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Show that the following are equivalent for a linear transformation \(T: V \rightarrow W\). 1\. ker \(T=V\) 2\. \(\operatorname{im} T=\\{\mathbf{0}\\}\) 3\. \(T=0\)

Short Answer

Expert verified
The three statements are equivalent: \( \text{ker}(T) = V \), \( \text{im}(T) = \{ \mathbf{0} \} \), and \( T = 0 \).

Step by step solution

01

Understanding the Kernel

The kernel of a linear transformation, \( \text{ker}(T) \), is the set of vectors in \( V \) that map to the zero vector in \( W \). If \( \text{ker}(T) = V \), it means every vector \( \mathbf{v} \in V \) maps to the zero vector \( \mathbf{0} \in W \) under \( T \). Thus, \( T(\mathbf{v}) = \mathbf{0} \) for all \( \mathbf{v} \in V \).
02

Interpreting the Image

The image of a linear transformation, \( \text{im}(T) \), is the set of all vectors \( T(\mathbf{v}) \) for \( \mathbf{v} \in V \). If \( \text{im}(T) = \{ \mathbf{0} \} \), it means the only vector that \( T \) maps any vector of \( V \) to is the zero vector in \( W \). This implies that for every vector \( \mathbf{v} \in V \), \( T(\mathbf{v}) = \mathbf{0} \).
03

Showing \( T = 0 \)

The statement \( T = 0 \) refers to the zero transformation, which maps every vector in \( V \) to the zero vector in \( W \). If \( T(\mathbf{v}) = \mathbf{0} \) for every \( \mathbf{v} \in V \), this is equivalent to saying \( T \) is the zero transformation.
04

Equivalence Proof

Now, consider the equivalence of these statements. If \( \text{ker}(T) = V \), then \( T(\mathbf{v}) = \mathbf{0} \) for all \( \mathbf{v} \in V \) which implies \( T = 0 \), thus \( \text{im}(T) = \{ \mathbf{0} \} \). Conversely, if \( \text{im}(T) = \{ \mathbf{0} \} \), then \( T = 0 \), which implies \( \text{ker}(T) = V \). Similarly, if \( T = 0 \), then \( \text{ker}(T) = V \) and \( \text{im}(T) = \{ \mathbf{0} \} \). Therefore, the three statements are equivalent.

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Key Concepts

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

Kernel of a Transformation
The kernel of a transformation, denoted as \( \text{ker}(T) \), is crucial in understanding the behavior of a linear transformation \( T: V \rightarrow W \). This set contains all vectors from the vector space \( V \) that are transformed into the zero vector in the target space \( W \). Why is the kernel important? Imagine having a system where certain elements completely lose their identity and become neutral or insignificant—this is similar to what happens in \( \text{ker}(T) \).
  • If \( \text{ker}(T) = V \), it implies that every vector in \( V \) maps to the zero vector. It gives us a specific insight about the transformation: \( T \) behaves identically across all inputs in \( V \).
  • This scenario indicates a reduction to the zero vector, suggesting that \( T \) could possibly be a zero transformation itself.
Image of a Transformation
The image of a transformation, \( \text{im}(T) \), reveals where vectors from space \( V \) land in space \( W \) after applying \( T \). It gives a picture of all possible outputs of the transformation.For \( \text{im}(T) = \{ \mathbf{0} \} \), the significance is profound:
  • No matter which vector from \( V \) is chosen, \( T \) always outputs the zero vector in \( W \).
  • This suggests the transformation has no effect in transferring or transforming information from \( V \) to any other vector in \( W \) except for the zero vector.
In essence, if the image contains only the zero vector, it indicates that the transformation does not "spread out" vectors in \( W \), aligning it closely with the characteristics of a zero transformation.
Zero Transformation
A zero transformation is a special type of linear transformation \( T \) where every vector in its domain \( V \) is mapped to the zero vector in the codomain \( W \). Simply put, \( T(\mathbf{v}) = \mathbf{0} \) for all \( \mathbf{v} \in V \).Why is this significant?
  • The zero transformation is like pressing a universal 'reset' button for vectors—no matter what you input, you always get a zero output.
  • This property directly implies that both the kernel and the image of the transformation have predictable forms, as we've discussed.
Recognizing a zero transformation is essential for solving equations and proving equivalences, as we'll see in the equivalence proof section.
Equivalence Proof
The concept of equivalence proof in linear transformations is about showing different statements lead to the same conclusion or truth about a transformation. Here's how these three statements interlink neatly:
  • If \( \text{ker}(T) = V \), it means \( T(\mathbf{v}) = \mathbf{0} \) for every \( \mathbf{v} \in V \), showing that the transformation doesn't provide any new information or change, i.e., \( T = 0 \).
  • With \( T = 0 \), the output for every vector is the zero vector, leading to \( \text{im}(T) = \{ \mathbf{0} \} \), since the only possible image is the zero vector.
  • When \( \text{im}(T) = \{ \mathbf{0} \} \), it indicates that \( T(\mathbf{v}) = \mathbf{0} \), aligning with the zero transformation definition and confirming that \( \text{ker}(T) = V \).
These equivalences provide a comprehensive understanding that both the kernel containing the entirety of \( V \), the image being only \{\mathbf{0}\}, and the zero transformation are indeed the same condition manifested in different forms.

