/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 For each of the following linear... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each of the following linear transformations \(T\), find im \(T\) and ker \(T\). (a) \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) given by \(T((x, y))=(x+y, 0) ;\) (b) \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) given by \(T((x, y))=(x+y, x-y)\); (c) \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) given by \(T((x, y, z))=(x, y, y)\) (d) \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) given by \(T((x, y, z))=(2 x-y+z,-x+3 y+5 z, 10 x-9 y-7 z)\) (e) \(T: M_{2,2} \rightarrow \mathbb{R}^{3}\) given by \(T\left(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\right)=(a+b-c, b+d, a-c-d)\) (f) \(T: P_{2} \rightarrow P_{2}\) given by \(T\left(a_{0}+a_{1} x+a_{2} x^{2}\right)=a_{1}+2 a_{2} x\).

Short Answer

Expert verified
(a) im \( T = \) span\((1, 0)\), ker \( T = \) span\((-1, 1)\). (b) im \( T = \mathbb{R}^2\), ker \( T = \{(0, 0)\}\). (c) im \( T = \) span\((1, 0, 0), (0, 1, 1)\), ker \( T = \) span\((0, 0, 1)\). (d) im \( T = \mathbb{R}^3\), ker \( T = \{0\}\). (e) im \( T = \mathbb{R}^3\), ker \( T = \) span\(\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}\). (f) im \( T = \) span\(1, x\), ker \( T = \) span\(1\).

Step by step solution

01

Understanding Transformation (a)

We are given that the transformation is defined by \(T((x, y)) = (x+y, 0)\). This means any input vector \((x, y)\) is mapped to \((x+y, 0)\).
02

Kernel of T (a)

The kernel \( \ker T \) consists of all vectors \((x, y)\) such that \( T((x, y)) = (0, 0) \). This gives the equation \( (x+y=0, 0=0) \). Therefore, the kernel is given by \( x + y = 0 \). The kernel is spanned by the vector \((-1, 1)\).
03

Image of T (a)

The image \( \text{im} T \) consists of all possible outputs of the transformation. Since the second component is always 0, \( \text{im} T \) consists of vectors of the form \((x, 0)\), i.e., the x-axis in \( \mathbb{R}^2 \). It is spanned by \((1, 0)\).
04

Understanding Transformation (b)

Transformation \( T \) is defined by \( T((x, y)) = (x+y, x-y) \). This maps \((x, y)\) into a new vector by these components.
05

Kernel of T (b)

For \( \ker T \), set \( T((x, y)) = (0, 0) \). Solve the equations \( x+y=0 \) and \( x-y=0 \). From these, both lead to \( x = 0 \) and \( y = 0 \). Thus, \( \ker T = \{ (0, 0) \} \).
06

Image of T (b)

The transformation \( T((x, y)) = (x+y, x-y) \) can achieve any vector in \( \mathbb{R}^2 \). Specifically, by choosing \( x = 1, y = 0 \) and \( x = 0, y = 1 \), we find that \( \text{im} T \) is spanned by \((1, 1)\) and \((1, -1)\), which form a basis for \( \mathbb{R}^2 \).
07

Understanding Transformation (c)

The transformation \( T: \mathbb{R}^3 \to \mathbb{R}^3 \) is given by \( T((x, y, z)) = (x, y, y) \), mapping a 3D vector to another 3D vector of that form.
08

Kernel of T (c)

Find \( (x, y, z) \) such that \( T((x, y, z)) = (0, 0, 0) \). This requires that \( x = 0 \) and \( y = 0 \). Any \( z \) is allowed, so \( \ker T \) is the set \{(0, 0, z) | z \in \mathbb{R}\} which is spanned by \( (0, 0, 1) \).
09

Image of T (c)

\( \text{im} T \) consists of all vectors of form \((x, y, y)\), meaning any value for \( x \) and \( y \). Therefore, it is spanned by \((1, 0, 0)\) and \((0, 1, 1)\).
10

Understanding Transformation (d)

The transformation \( T: \mathbb{R}^3 \to \mathbb{R}^3 \) is defined by mapping \( T((x, y, z)) = (2x - y + z, -x + 3y + 5z, 10x - 9y - 7z) \).
11

Kernel of T (d)

Set \( T((x, y, z)) = (0, 0, 0) \), which gives 3 linear equations in 3 unknowns. Solving these equations can be done by setting up and row-reducing the corresponding augmented matrix:\[ \begin{bmatrix} 2 & -1 & 1 & | & 0 \ -1 & 3 & 5 & | & 0 \ 10 & -9 & -7 & | & 0 \end{bmatrix} \].After row-reduction, find the solutions that describe \( \ker T \). Calculations show \( \ker T = \{0\} \).
12

Image of T (d)

The rank of the matrix from the transformation shows us the image. After reducing it finds its rank to be 3, indicating \( \text{im} T = \mathbb{R}^3 \), thus spanning the entire space.
13

