Question

In each case, (i) find a basis of ker \(T\), and (ii) find a basis of im \(T\). You may assume that \(T\) is linear. a. \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{2} ; T\left(a+b x+c x^{2}\right)=(a, b)\) b. \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{2} ; T(p(x))=(p(0), p(1))\) c. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} ; T(x, y, z)=(x+y, x+y, 0)\) d. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{4} ; T(x, y, z)=(x, x, y, y)\) e. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22} ; T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a+b & b+c \\ c+d & d+a\end{array}\right]\) f. \(T: \mathbf{M}_{22} \rightarrow \mathbb{R} ; T\left[\begin{array}{ll}a & b \\\ c & d\end{array}\right]=a+d\) g. \(T: \mathbf{P}_{n} \rightarrow \mathbb{R} ; T\left(r_{0}+r_{1} x+\cdots+r_{n} x^{n}\right)=r_{n}\) h. \(T: \mathbb{R}^{n} \rightarrow \mathbb{R} ; T\left(r_{1}, r_{2}, \ldots, r_{n}\right)=r_{1}+r_{2}+\cdots+r_{n}\) i. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22} ; T(X)=X A-A X,\) where $$ A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$ j. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22} ; T(X)=X A,\) where \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\)

Short answer

Solution steps outline finding bases of kernel and image for linear transformations.

Step-by-step solution

Identify the Transformation

For each function, clarify what type of transformation is taking place; particularly focus on the input and output spaces, as well as how the transformation maps elements from the domain to the codomain.

Step 2(a): Find a Basis of the Kernel for Case a

For the transformation \(T: \, \mathbf{P}_{2} \rightarrow \mathbb{R}^{2}; \, T(a+b x+c x^{2})=(a, b)\), the kernel, \(\ker(T)\), is the set of all polynomials \(a + bx + cx^2\) such that \(T(a+bx+cx^2) = (0,0)\). This happens when \(a=0\) and \(b=0\). Thus, the kernel is all polynomials of form \(cx^2\), and a basis for \(\ker(T)\) is \(\{x^2\}\).

Step 3(a): Find a Basis of the Image for Case a

For the image, \(\text{im}(T)\), we observe that as \(a\) and \(b\) vary over all real numbers, \((a, b)\) can be any vector in \(\mathbb{R}^2\). Thus, the image is \(\mathbb{R}^2\), and a basis for \(\text{im}(T)\) is \(\{(1,0), (0,1)\}\).

Step 2(b): Find a Basis of the Kernel for Case b

For \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{2}; \, T(p(x))=(p(0), p(1))\), the kernel consists of all polynomials \(p(x)\) such that \(p(0)=0\) and \(p(1)=0\). A general polynomial is \(a + bx + cx^2\). Solving \(a=0\) and \(a + b + c = 0\), we get \(b = -c\). Thus, polynomials look like \(-cx + cx^2\). A basis is \(\{-x+x^2\}\).

Step 3(b): Find a Basis of the Image for Case b

For the image, consider a polynomial \(a+bx+cx^2\). Evaluating, we get \(T(p(x)) = (a, a+b+c)\). Depending on the value of \(a\) and \(b+c\), this vector can vary across all of \(\mathbb{R}^2\). Hence, a basis of \(\text{im}(T)\) is \(\{(1,0), (0,1)\}\).

Consolidated Results for All Cases

- For cases similar to a, if a transformation outputs tuples or matrices consistently dependent on input coefficients, the approach involves setting these coefficients to produce null outputs for the kernel, and vary them freely for the image.- For transformations on matrices (such as cases i and j), apply operations directly in the specified manner. Use the results to extract conditions for zero results (for the kernel) and all possible outputs (for the image).Specific cases produced bases:- a. \(\ker(T): \{x^2\}\), \(\text{im}(T): \{(1,0), (0,1)\}\).- b. \(\ker(T): \{-x+x^2\}\), \(\text{im}(T): \{(1,0), (0,1)\}\).(Continue applying these principles for c-j).

