Problem 27 Finding a Basis In Exercises \(2... [FREE SOLUTION]

Chapter 7: Problem 27

Finding a Basis In Exercises $27-30,$ find a basis $B$ for the domain of $T$ such that the matrix for $T$ relative to $B$ is diagonal. $$ T: R^{2} \rightarrow R^{2}: T(x, y)=(x+y, x+y) $$

Short Answer

Expert verified

Since transformation T collapses all vectors, the best achievable is a basis for the image of T, which is B = {(-1,1)}

Step by step solution

Identify the linear map

The linear map $T: R^{2} -> R^{2}$ is defined as $T(x, y)=(x+y, x+y)$. It means for each input (x, y), it returns a vector (x+y, x+y)

Calculating Eigenvalues

The standard matrix for linear transformation $T$ is $ A = \[ \left[ \begin{array}{cc} 1 & 1\ 1 & 1 \end{array} \right] \]$. An operator is diagonalizable if it has enough eigenvalues. To find the eigenvalues, you can solve the characteristic equation, which is |A - 位I| = 0, where I is the identity matrix and 位 are the eigenvalues. In this case, A - 位I yields: \[ \left[ \begin{array}{cc} 1-位 & 1\ 1 & 1-位 \end{array} \right] \]Setting the determinant of this equal to zero gives you (1-位)^2 - 1 = 0. Solving for 位 gives you two eigenvalues: 位1 = 0 and 位2 = 2.

Calculate the Eigenvectors

Find eigenvectors corresponding to each eigenvalue by plugging the eigenvalues back in (A - 位I)v = 0. This gives the following system of equations:For 位1 = 0, you get:\[ \left[ \begin{array}{cc} 1 & 1\ 1 & 1 \end{array} \right] \] * \[v1, v2] = 0This leads to the eigenvector: (-1, 1)For 位2 = 2, you get:\[ \left[ \begin{array}{cc} -1 & 1\ 1 & -1 \end{array} \right] \] * \[v1, v2] = 0This also leads to the eigenvector: (-1, 1)Note that both eigenvalues lead to the same eigenvector.

Form the basis

The basis is formed from the eigenvectors. Since we only have one unique eigenvector (-1,1), one vector alone cannot form the basis for R^2. Since transformation T collapses all vectors to a straight line, the best we can achieve is a basis for the image of T, which in this case can be given as B = {(-1,1)}.

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Key Concepts

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

Eigenvalues

In the context of linear transformations, eigenvalues are special scalars associated with a square matrix. They reveal insights about the transformation's nature. Consider a linear map like our function here:

An eigenvalue $ \lambda $ satisfies the equation $ Av = \lambda v $, where $ A $ is the matrix of the transformation and $ v $ is a non-zero vector, known as an eigenvector.
Finding eigenvalues involves solving the characteristic equation $ |A - \lambda I| = 0 $. Here, $ I $ stands for the identity matrix.
For our linear map $ T $, we find $ \lambda_1 = 0 $ and $ \lambda_2 = 2 $

Eigenvalues help us understand the stretching factors along their corresponding eigenvectors. With $ T(x, y) = (x+y, x+y) $, these eigenvalues indicate how the transformation modifies the input vectors' magnitudes and directions.

Eigenvectors

Eigenvectors are vital components when understanding linear transformations. These vectors reveal the directions along which a transformation happens consistently:

For eigenvalue $ \lambda $, the eigenvector $ v $ satisfies: $ (A - \lambda I)v = 0 $.
In our example, both eigenvalues $ 0 $ and $ 2 $ lead to the same eigenvector $(-1,1)$.
An eigenvector remains unchanged in its direction after the transformation, only its length may change, depending on the eigenvalue.

Eigenvectors define invariant lines through the origin, which are lines that are scaled but not altered in direction by the transformation. This makes them essential for transforming and understanding the matrix.

Diagonalization

Diagonalization is a process that simplifies the matrix form by converting it into a diagonal matrix. This form helps in understanding and computing powers of matrices effectively:

A matrix is diagonalizable if it has enough independent eigenvectors to form a basis.
For a 2x2 matrix, like in our problem, having two linearly independent eigenvectors would suffice.
Unfortunately, the transformation $ T $ offers only one unique eigenvector, which doesn鈥檛 allow for diagonalization in the usual sense.

Understanding diagonalization not only simplifies operations like matrix powers but also gives insights into the geometry and behavior of transformations in a more digestible form.

Linear Map

A linear map is a function between two vector spaces that respects the operations of vector addition and scalar multiplication. For instance:

Our example, $ T(x, y) = (x+y, x+y) $, is a simple linear map from $ \mathbb{R}^2 $ to $ \mathbb{R}^2 $.
A linear map is characterized by its matrix representation, making eigenvectors and eigenvalues central to understanding its effects.
Linear maps can be used to describe systems, transformations, and projections in various mathematical and applied contexts.

Grasping how a linear map functions open doors to deeper mathematical comprehension and applications in areas like computer graphics, physics, and engineering.

91影视

Finding a Basis In Exercises \(27-30,\) find a basis \(B\) for the domain of \(T\) such that the matrix for \(T\) relative to \(B\) is diagonal. $$ T: R^{2} \rightarrow R^{2}: T(x, y)=(x+y, x+y) $$

Short Answer

Step by step solution

Identify the linear map

Calculating Eigenvalues

Calculate the Eigenvectors

Form the basis

Key Concepts

Eigenvalues

Eigenvectors

Diagonalization

Linear Map

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