Understanding Transformation (e)

The transformation maps a matrix to \( \mathbb{R}^3 \) given by \( T\left(\begin{bmatrix} a & b \ c & d \end{bmatrix}\right) = (a+b-c, b+d, a-c-d) \).
14

Kernel of T (e)

Set \( T\left(\begin{bmatrix} a & b \ c & d \end{bmatrix}\right) = (0, 0, 0) \) and solve \( a+b-c = 0 \), \( b+d = 0 \), \( a-c-d = 0 \). Solving simultaneously gives: \( b = -d \), \( a = c \). Hence, \( a \), and \( d \) are free, so \( \ker T \) is spanned by \( \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 0 & -1 \end{bmatrix} \).
15

Image of T (e)

\( \text{im} T \) is found from forming and reducing the matrix using the basis elements. Solving shows rank 3, thus \( \text{im} T = \mathbb{R}^3 \) with no restrictions spanning \( \mathbb{R}^3 \).
16

Understanding Transformation (f)

The transformation \( T: P_2 \to P_2 \) is given by mapping \( T(a_0 + a_1 x + a_2 x^2) = a_1 + 2a_2 x \).
17

Kernel of T (f)

Determine \( a_0 + a_1 x + a_2 x^2 \) such that \( T(a_0 + a_1 x + a_2 x^2) = 0 \), leading to the system \( a_1 = 0 \) and \( a_2 = 0 \). Thus, \( a_0 \) is free, and \( \ker T \) is spanned by \(1\).
18

Image of T (f)

Since the transformation outputs \( a_1 + 2a_2 x \), the image is a subspace of \( P_2 \) spanned by \(1, x\), indicating a full span over degree 1 polynomials.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kernel of a Transformation
In the study of linear transformations, one primary focus is understanding the kernel of a transformation. The kernel, or null space, is a set of input vectors that, when transformed, result in the zero vector. For a transformation \( T: V \rightarrow W \), where \( V \) and \( W \) are vector spaces, the kernel is defined as:
  • \( \ker T = \{ \mathbf{v} \in V \ | \ T(\mathbf{v}) = \mathbf{0} \} \) \
Finding the kernel involves solving for when the output of the transformation becomes zero. This process essentially reduces to solving a system of equations based on the transformation's mapping rule.
For example, if given \( T((x, y)) = (x+y, 0) \), we find the kernel by setting \( (x+y, 0) = (0, 0) \), ultimately revealing that \( x+y=0 \). Thus, the kernel is spanned by vectors of the form \((-1, 1)\). In other words, any vector on the line \( x+y=0 \) forms the kernel.
The kernel offers insight into how a transformation compresses or reduces dimensionality, as it encompasses vectors mapped to zero. It's a powerful tool to understand when and why certain mappings send elements to trivial results.
Image of a Transformation
The image of a transformation is another crucial concept. It describes the range, or set of all possible outputs, of a transformation applied to a vector space. For a transformation \( T: V \rightarrow W \), the image is given by:
  • \( \text{im} \, T = \{ \mathbf{w} \in W \ | \ \exists \mathbf{v} \in V, \text{ such that } T(\mathbf{v}) = \mathbf{w} \} \)
Determining the image involves expressing all possible outputs that the transformation rule can produce. This is often found by expressing a set of linearly independent vectors that span the transformed space.
Take, for instance, the transformation \( T((x, y)) = (x+y, x-y) \). To find its image, observe that any vector of form \((a, b)\) in \( \mathbb{R}^2 \) can be formed. Specifically, \( (1, 1) \) and \( (1, -1) \) serve as a basis for the range, fully spanning \( \mathbb{R}^2 \).
The image provides critical understanding of the transformation's behavior over an input space, indicating whether it fully covers the output space or is constrained to a subspace.
Basis of a Vector Space
A basis is a critical concept that connects deeply to both the kernel and image of a transformation. A basis forms a foundation from which every vector in a vector space can be uniquely represented. In mathematical terms, a basis for a vector space \( V \) is a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \} \) that are linearly independent and span \( V \).
  • Linearly independent means no vector in the set can be written as a linear combination of the others.
  • To span \( V \), every vector in \( V \) can be expressed as a linear combination of the basis vectors.
To find a basis, one typically selects vectors that satisfy these conditions. Looking at examples of transformations can clarify how basis vectors for the kernel or image are identified.
Consider a transformation \( T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \) like \( T((x, y, z)) = (x, y, y) \). The image of this transformation is spanned by the vectors \((1, 0, 0)\) and \((0, 1, 1)\), making these vectors a basis for the image. The kernel, on the other hand, is for vectors of form \((0, 0, z)\), with a basis provided by \((0, 0, 1)\).
Understanding the basis of a space is vital as it reveals the intrinsic dimensional characteristics of transformations, confirming which spaces they transform to or from efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.