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Most popular questions from this chapter

In each case, find a linear transformation with the given properties and compute \(T(\mathbf{v})\) $$ \begin{array}{l} \text { a. } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} ; T(1,2)=(1,0,1) \\ \quad T(-1,0)=(0,1,1) ; \mathbf{v}=(2,1) \\ \text { b. } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} ; T(2,-1)=(1,-1,1) \\\ \quad T(1,1)=(0,1,0) ; \mathbf{v}=(-1,2) \\ \text { c. } T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{3} ; T\left(x^{2}\right)=x^{3}, T(x+1)=0 \\ \quad T(x-1)=x ; \mathbf{v}=x^{2}+x+1 \\ \text { d. } T: \mathbf{M}_{22} \rightarrow \mathbb{R} ; T\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]=3, T\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=-1, \\ \quad T\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]=0=T\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] ; \mathbf{v}=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \end{array} $$

Let \(A\) and \(B\) be matrices of size \(p \times m\) and \(n \times q .\) Assume that \(m n=p q .\) Define \(R: \mathbf{M}_{m n} \rightarrow \mathbf{M}_{p q}\) by \(R(X)=A X B\) a. Show that \(\mathbf{M}_{m n} \cong \mathbf{M}_{p q}\) by comparing dimensions. b. Show that \(R\) is a linear transformation. c. Show that if \(R\) is an isomorphism, then \(m=p\) and \(n=q .\) [Hint: Show that \(T: \mathbf{M}_{m n} \rightarrow \mathbf{M}_{p n}\) given by \(T(X)=A X\) and \(S: \mathbf{M}_{m n} \rightarrow \mathbf{M}_{m q}\) given by \(S(X)=X B\) are both one-to-one, and use the dimension theorem.]

Let \(V\) and \(W\) be finite dimensional vector spaces. a. Show that \(\operatorname{dim} W \leq \operatorname{dim} V\) if and only if there exists an onto linear transformation \(T: V \rightarrow W\). [Hint: Theorem 6.4.1 and Theorem 7.1.3.] b. Show that \(\operatorname{dim} W \geq \operatorname{dim} V\) if and only if there exists a one-to-one linear transformation \(T: V \rightarrow W\). [Hint: Theorem 6.4.1 and Theorem 7.1.3.]

Let \(T: V \rightarrow V\) be a linear transformation. Show that \(T^{2}=1_{V}\) if and only if \(T\) is invertible and \(T=T^{-1}\)

Show that the following conditions are equivalent for a linear transformation \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\). 1\. \(\operatorname{tr}[T(A)]=\operatorname{tr} A\) for all \(A\) in \(\mathbf{M}_{22}\). $$ \text { 2. } T\left[\begin{array}{ll} r_{11} & r_{12} \\ r_{21} & r_{22} \end{array}\right]=r_{11} B_{11}+r_{12} B_{12}+r_{21} B_{21}+ $$ \(r_{22} B_{22}\) for matrices \(B_{i j}\) such that \(\operatorname{tr} B_{11}=1=\operatorname{tr} B_{22}\) and \(\operatorname{tr} B_{12}=0=\operatorname{tr} B_{21}\)

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