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Let \(T: V \rightarrow V^{\prime}\) be linear with kernel \(W\) and image \(U\). Show that the quotient space \(V / W\) is isomorphic to \(U\) under the mapping \(\theta: V / W \rightarrow U\) defined by \(\theta(v+W)=T(v)\). Furthermore, show that \(T=i \circ \theta \circ \eta\), where \(\eta: V \rightarrow V / W\) is the natural mapping of \(V\) into \(V / W\) (i.e., \(\eta(v)=v+W\), and \(i: U \hookrightarrow V^{\prime}\) is the inclusion mapping (i.e., \(i(u)=u\) ). (See diagram.)

Short Answer

Expert verified
In conclusion, to show that the quotient space \(V/W\) is isomorphic to \(U\) under the mapping \(\theta\), we completed the following steps: 1. Showed that \(\theta\) is well-defined. 2. Proved that \(\theta\) is a linear transformation. 3. Demonstrated that \(\theta\) is one-to-one. 4. Established that \(\theta\) is onto. Lastly, we proved that \(T = i \circ \theta \circ \eta\), confirming that the quotient space \(V/W\) is indeed isomorphic to \(U\) under the mapping \(\theta\).

Step by step solution

01

Show that theta is well-defined

If two elements of v+W are equal, then their images under theta should also be equal. So, let v1, v2 ∈ V such that (v1+W) = (v2+W). Then, v1 - v2 ∈ W. Since W is the kernel of T, we have T(v1 - v2) = 0. Thus, T(v1) - T(v2) = 0, and T(v1) = T(v2). Therefore, theta is well-defined.
02

Show that theta is a linear transformation

To prove this, we need to show that theta respects scalar multiplication and addition. Let x, y ∈ V and k ∈ the field F. Then: 1. θ((x+y)+W) = T(x+y) = T(x) + T(y) = θ(x+W) + θ(y+W). 2. θ(kx+W) = T(kx) = kT(x) = kθ(x+W). So, theta is a linear transformation.
03

Prove that theta is one-to-one

Suppose θ(x+W) = θ(y+W) for some x, y ∈ V. Then, T(x) = T(y) ⟹ T(x-y) = 0. Since W is the kernel of T, x-y ∈ W. This means x+ W = y+ W, or the two cosets are the same. Thus, theta is one-to-one.
04

Prove that theta is onto

Let u ∈ U. Since u is in the image of T, there exists v ∈ V such that T(v) = u. Hence, θ(v+W) = T(v) = u. Since u is arbitrary, theta is onto.
05

Prove that T = i * theta * eta

We need to show that for any v ∈ V, T(v) = i(θ(η(v))). By definition, η(v) = v + W. So, θ(η(v)) = θ(v+W) = T(v). Now, i(u) = u for all u ∈ U, so i(θ(η(v))) = T(v). Therefore, T = i * theta * eta. In conclusion, we've shown that the quotient space V/W is isomorphic to U under the mapping theta and that T = i * theta * eta.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Spaces
Quotient spaces in linear algebra are fascinating because they give us a way to "simplify" vector spaces by factoring out a subspace. Imagine you have a vector space \( V \) and a subspace \( W \). The quotient space \( V / W \) is formed by taking "cosets" of \( W \) in \( V \). Each coset is essentially a collection of vectors that differ by an element of \( W \).
To better understand this, think of \( V \) as a collection of different paths you can take, and \( W \) as obstacles or detours on your path. The quotient space \( V / W \) represents different paths you can take while ignoring the obstacles.
This concept is crucial because the original exercise demonstrates how such a quotient space \( V / W \) is isomorphic to the image \( U \) of a linear transformation \( T \). An isomorphism, in this context, means there is a one-to-one correspondence between elements of \( V / W \) and \( U \), preserving the vector space structure.
Linear Maps
Linear maps (or linear transformations) are functions between vector spaces that respect vector space operations like addition and scalar multiplication. If you have a map \( T: V \to V' \), and \( V \) and \( V' \) are vector spaces, \( T \) is linear if for all vectors \( x, y \in V \) and scalars \( k \): - \( T(x + y) = T(x) + T(y) \)- \( T(kx) = kT(x) \)This means linear maps are kind of predictable since they maintain the structure of vector spaces across the transformation.
In the context of the exercise, the linear map \( T \) was crucial because it preserved vector space properties when mapping from \( V \) to \( V' \). The map \( \theta \), defined from the quotient space \( V / W \) to the image \( U \), is also linear, as evidenced by its respect for addition and scalar multiplication. The exercise further shows using \( \theta \), \( \eta \) (natural mapping from \( V \) to \( V/W \)), and \( i \) (inclusion map from \( U \) to \( V' \)), that linear maps compose predictably. Indeed, \( T \) can be expressed as \( i \circ \theta \circ \eta \).
Kernel and Image
The kernel and image of a linear transformation are critical concepts that help us understand the structure of a linear map. - **Kernel** \( \text{ker}(T) \) is the set of all vectors \( v \in V \) such that \( T(v) = 0 \). It's like finding which vectors become "invisible" (mapped to zero) when transformed.
- **Image** \( \text{im}(T) \) is the set of all vectors that \( T \) maps to in \( V' \). It includes all the vectors you can "reach" from \( V \) using \( T \).In the exercise given, \( W \) is the kernel of the map \( T \), and the image \( U \) is where \( T \) sends all the vectors from \( V \). The isomorphism between the quotient space \( V / W \) and the image \( U \) shows a tight relationship between these concepts. When the kernel is used to define a quotient space, it helps simplify the work by "factoring out" redundant elements, leaving behind a space that behaves just like the image. Hence, understanding kernel and image is key to navigating through linear transformations efficiently.

