Intersections & Spans
Lemma 1:
Let U and V be subspaces of an 𝔽-vector space W (where field of scalars 𝔽 is either ℚ or ℝ or ℂ). Their intersection
\[
U \cap V = \left\{ {\bf v}\,:\, {\bf v} \in U \quad\mbox{and}\quad {\bf v} \in V \right\}
\]
is a subspace of W.
The zero vector is in both U and V, so it is in their intersection U ∩ V. Thus, U∩V is nonempty.
If v_{1}, v_{2} ∈ U∩V and
k is an arbitrary scalar, then kv_{1} +
v_{2} belongs to both U and V by assumption.
Example 4:
Let ℙ_{3} be the set of all polynomials of degree 3 or less.
It is a vector space. Consider another vector space ℙ_{2} of
polynomials of degree up to 2. It is a vector space as well. Obviously,
ℙ_{2} is a subspace of ℙ_{3}.
We raise here the following question: “can a vector space V be written
as a finite union of proper subspaces”? We show in the following example that
it is impossible when scalars are either real numbers or complex numbers.
End of Example 4
Lemma 2:
The union of two subspaces is a subspace if and only if one of the subspaces is contained in the other.
If U∪V is a subspace, but neither U nor V is
contained in the other, then we can choose u ∈ U\V
and v &isin V\U. By assumption, w = u
+ v ∈ U∪V, so w belongs to either
U or V. In the former case, v = w - u
&isin V, contradiction. Latter case, similarly.
Example 1:
Let ℙ be the set of polynomials; it has two subspaces ℙ_{even} and ℙ_{odd}. However, x² ∈
ℙ_{even} and x ∈ ℙ_{odd}, but x + x² ∉ ℙ_{even}∪ℙ_{odd}.
Example 2:
In the real vector space ℝ² (considered as the xy-plane), the
x axis X = { [x, 0]^{T} : x ∈ ℝ } and y axis Y = { [0, y]^{T} : y ∈ ℝ } are subspaces, but their union is not a subspace of ℝ² because [1, 0]^{T} + [0, 1]^{T} = [1, 1]^{T} ∉ X ∪ Y.
The linear span
(also called just span) of a set of vectors in a vector space is the
intersection of all linear subspaces which each contain every vector in that
set. Alternatively, the span of a set S of vectors may be defined as
the set of all finite linear combinations of elements of S.
The linear span of a set of vectors is therefore a vector space.
Example 3:
Suppose that a vector space V is a union of two proper subspaces:
V = V_{1} ∪V_{2} as a union of two
proper subspaces. By hypothesis, one can find two non-zero vectors
v_{1} ∈ V_{1}\V_{2} and
v_{2} ∈
V_{2}\V_{1}. The relation
v_{1} + v_{2} ∈ V_{1} leads
to the contradiction: v_{2} = (v_{1} + v_{2}) - v_{1} ∈ V_{1} while supposing
v_{1} + v_{2} ∈ V_{2} leads
to the contradiction: v_{1} = (v_{1} + v_{2}) - v_{2} ∈ V_{2}. Therefore, a
vector space can never be written as a union of two proper subspaces.
■
Let U and V be subspaces of an 𝔽-vector space W. The sum of U and V is the subspace span(U∪V). It is denoted by U + V.
The above theorem ensures us that any pair of subspaces V and W of a finite dimensional vector space U has a finite dimensional sum
\[
V + W = \mbox{span} \left\{ V\cup W \right\} = \left\{ {\bf v} + {\bf w} \,
\big| \, {\bf v} \in V, \ {\bf w} \in W \right\} .
\]
Therefore, a sum of two subspaces is the set of all possible sums v +
w of all possible vectors from each subspace.
The sum of two subspaces is a subspace, and it is contained inside any subspace
that contains V ∪ W. One also can say that V +
W is the subspace generated by V and W.
This actually gives a clearer idea of its definition. n practice,
V + W contains any linear combination of elements drawn from
V and W. We can also think of V + W as the
intersection of all (typically infinitely many) subspaces containing both
V and W.
