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In each case, show that \(T\) is self-inverse, that is: \(T^{-1}=T\). a. \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4} ; T(x, y, z, w)=(x,-y,-z, w)\) b. \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} ; T(x, y)=(k y-x, y), k\) any fixed number c. \(T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n} ; T(p(x))=p(3-x)\) d. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22} ; T(X)=A X\) where \(A=\frac{1}{4}\left[\begin{array}{ll}5 & -3 \\ 3 & -5\end{array}\right]\)

Short Answer

Expert verified
All given transformations are self-inverse.

Step by step solution

01

Verify T is self-inverse for part (a)

For the transformation \(T: \mathbb{R}^4 \rightarrow \mathbb{R}^4\) defined by \(T(x, y, z, w) = (x, -y, -z, w)\), apply the transformation twice:Apply \(T\) to \((x, y, z, w) \) to get \((T(x, y, z, w)) = (x, -y, -z, w)\).Now, apply \(T\) again: \[T(T(x, y, z, w)) = T(x, -y, -z, w) = (x, -(-y), -(-z), w) = (x, y, z, w).\]Thus, \(T(T(v)) = v\), hence \(T^{-1} = T\) for part (a).
02

Verify T is self-inverse for part (b)

Consider \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) defined by \(T(x, y) = (k y - x, y)\).Apply \(T\) to \((x, y)\): - Result: \((k y - x, y)\).Apply \(T\) again on \((k y - x, y)\): \[T(k y - x, y) = (k y - (k y - x), y) = (x, y).\]Since \(T(T(v)) = v\), \(T^{-1} = T\) for part (b).
03

Verify T is self-inverse for part (c)

Consider \(T: \mathbf{P}_n \rightarrow \mathbf{P}_n\) defined by \(T(p(x)) = p(3-x)\).Apply \(T\) to a polynomial \(p(x)\) and then apply \(T\) to the result:1. \(T(p(x)) = p(3-x)\).2. \(T(T(p(x))) = T(p(3-x)) = p(3-(3-x)) = p(x)\).Thus, \(T(T(v)) = v\), indicating \(T^{-1} = T\) for part (c).
04

Verify T is self-inverse for part (d)

Given \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) defined by \(T(X) = AX\) with \[A = \frac{1}{4}\begin{bmatrix} 5 & -3 \ 3 & -5 \end{bmatrix}.\]Check if \(A^2 = I\):Calculate \[A^2 = A \cdot A = \frac{1}{4}\begin{bmatrix} 5 & -3 \ 3 & -5 \end{bmatrix} \cdot \frac{1}{4}\begin{bmatrix} 5 & -3 \ 3 & -5 \end{bmatrix} = \frac{1}{16}\begin{bmatrix} 34 & 0 \ 0 & 34 \end{bmatrix} = I.\]Hence, \(A^2 = I\), so \(T^{-1} = T\) for part (d).

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Key Concepts

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

Self-inverse matrix
A self-inverse matrix, also known as an involutory matrix, is a matrix that is its own inverse. This means that when the matrix is multiplied by itself, it yields the identity matrix. Mathematically, if matrix \(A\) is self-inverse, then \(A^2 = I\), where \(I\) is the identity matrix of the same size.
  • The concept is important because it implies certain properties of transformations in vector spaces, which can simplify computations and provide insights into the structure of linear transformations.
  • For instance, a transformation represented by a self-inverse matrix will undo itself when applied twice consecutively.
  • Understanding self-inverse matrices can aid in solving linear equations efficiently.
To verify if a matrix is self-inverse, you multiply the matrix by itself and check if the result is an identity matrix. This concept applies not only to matrices representing linear transformations but also to transformations represented in other mathematical structures, such as vectors and polynomials.
Real vector spaces
Real vector spaces are collections of vectors where vector addition and scalar multiplication are defined and satisfy certain axioms. These spaces are over the field of real numbers, and examples include spaces like \(\mathbb{R}^n\), which consists of all n-tuples of real numbers.
  • A vector space must satisfy axioms such as associativity, distributivity, and the existence of a zero vector.
  • These axioms ensure that operations within the vector space are predictable and consistent.
  • Transformations, such as those described in the exercises, map elements from one vector space to another or onto itself.
One interesting property of transformations in real vector spaces is that they can be explored through different bases, helping solve geometric and algebraic problems by changing the perspective. Real vector spaces provide a foundation for understanding linear independence, basis, dimension, and span, which are crucial for advanced studies in linear algebra.
Polynomial transformations
Polynomial transformations involve mapping polynomials from one form to another according to certain rules. In the context of the exercise, a transformation like \(T(p(x)) = p(3-x)\) is a specific polynomial transformation that modifies the input polynomial by evaluating it at a shifted argument.
  • This type of transformation can be visualized as a horizontal reflection around a vertical line, shifting the polynomial's profile in a predictable manner.
  • The transformation maintains the degree of the polynomial, indicating a structural preservation within the polynomial span.
  • Such transformations can reveal symmetries and aid in solving polynomial equations more efficiently.
Understanding polynomial transformations is essential because polynomials are fundamental in both theoretical and applied mathematics, including fields such as calculus, dynamics, and systems analysis.
Matrix operations
Matrix operations include processes such as addition, subtraction, multiplication, and taking inverses. These operations follow specific rules and are used extensively to execute transformations, solve systems of linear equations, and manage data in computational applications.
  • Matrix multiplication is not commutative, meaning that \(AB eq BA\) in general, but it is associative and distributive over addition.
  • To find the inverse of a matrix, if it exists, one must ensure the matrix is square and has full rank, meaning it is non-singular.
  • The identity matrix serves as a multiplicative neutral element, making it a key concept in understanding inverses; for a matrix \(A\), its inverse \(A^{-1}\) satisfies \(AA^{-1} = I\).
Practicing these operations is essential for developing fluency in linear algebra, which is crucial for fields ranging from computer graphics to statistical data analysis. Mastery involves computed efficiency and theoretical understanding, enabling profound insights into both concrete problems and abstract mathematics.

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Most popular questions from this chapter

In each case, show that \(T\) is an isomorphism by defining \(T^{-1}\) explicitly. a. \(T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n}\) is given by \(T[p(x)]=p(x+1)\). b. \(T: \mathbf{M}_{n n} \rightarrow \mathbf{M}_{n n}\) is given by \(T(A)=U A\) where \(U\) is invertible in \(\mathbf{M}_{n n}\)

If \(T: V \rightarrow W\) is a linear transformation, show that \(T\left(\mathbf{v}-\mathbf{v}_{1}\right)=T(\mathbf{v})-T\left(\mathbf{v}_{1}\right)\) for all \(\mathbf{v}\) and \(\mathbf{v}_{1}\) in \(V\).

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

In each case, show that \(T^{6}=1_{R^{4}}\) and so determine \(T^{-1}\). $$ \begin{array}{l} \text { a. } T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4} ; T(x, y, z, w)=(-x, z, w, y) \\ \text { b. } T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4} ; T(x, y, z, w)=(-y, x-y, z,-w) \end{array} $$

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

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