KindiakMath

Creating New Vector Spaces

Let K be any set and \mathbb K be any field. Recall that the function space \mathcal F(K, \mathbb K) forms a vector space over \mathbb K. Furthermore, given any vector space V over \mathbb K, the function space \mathcal F(K, V) forms a vector space over K. In this manner, we can create many vector spaces.

But even if we restrict our attention to just one vector space V over K, we can create many, many vector spaces.

For any \mathbf v \in V, define

\mathbb K\{\mathbf v\} := \{c\mathbf v : c \in \mathbb K\}.

It should seem intuitive that \mathbb K\{\mathbf v\} forms a vector space over V. In this case, we would call \mathbb K\{\mathbf v\} a subspace of V.

Definition 1. We say that W \subseteq V is a subspace of V if W forms a vector space over \mathbb K.

However, this definition requires us to verify all 8 or more conditions of a vector space—this would be an incredibly arduous task. Is there a short-cut to determine this result? Thankfully, the answer is yes.

Theorem 1. For any W \subseteq V, W is a subspace of V if and only if the following three conditions are satisfied:

  • \mathbf 0 \in W,
  • For any \mathbf u, \mathbf v \in W, \mathbf u + \mathbf v \in W,
  • For any \mathbf v \in W and c \in \mathbb K, c \mathbf v \in W.

Proof Sketch. Apart from these closure properties, all other properties are guaranteed by the definition of vector spaces.

Example 1. For any \mathbf v \in V, \mathbb K\{ \mathbf v \} is a subspace of V.

Proof. We verify the three properties of a vector space, and suppose \mathbf v \neq \mathbf 0 for nontriviality.

  • Firstly, \mathbf 0 = 0 \mathbf v \in \mathbb K\{ \mathbf v \}.
  • Secondly, for any c, d \in \mathbb K, c \mathbf v + d \mathbf v = (c+d)\mathbf v \in \mathbb K\{ \mathbf v\}.
  • Finally for c, d \in \mathbb K, c(d \mathbf v) = (cd) \mathbf v \in \mathbb K\{\mathbf v \}.

Therefore, \mathbb K\{ \mathbf v \} is a subspace of V. In fact, if \mathbf v \neq \mathbf 0, then \mathbb K\{ \mathbf v \} is isomorphic to \mathbb K as vector spaces (more on isomorphisms in a future post). We call \mathbb K\{ \mathbf v \} a 1-dimensional subspace of V.

Theorem 2. Let W_1 , W_2 \subseteq V be subspaces. Then W_1 \cap W_2 is subspace of V (and similarly, W).

Proof. We verify the three identities.

  • Since W_1, W_2 are subspaces, \mathbf 0 \in W_1 and \mathbf 0 \in W_2, so that \mathbf 0 \in W_1 \cap W_2.
  • Fix \mathbf u, \mathbf v \in W_1 \cap W_2 \subseteq W_1. Since, \mathbf u \in W_1 and \mathbf v \in W_1, we have \mathbf u + \mathbf v \in W_1. Symmetrically, since W_1 \cap W_2 = W_2 \cap W_1 \subseteq W_2, \mathbf u + \mathbf v \in W_2. Therefore, \mathbf u + \mathbf v \in W_1 \cap W_2.
  • Fix \mathbf v \in W_1 \cap W_2 \subseteq W_1 and c \in \mathbb K. Since, \mathbf v \in W_1, we have c \mathbf v \in W_1. Symmetrically, since W_1 \cap W_2 = W_2 \cap W_1 \subseteq W_2, \mathbf u + \mathbf v \in W_2. Therefore, \mathbf u + \mathbf v \in W_1 \cap W_2.

In general, W_1 \cup W_2 does not form a vector space. For a concrete example, recall that \mathbb R^2 = \mathcal F(\{1, 2\}, \mathbb R). For each i = 1, 2, define \mathbf e_i := \mathbb I_{ \{i\} }. Then clearly, \mathbf e_i = 1 \cdot \mathbf e_i \in \mathbb R\{\mathbf e_i \} , but \mathbf e_1 + \mathbf e_2 \notin \mathbb R\{\mathbf e_1 \} \cup \mathbb R\{\mathbf e_2 \}.

The correct generalisation would be direct sum. In fact, we can characterise subspaces in terms of sums of subsets.

Theorem 3. For subsets U, W \subseteq V, and K \subseteq \mathbb K, define the subsets

U + W := \{\mathbf u + \mathbf w : \mathbf u \in U, \mathbf w \in W\},\quad K(W) := \{c \mathbf w : c \in K, \mathbf w \in W\}.

Then W is a subspace of V if and only if W + W \subseteq W and \mathbb K (W) \subseteq W. Furthermore, if U and V are subspaces of V, then U + W is a subspace of V.

Proof. It is not hard to see that if U,W are subspaces of V, then

(U+W) + (U+W) \subseteq (U+U) + (W+W) \subseteq U + W.