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Most popular questions from this chapter

Let \(W=Z(v, T),\) and suppose the \(T\) -annihilator of \(v\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial of degree \(d\). Show that \(f(T)^{3}(W)\) is a cyclic subspace generated by \(f(T)^{2}(v)\) and that it has dimension \(d(n-s)\) if \(n>s\) and dimension 0 if \(n \leq s\).

Suppose \(E: V \rightarrow V\) is a projection. Prove (i) \(I-E\) is a projection and \(V=\operatorname{Im} E \oplus \operatorname{Im}(I-E)\), (ii) \(I+E\) is invertible (if \(1+1 \neq 0\) ).

Let \(W\) be invariant under \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\). Prove \(W\) is also invariant under \(T_{1}+T_{2}\) and \(T_{1} T_{2}\).

Let \(A=\left[\begin{array}{lllll}0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{lllll}0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\). The reader can verify that \(A\) and \(B\) are both nilpotent of index 3 ; that is, \(A^{3}=0\) but \(A^{2} \neq 0\), and \(B^{3}=0\) but \(B^{2} \neq 0\). Find the nilpotent matrices \(M_{A}\) and \(M_{B}\) in canonical form that are similar to \(A\) and \(B\), respectively. Because \(A\) and \(B\) are nilpotent of index \(3, M_{A}\) and \(M_{B}\) must each contain a Jordan nilpotent block of order 3 , and none greater then 3 . Note that \(\operatorname{rank}(A)=2\) and \(\operatorname{rank}(B)=3\), so nullity \((A)=5-2=3\) and nullity \((B)=5-3=2\). Thus, \(M_{A}\) must contain three diagonal blocks, which must be one of order 3 and two of order 1 ; and \(M_{B}\) must contain two diagonal blocks, which must be one of order 3 and one of order 2 . Namely, $$ M_{A}=\left[\begin{array}{lllll} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \quad \text { and } \quad M_{B}=\left[\begin{array}{lllll} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$

Suppose \(E: V \rightarrow V\) is linear and \(E^{2}=E .\) Show that (a) \(E(u)=u\) for any \(u \in \operatorname{Im} E\) (i.e., the restriction of \(E\) to its image is the identity mapping); (b) \(V\) is the direct sum of the image and kernel of \(E: V=\operatorname{Im} E \oplus \operatorname{Ker} E ;\) (c) \(E\) is the projection of \(V\) into \(\operatorname{Im} E\), its image. Thus, by the preceding problem, a linear mapping \(T: V \rightarrow V\) is a projection if and only if \(T^{2}=T ;\) this characterization of a projection is frequently used as its definition. (a) If \(u \in \operatorname{Im} E\), then there exists \(v \in V\) for which \(E(v)=u\); hence, as required, $$ E(u)=E(E(v))=E^{2}(v)=E(v)=u $$ (b) Let \(v \in V\). We can write \(v\) in the form \(v=E(v)+v-E(v)\). Now \(E(v) \in \operatorname{Im} E\) and, because $$ E(v-E(v))=E(v)-E^{2}(v)=E(v)-E(v)=0 $$ \(v-E(v) \in \operatorname{Ker} E .\) Accordingly, \(V=\operatorname{Im} E+\operatorname{Ker} E\) Now suppose \(w \in \operatorname{Im} E \cap \operatorname{Ker} E .\) By \((i), E(w)=w\) because \(w \in \operatorname{Im} E .\) On the other hand, \(E(w)=0\) because \(w \in \operatorname{Ker} E\). Thus, \(w=0\), and so \(\operatorname{Im} E \cap \operatorname{Ker} E=\\{0\\} .\) These two conditions imply that \(V\) is the direct sum of the image and kernel of \(E\). (c) Let \(v \in V\) and suppose \(v=u+w\), where \(u \in \operatorname{Im} E\) and \(w \in\) Ker \(E\). Note that \(E(u)=u\) by (i), and \(E(w)=0\) because \(w \in \operatorname{Ker} E\). Hence, $$ E(v)=E(u+w)=E(u)+E(w)=u+0=u $$ That is, \(E\) is the projection of \(V\) into its image.

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