Calculus of subspaces

Theorem 1. The intersection of any collection of subspaces is a subspace.

Proof. If we use an index $ν$ to tell apart the members of the collection, so that the given subspaces are $ℳ_{ν}$ , let us write $ℳ = ⋂_{ν} ℳ_{ν} .$ Since every $ℳ_{ν}$ , contains $0$ , so does $ℳ$ , and therefore $ℳ$ is not empty. If $x$ and $y$ belong to $ℳ$ (that is, to all $ℳ_{ν}$ ), then $α x + β y$ belongs to all $ℳ_{ν}$ , and therefore $ℳ$ is a subspace. ◻

To see an application of this theorem, suppose that $𝒮$ is an arbitrary set of vectors (not necessarily a subspace) in a vector space $𝒱$ . There certainly exist subspaces $ℳ$ containing every element of $𝒮$ (that is, such that $𝒮 \subset ℳ$ ); the whole space $𝒱$ is, for example, such a subspace. Let $ℳ$ be the intersection of all the subspaces containing $𝒮$ ; it is clear that $ℳ$ itself is a subspace containing $𝒮$ . It is clear, moreover, that $ℳ$ is the smallest such subspace; if $𝒮$ is also contained in the subspace $𝒩$ , $𝒮 \subset 𝒩$ , then $ℳ \subset 𝒩$ . The subspace $ℳ$ so defined is called the subspace spanned by $𝒮$ or the span of $𝒮$ . The following result establishes the connection between the notion of spanning and the concepts studied in Sections 5–9.

Theorem 2. If $𝒮$ is any set of vectors in a vector space $𝒱$ and if $ℳ$ is the subspace spanned by $𝒮$ , then $ℳ$ is the same as the set of all linear combinations of elements of $𝒮$ .

Proof. It is clear that a linear combination of linear combinations of elements of $𝒮$ may again be written as a linear combination of elements of $𝒮$ . Hence the set of all linear combinations of elements of $𝒮$ is a subspace containing $𝒮$ ; it follows that this subspace must also contain $ℳ$ . Now turn the argument around: $ℳ$ contains $ℳ$ and is a subspace; hence $ℳ$ contains all linear combinations of elements of $𝒮$ . ◻

We see therefore that in our new terminology we may define a linear basis as a set of linearly independent vectors that spans the whole space.

Our next result is an easy consequence of Theorem 2; its proof may be safely left to the reader.

Theorem 3. If $ℋ$ and $𝒦$ are any two subspaces and if $ℳ$ is the subspace spanned by $ℋ$ and $𝒦$ together, then $ℳ$ is the same as the set of all vectors of the form $x + y$ , with $x$ in $ℋ$ and $y$ in $𝒦$ .

Prompted by this theorem, we shall use the notation $ℋ + 𝒦$ for the subspace $ℳ$ spanned by $ℋ$ and $𝒦$ . We shall say that a subspace $𝒦$ of a vector space $𝒱$ is a complement of a subspace $ℋ$ if $ℋ \cap 𝒦 = 𝒪$ and $ℋ + 𝒦 = 𝒱$ .

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