The intersection V ∩ W of two subspaces is always a subspace of their embedding space U. So any basis for V ∩ W can be extended to a basis for V; it can be extended to a basis for W.
Theorem 3:
Let V and W be a subspaces of a finite dimensional vector space
U. Then
\[
\mbox{dim}\left( V \cap W \right) + \mbox{dim} \left( V + W \right) =
\left( V \right) + \left( W \right) .
\]
Let k = dim(V∩W). Since V∩W is a
subspace of both V and W, the preceding theorem ensures that
k ≤ dim(V) and k ≤ dim(W). Let p =
dim(V) - k and q = dim(W) - k. Let
v_{1}, v_{2}, ... , v_{k} be a
basis for V∩W. Then there are vectors
u_{1}, u_{2}, ... , u_{p} and
w_{1}, w_{2}, ... , w_{q} such
that
\[
\left\{ {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_k , {\bf u}_1 , {\bf u}_2 ,
\ldots , {\bf u}_p \right\}
\]
is a basis for V and
\[
\left\{ {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_k , {\bf w}_1 , {\bf w}_2 ,
\ldots , {\bf w}_q \right\}
\]
is a basis for W. We must show that
\[
\mbox{dim} \left( V + W \right) = (p+k) + (q+k) -k = k+p+q .
\]
Since every vector in V+W is the sum of a vector in V
and a vector in W, the span of
\[
\left\{ {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_k , {\bf u}_1 , {\bf u}_2 ,
\ldots , {\bf u}_p , {\bf w}_1 , {\bf w}_2 ,
\ldots , {\bf w}_q \right\}
\]
is V+W. It suffices to show that the above list of vectors is
linearly independent. Suppose that
\[
\sum_{i=1}^k a_i {\bf v}_i + \sum_{i=1}^p b_i {\bf u}_i +
\sum_{i=1}^q c_i {\bf w}_i = {\bf 0} .
\]
Then
\[
\sum_{i=1}^k a_i {\bf v}_i + \sum_{i=1}^p b_i {\bf u}_i =
\sum_{i=1}^q \left( -c_i \right) {\bf w}_i .
\]
The right-hand side is in W and its left-hand side is in V, so
both sides are in V ∩ W. Thus, there are scalars
d_{1}, d_{2}, ... , d_{k} such
that
\[
\sum_{i=1}^k a_i {\bf v}_i + \sum_{i=1}^p b_i {\bf u}_i =
\sum_{i=1}^k d_i {\bf v}_i .
\]
Consequently,
\[
\sum_{i=1}^k \left( a_i - d_i \right) {\bf v}_i + \sum_{i=1}^p b_i {\bf u}_i =
{\bf 0} .
\]
The linear independence of vectors v_{i} and
u_{i} ensures that
\[
b_1 = b_2 = \cdots = b_p =0
\]
and it follows that
\[
\sum_{i=1}^k a_i {\bf v}_i + \sum_{i=1}^p b_i {\bf u}_i = {\bf 0} .
\]
The linear independence of another list v_{i} and
w_{i} ensures that
\[
c_1 = c_2 = \cdots = c_q =0
\]
Therefore, we conclude that the required list is linearly independent.
Theorem 4:
Let V and W be a subspaces of a finite dimensional vector space
U and let k be a positive integer.
- If dim(V) + dim(W) > dim(U), then V∩W contains a nonzero vector.
- If dim(V) + dim(W) ≥ dim(U) + k, then V∩W contains k linearly independent vectors.
The assertion (a) is the case k = 1 of the assertion (b). Under the
hypothesis in (b)
\[
\mbox{dim} \left( V \cap W \right) = \mbox{dim} (V) + \mbox{dim}(W) -
\mbox{dim} \left( V+W \right) \ge \mbox{dim} (V) + \mbox{dim}(W) -
\mbox{dim} (U) \ge k ,
\]
so V ∩ W has a basis comprising at least k vectors.
