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Chapter 7: Problem 15

Define \(T: \mathbf{P}_{n} \rightarrow \mathbb{R}\) by \(T[p(x)]=\) the sum of all the coefficients of \(p(x)\). a. Use the dimension theorem to show that \(\operatorname{dim}(\operatorname{ker} T)=n\) b. Conclude that \(\left\\{x-1, x^{2}-1, \ldots, x^{n}-1\right\\}\) is a basis of \(\operatorname{ker} T\)

Short Answer

Expert verified
\(\operatorname{dim}(\ker(T)) = n\) and \(\{x-1, x^2-1, \ldots, x^n-1\}\) is a basis of \(\ker(T)\).

Step by step solution

01

Understanding the Transformation T

The transformation \(T: \mathbf{P}_{n} \rightarrow \mathbb{R}\) is defined by \(T[p(x)] =\) the sum of all coefficients of the polynomial \(p(x)\). For example, if \(p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0\), then \(T[p(x)] = a_n + a_{n-1} + \ldots + a_0\).
02

Define the Kernel of T

The kernel of \(T\), denoted \(\operatorname{ker}(T)\), consists of all polynomials \(p(x) \in \mathbf{P}_n\) such that \(T[p(x)] = 0\). This means \(a_n + a_{n-1} + \ldots + a_0 = 0\).
03

Apply the Dimension Theorem

According to the dimension theorem, the dimension of the kernel of a linear transformation \(T\) from a vector space \(V\) to another vector space \(W\) is given by \(\operatorname{dim}(\ker(T)) = \operatorname{dim}(V) - \operatorname{dim}(\operatorname{Im}(T))\). For polynomials of degree \(n\), \(\operatorname{dim}(\mathbf{P}_n) = n+1\).
04

Find the Dimension of Image of T

The image of \(T\) is the set of all possible sums of coefficients and is a subset of \(\mathbb{R}\). Since there is at least one polynomial, like \(p(x) = 1\), which maps to a non-zero element of \(\mathbb{R}\) (i.e., \(T[1] = 1\)), \(\operatorname{dim}(T) = 1\).
05

Calculate the Dimension of the Kernel

Using the dimension theorem, \(\operatorname{dim}(\operatorname{ker}(T)) = \operatorname{dim}(\mathbf{P}_n) - \operatorname{dim}(\text{Im}(T)) = (n+1) - 1 = n\).
06

Identify the Basis of Kernel

The basis of \(\operatorname{ker(T)}\) consists of polynomials for which the sum of their coefficients is zero. Given that \(x-1, x^2-1, \ldots, x^n-1\) yield sums of coefficients equal to zero, they form linearly independent vectors and span \(\operatorname{ker(T)}\). Therefore, \(\{x-1, x^2-1, \ldots, x^n-1\}\) forms a basis of \(\operatorname{ker(T)}\).

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Key Concepts

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

Polynomials
Polynomials are key elements in mathematics and serve as the foundation for understanding various concepts in linear algebra.
They are expressions involving variables with different powers and coefficients. For example, a simple polynomial can be written as \[p(x) = a_nx^n + a_{n-1}x^{n-1} + ext{...} + a_0\]where each term is a product of a coefficient and a power of the variable.
  • Variables: The symbols (often "x" or "y") that represent the unknown values.
  • Coefficients: The numbers that multiply the variables, indicating how much each term contributes to the value of the polynomial.
  • Degree: The highest power of the variable, which determines the behavior and characteristics of the polynomial. For instance, the polynomial above has a degree of \(n\).
Polynomials can be added, subtracted, multiplied, and divided, much like regular numbers, and play a crucial role in defining functions and transformations.
Vector Spaces
Vector spaces are mathematical structures formed by vectors that can be added together and multiplied by scalars (numbers). They encompass many familiar mathematical objects, including polynomials. A vector space must satisfy certain properties:
  • Closure Under Addition: Adding two vectors in the space results in another vector in the space.
  • Closure Under Scalar Multiplication: Multiplying a vector by a scalar results in another vector in the space.
  • Contains a zero vector: A vector that acts as an additive identity (adding it does not change other vectors).
  • Contains additive inverses: For every vector, there is another vector that negates it when added together.
Understanding vector spaces help us in understanding linear transformations, as they map elements from one vector space to another.
Dimension Theorem
The dimension theorem is a fundamental result in linear algebra that relates the dimensions of certain vector subspaces. It states that if \(T\) is a linear transformation from a vector space \(V\) to another vector space \(W\), then:\[\text{dim}(V) = \text{dim}(\ker(T)) + \text{dim}(\text{Im}(T))\]Here's a breakdown of the elements:
  • \(\text{dim}(V)\): The dimension of the complete vector space \(V\), which is the number of vectors in a basis of \(V\).
  • \(\text{dim}(\ker(T))\): The dimension of the kernel of \(T\), consisting of vectors that \(T\) maps to zero.
  • \(\text{dim}(\text{Im}(T))\): The dimension of the image of \(T\), which corresponds to all possible outputs of \(T\).
This theorem helps us understand how a linear transformation affects the structure of vector spaces.
Kernel and Image
The kernel and image are critical concepts when dealing with linear transformations. They describe where and how the input (from one vector space) is mapped under the transformation.- **Kernel (\(\ker(T)\))**: The set of all vectors in the domain that are transformed to the zero vector in the codomain. It describes "what maps to zero." - For the transformation \(T: \mathbf{P}_n \rightarrow \mathbb{R}\), the kernel includes polynomials whose coefficients add up to zero. - The dimensions of the kernel help in determining "space lost" in the transformation.- **Image (\(\text{Im}(T)\))**: All vectors in the codomain that are the output of the transformation with some input vector from the domain. It is essentially "what you can reach with \(T\)." - For our transformation, the image is a subset of \(\mathbb{R}\), representing achievable sums.Understanding these concepts helps explain how transformations affect objects and maintain, lose, or change essential structural properties.

