The objective of this section is to show that every subspace of a vector space V is the kernel of a linear transformation defined on V.

Megan, page 10 about SLn

Quotient spaces

Let V be a vector space with subspace W. A W-coset is a set of the form \[ {\bf v} + W = \left\{ {\bf v} + {\bf w} \ : \ {\bf w} \in W \right\} , \] where v is any vector in V. This set is called an affine subspace a coset for v and W.
Using this definition, we define equivalent relation between elements of V.
Let V be a vector space and W a subspace. We can define an equivalence relation on V (with we’ll denote ∼ for now) by v ∼ u if v − u ∈ W.
   
Example 1: Let W    ■
End of Example 1
   
Example 1: Let G = GLn(𝔽) and let H = SLn(𝔽). For gi in G, we have g1 H = g2 H if and only if g1−1 g2 is in H, that is, if and only if det(g1−1 g2 ) = 1. We have \[ 1 = \det\left( g_1^{−1} g+2 \right) = \det\left( g_1^{−1} \right) \det _g2 = \left(\det g_1 \right)^{−1} \det g_2 . \] It follows that g1 H = g2 H if and only if det g1 = det g2 . From there, we see that ϕ : G/H → F∗ given by ϕ(g) = det g is well-defined and injective. That ϕ is a group homomorphism is immediate by the fact that the determinant mapping on G is a group homomorphism. We leave it as an easy exercise to show that ϕ is  surjective. This proves it is an isomorphism of groups, thus, that G/H ∼ = 𝔽*.

Quotient groups can seem mysterious but Theorem 10.22 tells a straightforward tale. The partition on GLn (F) induced by the left (or right) SLn (F)-cosets is according to determinant: Two matrices are in the same SLn (F)-coset if and only if they have the same determinant. The thing distinguishing one element from another in GLn(𝔽)/SLn(𝔽) is a nonzero element in the field.    ■

End of Example 1

Lemma 1: Let V be a vector space with subspace W. If v and x belong to V, then the following statements are logically equivalent:

  1. v is in x + W;
  2. v − x is in W;
  3. x − v is in W;
  4. x is in v + W;
  5. x + W = v + W.
   
Example 2: Suppose V, W, v, and x are as hypothesized.    ■
End of Example 2

Theorem 1:

   
Example 3: Let W    ■
End of Example 3

Theorem 2: If V is a vector space with subspace W, then V/W is a vector space with addition defined by \[ ({\bf v}_1 + W ) + ({\bf v}_2 + W ) := ({\bf v}_1 + {\bf v}_2 ) + W \] and scaling defined by \[ k \left( {\bf v} + W \right) := k\,{\bf v} + W. \]

   
Example 4: Let W    ■
End of Example 4
Let V be a vector space with subspace W. The quotient space of V by W is the set V/W with addition and scaling determined by adding and scaling coset representatives.
We read V/W as V mod W.

   
Example 5: Let ℤ = { 0, ±1, ±2, … } be the set of all integers We demote by (2ℤ) its subspace of even bintegers with respect to ℤ; it is also subspace over field (2ℤ) itself.

Let ℤ₂ = ℤ/(2ℤ) be a quotient space, which is more appropriate to denote by GF(2). It is the finite field with two elements (GF is the initialism of Galois field, another name for finite fields). GF(2) is the field with the smallest possible number of elements, and is unique if the additive identity and the multiplicative identity are denoted respectively 0 and 1, as usual. GF(2) can be identified with the field of the integers modulo 2, that is, the quotient ring of the ring of integers ℤ by the ideal 2ℤ of all even numbers: GF(2) = ℤ/(2ℤ).

The multiplication of GF(2) is again the usual multiplication modulo 2 (see the table below), and on boolean variables corresponds to the logical AND operation.

+ 0 1
0 0 1
1 1 0
If the elements of GF(2) are seen as boolean values, then the addition is the same as that of the logical XOR operation. Since each element equals its opposite, subtraction is thus the same operation as addition.

The multiplication of GF(2) is again the usual multiplication modulo 2 (see the table below), and on boolean variables corresponds to the logical AND operation.

× 0 1
0 0 0
1 0 1
   ■
End of Example 5
   
Example 6: Let W    ■
End of Example 6

Corollary 1: Let V be a vector space with subspace, W. If dim V = n and dim W = k, then dim V/W = n − k.

Theorem 3: Every subspace of a vector space V is the kernel of a linear transformation on V.

Let V be a vector space with subspace W. Define L : V ⇾ V/W by L(v) = v + W . Given v1 , v2 in V and c in the underlying field, we have \[ L(cv₁ + v₂) = (cv₁ + v₂) + W = (cv₁ + W) + (v₂ + W) = c L(v₁) + L(v₂). \] This shows that L is a linear transformation. Since L(v) = W = 0V/W if and only if v is in W, W = Ker L.
The mapping L : V ⇾ V/W described in the proof of Theorem 3 is called the canonical mapping V ⇾ V/W. When we say that a mapping is canonical, we mean that it is defined without reference to a coordinate system, that is, a basis. This idea comes up frequently in the sequel.

   

Example 7: Let W    ■
End of Example 7

Theorem 4: Let V be a vector space over a field 𝔽 and W be a subspace of V. Then \[ W^{\ast} \cong V^{\ast} / W^0 , \] where W⁰ is the annihilator of W.

Given that W is a subspace of V. Suppose that φ ∈ V✶ and φ|W, the restriction of φ on W. Then it is straightforward to see that φ|W ∈ W✶. Now define a map f : V✶ ⇾ W✶ such that f(φ) = φ|W. It is clear that for any 𝑎, b ∈ 𝔽 and ϕ, ψ ∈ V✶, \[ f \left( a\,\phi + b\,\psi \right) = \left. \left( a\,\phi + b\,\psi \right) \right\vert_{W} = a\,f(\phi ) + b\,f(\psi ) . \] This shows that f is a vector space homomorphism. Now if ϕ ∈ Ker(f), then the restriction of ϕ to W must be zero, i.e., ϕ(w) = 0 for all w ∈ W or ϕ ∈ W⁰. Conversely, if ϕ ∈ W⁰, i.e., ϕ(w) = 0 for all w ∈ W, then ϕ|W = 0 and ϕ ∈ Ker(f). Hence, Ker(f) = W⁰.

Now we show that f is onto. Then we show that any given χ ∈ W⁰ is the restriction of some φ ∈ W⁰. Let { w₁, w₂, … , wm } be a basis of W. Then it can be extended to a basis of V, say { w₁, w₂, … , wm, u₁, u₂, … , ur }, where m + r = dimV. Hence, we can write V = W ⊕ U, where U is a subspace of V spanned by { u₁, u₂, … , ur }. Now for any χ ∈ W✶ define ξ ∈ V✶ such that for any v ∈ V, v = w + v and ξ(v) = χ(w), where w ∈ W, u ∈ U. Let v₁ = v₂ and suppose that v₁ = w₁ + u₁ and v₂ = w₂ + u₂, where w₁, w₂ ∈ W and u₁, u₂ ∈ u. This implies that w₁ = w₂ and u₁ = u₂. As χ is a linear map, we get χ(w₁) = χ(w₂). This implies that ξ(v₁) = ξ(v₂) so ξ is well defined. It can be seen that ξ is a linear functional whose restriction on W is χ, i.e., f(ξ) = ξ|W = χ. Hence, f is onto and by fundamental theorem of vector space homomorphism W✶ ≌ V✶/Ker(f), namely, W✶ ≌ V✶/W⁰.

 

 


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