Example 5:
A sum of subspaces can be less than the entire space. Inside of
P_{3}, the set of all polynomails of one variable of degree up
to 3. Let V be the subspace of linear polynomials \(
V = \left\{ a + b\, x \,\big|\, a , b ∈ \mathbb{R} \right\} \) and let
W be the subspace of purely-cubic polynomials\(
W = \left\{ c\, x^3 \,\big|\, c ∈ \mathbb{R} \right\} . \)
Then V + W is not all of P_{3}. Instead, it is
the subspace
\[
V + W = \mbox{span} \left\{ V\cup W \right\} = \left\{ a + b\, x + c\, x^3
\,\big|\, a , b , c \in \mathbb{R} \right\} . \qquad\blacksquare
\]
Spans
Theorem 2:
If S = { u_{1}, u_{2}, … , u_{r} } is a nonempty set of vectors in a vector space V, then
- The span of S (that is, the set of all possible linear combinations of the vectors in S) is a subspace of V.
- The set U = span(S) is the "smallest" subspace of V that contains all of the vectors from S in the sense that any other subspace that contains those vectors contains U.
Let U be the set of all possible linear combinations of the vectors in S. We must show that U is closed under vector addition and scalar multiplication. To prove closure under addition, let
\[
{\bf u} = c_1 {\bf u}_1 + c_2 {\bf u}_2 + \cdots + c_r {\bf u}_r \qquad\mbox{and} \qquad {\bf v} = k_1 {\bf u}_1 + k_2 {\bf u}_2 + \cdots + k_r {\bf u}_r
\]
be two vectors in U. It follows that their sum can be written as
\[
{\bf u} + {\bf v} = \left( c_1 + k_1 \right) {\bf u}_1 + \left( c_2 + k_2 \right) {\bf u}_2 + \cdots + \left( c_r + k_r \right) {\bf u}_r ,
\]
which is a linear combination of the vectors in S. Thus, U is closed under vector addition. Similarly, it can be shown that sclar multiplication is also closed.
Proof of part (b): Let W be any subspace of V that contains all of the vectors in S. Since W is closed under vector addition and scalar multiplication, it contains all linear combinations of the vectors from S and hence contains U.
Example 5:
In two dimensional Euclidean space ℝ², consider an arbitrary
nonzero vector v. Let U be the set of all real scalar multiples
of v. Geometrically, U is a line in the direction of vector
v through the origin. Then U is a subspace of ℝ².
Indeed, for any two vectors x = αv and y = βv, we have
\[
{\bf x} + {\bf y} = \alpha\,{\bf v} + \beta\,{\bf v} =
\left( \alpha + \beta \right) {\bf v} \in U .
\]
Also, for arbitrary scalar k, we have
\[
k\,{\bf x} = k\,\alpha\,{\bf v} = \left( k\,\alpha \right) {\bf v} \in U .
\]
Example 5:
Let U be the set of all vectors of the form (𝑎, b − 𝑎, 3b, 𝑎 - 2b), where 𝑎 and b are arbitrary scalas. That is, let U be the set
\[
U = \left\{ (a, b -a, 3b, a-2b) \, : \, a, b \in \mathbb{F} \right\}
\]
To show that U is a subspace of 𝔽^{4}, we write each vector from U as a column-vector:
\[
\begin{pmatrix}
a \\ b - a \\ 3b \\ a-2b \end{pmatrix} = a \begin{pmatrix} \phantom{-}1 \\ - a \\ \phantom{-}0 \\ \phantom{-}1 \end{pmatrix} + b \begin{pmatrix} \phantom{-}0 \\ \phantom{-}1 \\ \phantom{-}3 \\ -2 \end{pmatrix} = a {\bf v} + b {\bf u} ,
\]
where
\[
{\bf v} = \begin{pmatrix} \phantom{-}1 \\ - a \\ \phantom{-}0 \\ \phantom{-}1 \end{pmatrix} , \qquad {\bf u} = \begin{pmatrix} \phantom{-}0 \\ \phantom{-}1 \\ \phantom{-}3 \\ -2 \end{pmatrix} \,\in \mathbb{F}^4 .
\]
This calculation shows that U is the span of these two vectors v and u shown above.