Furthermore, \mathbb K(U + W) \subseteq \mathbb K (U) + \mathbb K (W) \subseteq U + W.

Let V be a vector space and W_1, W_2 \subseteq V be subspaces. Then any subspace U containing W_1 \cup W_2 must contain W_1 + W_2. In that sense, W_1 + W_2 is the smallest subspace that contains W_1 \cup W_2.

In particular, given nonzero vectors \mathbf v_1,\mathbf v_2 \in V, \mathbb K\{\mathbf v_1\} + \mathbb K\{\mathbf v_2\} will be the smallest vector space that contains \mathbb K\{\mathbf v_1\} \cup \mathbb K\{\mathbf v_2\}. In fact, more is true.

Lemma 1. For any subspace U \subseteq V such that \{\mathbf v_1, \mathbf v_2\} \subseteq U,

\mathbb K\{\mathbf v_1\} + \mathbb K\{\mathbf v_2\} \subseteq U + U \subseteq U.

Hence \mathbb K\{\mathbf v_1\} + \mathbb K\{\mathbf v_2\} is the smallest vector space that contains \{\mathbf v_1,\mathbf v_2\}.

More generally, if W \subseteq V is just a set, then V is a subspace of V that contains W. The smallest subspace that contains W will be the intersections of all subspaces that contain W. This is called the span of W.

Theorem 4. Let W \subseteq V. Let \mathcal S(W) denote the collection of subspaces of V that contain W. The span of W is then defined by

\displaystyle \mathrm{span}(W) := \bigcap_{U \in \mathcal S(W)} U.

Then \mathrm{span}(W) is the smallest subspace of V that contains W.

Proof. Exercise.

If W is a finite set, then we can write \mathrm{span}(W) as a combination of one-dimensional subspaces.

Theorem 5. For vectors \mathbf v_1,\dots,\mathbf v_k \in V, then

\displaystyle \mathrm{span}(\{\mathbf v_1,\dots,\mathbf v_k\}) = \mathbb K\{\mathbf v_1\} + \cdots + \mathbb K\{\mathbf v_k\}.

Proof. For the case k =2, the equality

\mathrm{span}(\{\mathbf v_1, \mathbf v_2\}) = \mathbb K\{\mathbf v_1\} + \mathbb K\{\mathbf v_2\}

holds since both sides of the proposed equality are the smallest subspace containing \{\mathbf v_1, \mathbf v_2\}. Apply induction on Lemma 1 to obtain the desired result.

In fact, we can even take the span of the empty set.

Example 2. \mathrm{span}(\emptyset) = \{\mathbf 0\}.

Proof. We leave it as an exercise to verify that \{\mathbf 0\} is a subspace that contains \emptyset. On the other hand, any subspace that contains \emptyset just refers to any subspace, which must contain \{\mathbf 0\}. Hence, \mathrm{span}(\emptyset) = \{\mathbf 0\}.

Finally, we obtain the usual definition for spans of sets in a vector space.

Corollary 1. For vectors \mathbf v_1,\dots,\mathbf v_k \in V,

\mathrm{span}(\{\mathbf v_1,\dots,\mathbf v_k\}) = \{c_1 \mathbf v_1 + \cdots + c_k \mathbf v_k : c_1,\dots,c_k \in \mathbb K\}.

Intuitively, any vector \mathbf v belongs to \mathrm{span}(\{\mathbf v_1,\dots,\mathbf v_k\}) if we can “cook” \mathbf v up by combining some recipe (c_1,\dots,c_k) with the ingredients:

\mathbf v = c_1 \mathbf v_1 + \cdots + c_k \mathbf v_k ,

where c_i denotes the amount of the ingredient \mathbf v_i used to cook up \mathbf v.

It might be tempting therefore to assume that subspaces of the form \mathbb K\{\mathbf v_1\} + \cdots + \mathbb K\{\mathbf v_k\} require all k ingredients \mathbf v_1, \dots, \mathbf v_k to generate. However, that is not always true. Observe that \mathbb K\{2\mathbf v\} \subseteq \mathbb K\{\mathbf v\} implies

\begin{aligned} \mathbb K\{\mathbf v\} &= \mathbb K\{\mathbf v\} + \{\mathbf 0\} \\ &\subseteq \mathbb K\{\mathbf v\} + \mathbb K\{2\mathbf v\} \\ &\subseteq \mathbb K\{\mathbf v\} + \mathbb K\{\mathbf v\} = \mathbb K\{\mathbf v\} \end{aligned}

so that \mathbb K\{\mathbf v\} + \mathbb K\{2\mathbf v\} = \mathbb K\{\mathbf v\} only requires one vector instead of two. How do know if we have hit the lowest possible number of ingredients? We need a fundamental tool called linear independence, which is our next topic of discussion.

—Joel Kindiak, 22 Feb 25, 2226H

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