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Ais an nxn matrix. Determine whether the statement below is true or false. Justify the answer. If Ax =x for some scalar A, then x is an eigenvector of A.

Ais an nxn matrix. Determine whether the statement below is true or false. Justify the answer. If Ax =x for some scalar A, then x is an eigenvector of A.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Eigenvectors and Eigenvalues: True or False? **Problem Statement:** Given that \( A \) is an \( n \times n \) matrix, determine whether the following statement is true or false. Justify the answer. --- **Statement:** If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \). --- **Options:** **A.** - The statement is false. The equation \( Ax = \lambda x \) is not used to determine eigenvectors. If \( \lambda x = 0 \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \). **B.** - The statement is true. If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \) because \( \lambda \) is an inverse of \( A \). **C.** - The statement is false. The condition that \( Ax = \lambda x \) for some scalar \( \lambda \) is not sufficient to determine if \( x \) is an eigenvector of \( A \). The vector \( x \) must be nonzero. **D.** - The statement is true. If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \) because the only solution to this equation is the trivial solution. --- ### Explanation: When assessing whether the given \( x \) is an eigenvector of matrix \( A \), we need to verify if it satisfies the eigenvector equation \( Ax = \lambda x \), where \( \lambda \) is a scalar (known as the eigenvalue). - **Option A** incorrectly explains the process as it denies the correct usage of the equation \( Ax = \lambda x \). - **Option B** is incorrect because it misinterprets the role of \( \lambda \) and says that \( \lambda \) is an inverse of \( A \), which is not accurate. - **Option C** provides a sound clarification by specifying that \( x \) must be non-zero. An eigenvector by definition cannot be a zero vector. - **Option D** inaccurately suggests exclusivity to the trivial solution, even though the equation itself classifies nonzero solutions as
Transcribed Image Text:### Eigenvectors and Eigenvalues: True or False? **Problem Statement:** Given that \( A \) is an \( n \times n \) matrix, determine whether the following statement is true or false. Justify the answer. --- **Statement:** If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \). --- **Options:** **A.** - The statement is false. The equation \( Ax = \lambda x \) is not used to determine eigenvectors. If \( \lambda x = 0 \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \). **B.** - The statement is true. If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \) because \( \lambda \) is an inverse of \( A \). **C.** - The statement is false. The condition that \( Ax = \lambda x \) for some scalar \( \lambda \) is not sufficient to determine if \( x \) is an eigenvector of \( A \). The vector \( x \) must be nonzero. **D.** - The statement is true. If \( Ax = \lambda x \) for some scalar \( \lambda \), then \( x \) is an eigenvector of \( A \) because the only solution to this equation is the trivial solution. --- ### Explanation: When assessing whether the given \( x \) is an eigenvector of matrix \( A \), we need to verify if it satisfies the eigenvector equation \( Ax = \lambda x \), where \( \lambda \) is a scalar (known as the eigenvalue). - **Option A** incorrectly explains the process as it denies the correct usage of the equation \( Ax = \lambda x \). - **Option B** is incorrect because it misinterprets the role of \( \lambda \) and says that \( \lambda \) is an inverse of \( A \), which is not accurate. - **Option C** provides a sound clarification by specifying that \( x \) must be non-zero. An eigenvector by definition cannot be a zero vector. - **Option D** inaccurately suggests exclusivity to the trivial solution, even though the equation itself classifies nonzero solutions as
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