}_1, \ldots, \mathbf{x}_n\right> = \left\{a_1 \cdot \mathbf{x}_1+\ldots+a_n \cdot \mathbf{x}_n | a_1, \ldots, a_n \in \mathbb{R}\right\}, $$

if we have a finite number of vectors. If we have an infinite set of vectors X, we can take all finite subsets of Yi ⊆ X, i.e. |Yi| = r for some r ∈ ℕ, and the linear wrapper of the set X would then be the union of all sets <Yi>.

Example

Staying with the popular vector space ℝ3. We choose a one-point set X1 = {[1,2,1]} and ask what is the linear cover of this set? Are all linear combinations of the vector [1,2,1]. The resulting wrapper will be of the form

$$ \left<X_1\right> = \left\{\left[a,2a,a\right]|a\in \mathbb{R}\right\} $$

So they will be vectors of the form [3,6,3],[8,16,8],[−1,−2,−1] etc. We can try to verify that the sum of some two vectors gives us a new vector of the same shape:

$$\begin{eqnarray} \left[3{,}6,3\right]+\left[3{,}6,3\right]&=&\left[6{,}12,6\right]\\ \left[3{,}6,3\right]+\left[-1,-2,-1\right]&=&\left[2{,}4,2\right]\\ \left[8{,}16,8\right]+\left[2{,}4,2\right]&=&\left[10{,}20,10\right] \end{eqnarray}$$

A second example might be the classical vectors X2 = {[1,0,0], [0,1,0]}. If we count all the linear combinations, we find that they have the shape [a, b, 0], where a, b, ∈ ℝ. This can be shown easily because

$$\left[a,b,0\right] = a \cdot \left[1{,}0,0\right] + b \cdot \left[0{,}1,0\right].$$

Thus, the vectors of this wrapper are, for example, [0,8,0], [14,15,0], etc. Formally, we would write it as follows:

$$ \left<X_2\right> = \left\{\left[a,b,0\right] | a,b \in \mathbb{R}\right\} $$

A linear wrapper as the smallest subspace

Consider a vector space V and some set of vectors X ⊆ V. The linear wrapper <X> is then a vector subspace of the space V, i.e. the wrapper <X> is a vector space.

We just need to show that the cover <X> is closed with respect to addition and multiplication. It surely is, since any vector x ∈ <X> arose as a linear combination of vectors from X.

The linear cover of the set X is also the smallest subspace that contains all the vectors from the set X. Why? The smallest vector subspace that contains the vectors from X, must also contain all linear combinations of the vectors from X - otherwise it would not be a vector space. We can assume by argument that there is some smaller vector subspace, let us denote it Y, which contains all the vectors from X, i.e. X ⊆ Y. Since, by assumption, Y is smaller than the envelope of <X>, there must be a vector x, which is in <X>, but not in Y, i.e. $\mathbf{x}\in\left<X\right> \wedge \mathbf{x}\notin Y$.

But in doing so, the vector x must be constructible as a linear combination of vectors from X, so there must exist vectors x1, …, xn∈ X and coefficients a1, …, an such that

$$ \mathbf{x}=a_1 \cdot \mathbf{x}_1 + \ldots + a_n \cdot \mathbf{x}_n $$

But if Y does not contain this vector x, and yet contains the vectors x1, …, xn, then Y cannot be a vector space, and hence not a subspace. Thus the cover of <X> is the smallest subspace that contains all the vectors of X.

Basic properties of a linear wrapper

  • It is always true that W ⊆ <W>. This should be obvious. The linear cover of the set W will always be equal to or larger than the set W. Since the linear envelope of <W> contains all linear combinations of vectors from W, then <W> must also contain all vectors from W, because if we take a vector x ∈ W, we get the combination a · x = x for a = 1.

    Example: the linear cover of the set W = {[1,0,0]} is the set

    $$\begin{eqnarray} \left<W\right> &=& \left\{\left[a,0{,}0\right],|,a\in \mathbb{R}\right\}\\. \end{eqnarray}$$

    Obviously, {[1,0,0]} ⊆ {[a,0,0],|,a∈ ℝ}.

  • It is always true that <W> = <<W>>. Once we compute a linear wrapper, the linear wrapper of that wrapper is the same wrapper. <W> contains all the linear combinations from W. If we were to count all the linear combinations from <W> again , we would not get any new ones.

  • Consider some vector space V and have two subsets of this space (not necessarily subspaces) W1 and W2, i.e. W1, W2 ⊆ V. If W1 ⊆ W2 holds at the same time, then <W1> ⊆ <W2> also holds.

    In other words: if we have two sets of vectors, one of which is smaller, then the smaller set of vectors has a smaller or equal linear envelope than the larger set.

    Example: let W1 = {[1,0,0]} and W2 = {[1,0,0], [0, 1, 0]}. We see that W1 ⊆ W2 holds. What would the linear envelopes be?

    $$\begin{eqnarray} \left<W_1\right> &=& \left\{\left[a,0{,}0\right],|,a\in \mathbb{R}\right\}\\ \left<W_2\right> &=& \left\{\left[a,b,0\right],|,a,b\in \mathbb{R}\right\}\\ \end{eqnarray}$$

    The set of vectors W2 "generated a larger envelope" than the set W1, which is what we expected. The set W2 contains all the vectors from the set W1, so all the linear vector combinations from W1 will also be generated from the vectors in W2.

    Another example: W3 = {[1,2,0]} and W2 = {[1,2,0], [5,10,0]}. At first glance, we can see that the vectors in W2 are dependent. Therefore, both sets will generate the same wrapper. Thus, although W1 ⊂ W2, so does <W1> = <W2>.