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Let \(T: V \rightarrow V^{\prime \prime}\) 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
We proved that the quotient space, \(V / W\), is isomorphic to \(U\) under the mapping \(\theta\) as follows: 1. Showed \(\theta\) is linear: \(\theta((v_1+v_2)+W) = \theta(v_1+W) + \theta(v_2+W)\) and \(\theta(cv_1+W) = c\theta(v_1+W)\). 2. Showed \(\theta\) is surjective: For each \(u \in U\), there exists a \(v+W \in V / W\) such that \(\theta(v+W)=u\). 3. Showed \(\theta\) is injective: If \(\theta(v_1+W) = \theta(v_2+W)\), then \(v_1+W = v_2+W\). Finally, we showed that \(T = i \circ \theta \circ \eta\) by proving that \(T(v) = (i\circ\theta\circ\eta)(v)\) for all \(v \in V\).

Step by step solution

01

饾渻 is Linear

First, we need to show that 饾渻 is linear. Let v鈧, v鈧 鈭 V and c 鈭 F be the field of scalars. Then, we have 饾渻((v鈧+W) + (v鈧+W)) and 饾渻(c(v鈧+W)). We can rewrite these expressions as 饾渻((v鈧+v鈧)+W) and 饾渻(cv鈧+W) to simplify them. Now, by the definition of 饾渻, we have: - 饾渻((v鈧+v鈧)+W) = T(v鈧+v鈧) - 饾渻(cv鈧+W) = T(cv鈧) Since T is linear, we have: - T(v鈧+v鈧) = T(v鈧) + T(v鈧) - T(cv鈧) = cT(v鈧) So, we can rewrite our expressions as: - 饾渻((v鈧+v鈧)+W) = 饾渻(v鈧+W) + 饾渻(v鈧+W) - 饾渻(cv鈧+W) = c饾渻(v鈧+W) Thus, 饾渻 is linear.
02

饾渻 is Surjective

Next, we need to show that 饾渻 is surjective, i.e., for each u 鈭 U, there exists a v+W 鈭 V/W such that 饾渻(v+W)=u. Since u 鈭 U and U is the image of T, there must be a v 鈭 V such that T(v)=u. Now, 饾渻(v+W)=T(v)=u by the definition of 饾渻. Therefore, 饾渻 is surjective.
03

饾渻 is Injective

Finally, we need to show 饾渻 is injective, i.e., if 饾渻(v鈧+W) = 饾渻(v鈧+W), then v鈧+W = v鈧+W. Suppose 饾渻(v鈧+W)=饾渻(v鈧+W), then we have T(v鈧)=T(v鈧). So, T(v鈧 - v鈧) = 0, which means (v鈧 - v鈧) must be in the kernel W of T. Now, since (v鈧 - v鈧) 鈭 W, we have (v鈧 - v鈧) + W = W, which is equivalent to v鈧+W = v鈧+W. Hence, 饾渻 is injective. Since 饾渻 is linear, surjective, and injective, V/W is isomorphic to U under 饾渻.
04

T = i鈭橉潨冣垬饾渹

We need to show that T(v) = (i鈭橉潨冣垬饾渹)(v) for all v 鈭 V. By the definitions of 饾渹 and 饾渻, we have: - 饾渹(v) = v+W - 饾渻(v+W) = T(v) Now, since i is an inclusion mapping, i(u)=u for all u 鈭 U. In this case, we have u = T(v). So, i(饾渻(v+W)) = i(T(v)) = T(v). Therefore, we have (i鈭橉潨冣垬饾渹)(v) = i(饾渻(饾渹(v))) = i(饾渻(v+W)) = T(v). So, T=i鈭橉潨冣垬饾渹, which completes the proof.

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Key Concepts

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

What Is a Quotient Space?
In linear algebra, the concept of a quotient space arises when you want to partition a vector space by a subspace. Imagine you have a vector space, denoted by V, and a subspace W contained within it. The quotient space, denoted as V/W, is essentially the set of all possible cosets of W in V. A coset of W is formed by taking a vector v from V and 'adding' it to W, which is denoted by v+W. This means that each element in V/W is not just a single vector, but a whole set of vectors that differ from one another by an element of W.

The creation of quotient spaces is a powerful tool because it allows us to study the structure of V in relation to W. In the context of our exercise, the quotient space V/W represents the congruence classes of vectors in V under the equivalence relation 'differing by an element of W'. Understanding quotient spaces gives us a method to 'factor out' the effects of the subspace W, leading us to investigate properties of V that are unaffected by W.
Understanding Mapping in Linear Algebra
The term 'mapping' in linear algebra refers to a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In the case of our exercise, the mapping is a function from the quotient space V/W to the image U, associating each coset v+W in V/W with a vector T(v) in U. To be considered linear (also known as a linear transformation), this mapping must satisfy two main properties: it must be additive and homogeneous. This means if you take two elements from V/W and 'add' them, their image under the mapping must be the sum of their images. Similarly, multiplying an element by a scalar before mapping must give the same result as mapping first and then multiplying by the scalar.

These properties are crucial for our mapping to be considered a linear transformation and for the quotient space V/W to be isomorphic to the image U. The mapping plays an essential role in understanding how different vector spaces can be related and transformed into one another.
Kernel of a Linear Transformation
The kernel of a linear transformation is a central concept in understanding how a transformation 'compresses' a vector space. For a linear transformation T from V to V'', the kernel is the set of all vectors in V that T maps to the zero vector in V''. It is denoted as kernel(T) or simply Ker(T). Mathematically, Ker(T) = {v in V | T(v) = 0}.

The kernel is a subspace of V. If Ker(T) only contains the zero vector, then T is injective, or one-to-one. In our exercise, the kernel is denoted by W which means any vector within this space will be 'ignored' or transformed into zero when applying T. The kernel helps us understand the 'loss of information' as V is mapped into V''. By examining the kernel, we can judge the level of injectivity of our linear transformation and ascertain the characteristics of our quotient space V/W.
Image of a Linear Transformation
Conversely to the kernel, the image of a linear transformation captures the 'output' of the transformation, containing all vectors in the codomain that T can reach from V. Denoted as image(T) or Im(T), the image is the set of all vectors in the codomain V'' to which at least one vector from the domain V is mapped. Mathematically, it's given by Im(T) = {T(v) | v in V}.

The image is a subspace of the codomain V'' and provides insight into the coverage of T. It can be thought of as a 'shadow' of V cast onto V'' by the 'light' of T. The larger the image, the more of V'' is 'illuminated' by V. In the context of the exercise, the image, called U, is isomorphic to the quotient space V/W, meaning there's a one-to-one correspondence between elements of U and cosets of W in V. This idea of the image is vital for understanding how a vector space is transformed and what aspects of its structure are preserved in the process.

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