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Most popular questions from this chapter

Show that each of the following functions is a linear transformation. a. \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} ; T(x, y)=(x,-y)\) (reflection in the \(x\) axis) b. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} ; T(x, y, z)=(x, y,-z)\) (reflection in the \(x\) - \(y\) plane) c. \(T: \mathbb{C} \rightarrow \mathbb{C} ; T(z)=\bar{z}\) (conjugation) d. \(T: \mathbf{M}_{m n} \rightarrow \mathbf{M}_{k l} ; T(A)=P A Q, P\) a \(k \times m\) matrix, \(Q\) an \(n \times l\) matrix, both fixed e. \(T: \mathbf{M}_{n n} \rightarrow \mathbf{M}_{n n} ; T(A)=A^{T}+A\) f. \(T: \mathbf{P}_{n} \rightarrow \mathbb{R} ; T[p(x)]=p(0)\) g. \(T: \mathbf{P}_{n} \rightarrow \mathbb{R} ; T\left(r_{0}+r_{1} x+\cdots+r_{n} x^{n}\right)=r_{n}\) h. \(T: \mathbb{R}^{n} \rightarrow \mathbb{R} ; T(\mathbf{x})=\mathbf{x} \cdot \mathbf{z}, \mathbf{z}\) a fixed vector in \(\mathbb{R}^{n}\) i. \(T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n} ; T[p(x)]=p(x+1)\) j. \(T: \mathbb{R}^{n} \rightarrow V ; T\left(r_{1}, \cdots, r_{n}\right)=r_{1} \mathbf{e}_{1}+\cdots+r_{n} \mathbf{e}_{n}\) where \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\right\\}\) is a fixed basis of \(V\) k. \(T: V \rightarrow \mathbb{R} ; T\left(r_{1} \mathbf{e}_{1}+\cdots+r_{n} \mathbf{e}_{n}\right)=r_{1},\) where \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\right\\}\) is a fixed basis of \(V\)

Let \(T: V \rightarrow V\) be a linear transformation where \(V\) is finite dimensional. Show that exactly one of (i) and (ii) holds: (i) \(T(\mathbf{v})=\mathbf{0}\) for some \(\mathbf{v} \neq \mathbf{0}\) in \(V\) (ii) \(T(\mathbf{x})=\mathbf{v}\) has a solution \(\mathbf{x}\) in \(V\) for every \(\mathbf{v}\) in \(V\).

In each case either prove the statement or give an example in which it is false. Throughout, let \(T: V \rightarrow W\) be a linear transformation where \(V\) and \(W\) are finite dimensional. a. If \(V=W,\) then ker \(T \subseteq \operatorname{im} T\). b. If \(\operatorname{dim} V=5, \operatorname{dim} W=3,\) and \(\operatorname{dim}(\operatorname{ker} T)=2\), then \(T\) is onto. c. If \(\operatorname{dim} V=5\) and \(\operatorname{dim} W=4,\) then \(\operatorname{ker} T \neq\\{\mathbf{0}\\}\). d. If ker \(T=V,\) then \(W=\\{\mathbf{0}\\}\). e. If \(W=\\{\mathbf{0}\\},\) then \(\operatorname{ker} T=V\). f. If \(W=V,\) and \(\operatorname{im} T \subseteq \operatorname{ker} T,\) then \(T=0\). g. If \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\}\) is a basis of \(V\) and \(T\left(\mathbf{e}_{1}\right)=\mathbf{0}=T\left(\mathbf{e}_{2}\right),\) then \(\operatorname{dim}(\operatorname{im} T) \leq 1\) h. If \(\operatorname{dim}(\operatorname{ker} T) \leq \operatorname{dim} W,\) then \(\operatorname{dim} W \geq \frac{1}{2} \operatorname{dim} V\). i. If \(T\) is one-to-one, then \(\operatorname{dim} V \leq \operatorname{dim} W\). j. If \(\operatorname{dim} V \leq \operatorname{dim} W,\) then \(T\) is one-to- one. k. If \(T\) is onto, then \(\operatorname{dim} V \geq \operatorname{dim} W\). 1\. If \(\operatorname{dim} V \geq \operatorname{dim} W,\) then \(T\) is onto. m. If \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{k}\right)\right\\}\) is independent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) is independent. n. If \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) spans \(V,\) then \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{k}\right)\right\\}\) spans \(W\)

Let \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\\}\) be the standard basis of \(\mathbb{R}^{2}\). Is it possible to have a linear transformation \(T\) such that \(T\left(\mathbf{e}_{1}\right)\) lies in \(\mathbb{R}\) while \(T\left(\mathbf{e}_{2}\right)\) lies in \(\mathbb{R}^{2}\) ? Explain your answer.

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 \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\)
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