Short Answer

Arguing geometrically, find all eigenvectors and eigenvalues of the linear transformations in Exercises 15 through 22. In each case, find an eigenbasis if you can, and thus determine whether the given transformation is diagonalizable.
Rotation through an angle of $180^{°}$ in $R^{2}$ .

So, A is diagonalizable.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of diagonalizable

A matrix is known as diagonalizable if it can be written in the form of $A = P D P^{- 1}$ .

Step 2: Checking whether the matrix diagonalizable

Rotation through an angle of $180^{°}$ will turn any vector v into -v , so role="math" localid="1659525298008" $A = - l_{2}$ , and all vectors $v \in R^{2}$ are eigenvectors of A with the corresponding eigenvalue being $λ = - 1$ . The very fact that $A = - l_{2}$ is already diagonal proves that it's diagonalizable.

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Linear Algebra With Applications

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Short Answer

Arguing geometrically, find all eigenvectors and eigenvalues of the linear transformations in Exercises 15 through 22. In each case, find an eigenbasis if you can, and thus determine whether the given transformation is diagonalizable.
Rotation through an angle of $180^{°}$ in $R^{2}$ .

Step by Step Solution

Step 1: Definition of diagonalizable

Step 2: Checking whether the matrix diagonalizable

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Linear Algebra With Applications

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Short Answer

Arguing geometrically, find all eigenvectors and eigenvalues of the linear transformations in Exercises 15 through 22. In each case, find an eigenbasis if you can, and thus determine whether the given transformation is diagonalizable.Rotation through an angle of 180° in R2.

Step by Step Solution

Step 1: Definition of diagonalizable

Step 2: Checking whether the matrix diagonalizable

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Arguing geometrically, find all eigenvectors and eigenvalues of the linear transformations in Exercises 15 through 22. In each case, find an eigenbasis if you can, and thus determine whether the given transformation is diagonalizable.
Rotation through an angle of $180^{°}$ in $R^{2}$ .