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Chapter 10: Problem 26

Let \(V\) and \(W\) be vector spaces over the same field \(F\). A function \(T: V \rightarrow W\) is said to be a linear transformation from \(V\) to \(W\) if for all \(c, d \in F\) and all \(u, v \in V\) we have $$ T(c u+v d)=c T(u)+d T(v) $$ Let \(T\) be a linear transformation. (a) Show that \(\operatorname{lm}(T)\), the image of \(T\), is a subspace of \(W\). (b) Show that Kern \(T=\\{v \in V \mid T(v)=0\\}\) is a subspace of \(V\).

Short Answer

Expert verified
Both the image and kernel of a linear transformation are subspaces.

Step by step solution

01

Define the Image of T

The image of a linear transformation \( T: V \rightarrow W \) is the set \( \operatorname{Im}(T) = \{ T(v) \, | \, v \in V \} \). We need to show that this set is a subspace of \( W \).
02

Verify Non-emptiness of Image

The zero vector of \( W \), denoted as \( 0_W \), should be in \( \operatorname{Im}(T) \). Since \( T(0_V) = 0_W \) for any linear transformation and \( 0_V \) is in \( V \), \( 0_W \in \operatorname{Im}(T) \), ensuring non-emptiness.
03

Closure Under Addition for \( \operatorname{Im}(T) \)

Take any two elements \( T(u) \) and \( T(v) \) in \( \operatorname{Im}(T) \). Then, \( T(u) + T(v) = T(u+v) \) because \( T \) is linear. As \( u+v \in V \), thus \( T(u+v) \in \operatorname{Im}(T) \), proving closure under addition.
04

Closure Under Scalar Multiplication for \( \operatorname{Im}(T) \)

For any scalar \( c \in F \) and for any vector \( T(u) \in \operatorname{Im}(T) \), we have \( cT(u) = T(cu) \) by linearity. Since \( cu \in V \), \( T(cu) \in \operatorname{Im}(T) \), showing closure under scalar multiplication.
05

Define Kernel of T

The kernel of a linear transformation \( T: V \rightarrow W \) is given by \( \operatorname{Ker}(T) = \{ v \in V \mid T(v) = 0_W \} \). We need to show this is a subspace of \( V \).
06

Verify Non-emptiness of Kernel

The zero vector \( 0_V \in V \) is in \( \operatorname{Ker}(T) \) because \( T(0_V) = 0_W \). Thus, \( \operatorname{Ker}(T) \) is non-empty.
07

Closure Under Addition for \( \operatorname{Ker}(T) \)

Let \( u, v \in \operatorname{Ker}(T) \), which means \( T(u) = 0_W \) and \( T(v) = 0_W \). Then \( T(u+v) = T(u) + T(v) = 0_W + 0_W = 0_W \). So, \( u+v \in \operatorname{Ker}(T) \), proving closure under addition.
08

Closure Under Scalar Multiplication for \( \operatorname{Ker}(T) \)

For any \( c \in F \) and any \( u \in \operatorname{Ker}(T) \), \( T(cu) = cT(u) = c \cdot 0_W = 0_W \). Therefore, \( cu \in \operatorname{Ker}(T) \), establishing closure under scalar multiplication.

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Key Concepts

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

Vector Spaces
Vector spaces are fundamental structures in linear algebra. They consist of a set of vectors, which are often denoted by capital letters like \( V \) or \( W \), accompanied by two operations: vector addition and scalar multiplication. To qualify as a vector space, the set and operations must satisfy specific properties.

There are key properties that define a vector space:
  • **Closure:** The sum of any two vectors in the space must also be a vector in the space. Similarly, the product of a vector by a scalar must also belong to the space.
  • **Associativity and Commutativity:** Vector addition must be associative and commutative, meaning the order in which you add vectors doesn鈥檛 affect the result.
  • **Existence of Identity and Inverse Elements:** There must be a zero vector that acts as an additive identity, and each vector should have an additive inverse.
  • **Distributivity:** Distributive laws must hold for both vectors and scalars over vector addition.
Understanding these properties is essential for working with linear transformations and proving whether certain subsets, like the kernel or image of a transformation, are themselves vector spaces known as subspaces.
Subspace
A subspace is a subset of a vector space that is itself a vector space, under the same operations of addition and scalar multiplication. For any subset \( S \) of a vector space \( V \) to be a subspace, it must satisfy three main criteria:

  • **Non-emptiness:** The subspace must contain the zero vector. The presence of the zero vector ensures that a subspace is non-empty and can serve as a base point for addition and scalar multiplication.
  • **Closed under Addition:** If you have any two vectors \( u \) and \( v \) in \( S \), their sum \( u + v \) must also be in \( S \).
  • **Closed under Scalar Multiplication:** If \( c \) is a scalar and \( u \) a vector in \( S \), then \( cu \) must also be in \( S \).
If all these conditions are met, \( S \) can be considered a subspace of \( V \). In the context of linear transformations, both the kernel and image of a transformation are examples of such subspaces.
Kernel of Transformation
The kernel of a linear transformation \( T: V \rightarrow W \) is a special subspace within \( V \). It consists of all vectors \( v \) in \( V \) such that \( T(v) = 0_W \). This means the kernel captures the vectors that are essentially "annihilated" by the transformation.

To demonstrate that the kernel is indeed a subspace, we consider the following:
  • **Non-emptiness:** The zero vector \( 0_V \) itself is always in the kernel, given that \( T(0_V) = 0_W \). Thus, the kernel is non-empty.
  • **Closure under Addition:** If \( u \) and \( v \) are in the kernel, then \( T(u) = 0_W \) and \( T(v) = 0_W \). Thus, \( T(u+v) = T(u) + T(v) = 0_W \), meaning \( u+v \) is also in the kernel.
  • **Closure under Scalar Multiplication:** For a scalar \( c \), if \( u \) is in the kernel, \( T(cu) = cT(u) = c imes 0_W = 0_W \), so \( cu \) is in the kernel.
By satisfying these properties, the kernel acts as a foundational subspace that offers insight into the nature of the transformation \( T \), essentially indicating where \( T \) maps vectors to zero.
Image of Transformation
The image of a linear transformation \( T: V \rightarrow W \) is the set of all vectors in \( W \) that can be expressed as \( T(v) \) for some vector \( v \) in \( V \). This set, expressed as \( \, \operatorname{Im}(T) = \{ T(v) \, | \, v \in V \} \, \), represents what \( V \) looks like when transformed by \( T \).

To prove that the image is a subspace of \( W \), the following must hold:
  • **Non-emptiness:** The zero vector \( 0_W \) is part of the image because \( T(0_V) = 0_W \). Hence, the image is non-empty.
  • **Closure under Addition:** If \( T(u) \) and \( T(v) \) are in the image, then their sum \( T(u) + T(v) = T(u+v) \) is also in the image, since \( u+v \) is in \( V \).
  • **Closure under Scalar Multiplication:** If \( c \) is any scalar and \( T(u) \) is in the image, then \( cT(u) = T(cu) \) and since \( cu \) is in \( V \), \( T(cu) \) belongs to the image.
By meeting these conditions, the image verifies its status as a subspace of \( W \). It effectively maps out all possible outputs \( W \) can achieve through the transformation \( T \), illustrating the reach of \( V \) once transformed.

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