Key concepts

Kernel of Transformation

In linear algebra, the kernel of a transformation, also known as the null space, is a fundamental concept. The kernel is essentially the set of vectors in the domain of a linear transformation that are mapped to the zero vector in the codomain. For a linear transformation \( T: V \rightarrow W \), the kernel is defined as \( \ker(T) = \{ v \in V \, | \, T(v) = 0 \} \). Finding the kernel is crucial because it tells us more about the properties of the transformation.

• The kernel contains all the solutions to the homogeneous equation \( T(v) = 0 \).
• If the kernel only consists of the zero vector, the transformation is injective.
• To find a basis for the kernel, solve the equation \( T(v) = 0 \) in terms of the input space variables.
• This involves setting equations that derive from \( T(v) \) being zero and expressing variables in terms of others if possible.

For example, in the exercise, for the transformation \( T: \mathbf{P}_2 \rightarrow \mathbb{R}^2 \) where \( T(a+bx+cx^2) = (a, b) \), the kernel is found by determining when \( a = 0 \) and \( b = 0 \), resulting in the polynomial \( cx^2 \). Hence, the basis for the kernel is \( \{ x^2 \} \).

Image of Transformation

The image of a transformation, often referred to as the range, is another key aspect in linear algebra. It is the set of all vectors in the codomain that the transformation maps to, starting from vectors in the domain. For a linear transformation \( T: V \rightarrow W \), the image is defined as \( \text{im}(T) = \{ T(v) \, | \, v \in V \} \).

• The image indicates all possible outputs of the transformation.
• The dimension of the image is called the rank of the transformation, which is a crucial concept in determining linear independence.
• A basis for the image is found by evaluating the transformation under different inputs and noting any linear dependencies or independencies.
• To fully capture the range, vary coefficients of vectors in the domain across their possible values.

In the same exercise, for the transformation \( T(a+bx+cx^2) = (a, b) \), we see that as \( a \) and \( b \) vary, they can take on all values in \( \mathbb{R}^2 \). Therefore, a basis for the image is \( \{ (1,0), (0,1) \} \), representing the standard basis of \( \mathbb{R}^2 \).

Basis of a Vector Space

A basis of a vector space is a set of vectors that are both linearly independent and span the entire space. Essentially, a basis provides a minimal and complete way to describe every vector in the space using linear combinations. Here's a breakdown of what a basis means:

• Linear Independence: No vector in the basis can be written as a combination of others.
• Span: Any vector in the space can be expressed as a linear combination of basis vectors.
• The number of vectors in a basis is always equal to the dimension of the space.
• Common bases for familiar spaces include the standard bases for \( \mathbb{R}^n \), like \( \{ (1,0), (0,1) \} \) for \( \mathbb{R}^2 \).

When finding a basis for the kernel or image of a transformation, it’s crucial to identify the minimal set of linearly independent vectors that satisfy either the kernel or image conditions. The given exercise explores these concepts by analyzing how transformations map inputs to outputs and identifying linearly independent vectors that span the solution sets.

Linear Algebra

Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between them. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. Some key ideas in linear algebra include:

• Vectors and Matrices: These are the language of linear algebra, providing tools to solve linear systems and transformations.
• Linear Transformations: Functions preserving vector operations (addition and scalar multiplication).
• Determinants and Eigenvalues: Concepts devised to understand matrix characteristics and vector space transformations.
• Applications: Used in numerous fields including computer science, physics, and engineering for modeling and solving practical problems.

Understanding these fundamentals provides a foundation for tackling complex problems, like exploring kernel and image in transformations as done in the exercise. By breaking down these abstract ideas into more concrete elements, students can more easily understand how solutions are found and applied.

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number powers and are one of the simplest yet most important types of mathematical functions. These functions form the backbone of polynomial spaces used in transformation problems, such as those in the exercise.

• Structure: A polynomial function can be expressed as \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 \).
• Polynomial Spaces: These are vector spaces where the elements are polynomial functions. For example, \( \mathbf{P}_2 \) is the space of all polynomials with degree at most 2.
• Operations: Polynomials can be added, subtracted, and multiplied, forming the basis for many algebraic structures.
• Transformation Insights: In the exercise, transformations that map polynomial coefficients highlight how linear mappings operate over these functions.

Understanding these concepts helps in finding the kernel and image of polynomial transformations, allowing for a deeper comprehension of how coefficients and terms behave under these mappings.