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True/False: If the statement is false, justify why it is false. (d) If X + 2 is a factor of the characteristic polynomial of A, then 2 is an eigenvalue of A. (e) In order for an n x n matrix A to be diagonalizable, A must has n distinct eigenvalues.

True/False: If the statement is false, justify why it is false. (d) If X + 2 is a factor of the characteristic polynomial of A, then 2 is an eigenvalue of A. (e) In order for an n x n matrix A to be diagonalizable, A must has n distinct eigenvalues.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**True/False: If the statement is false, justify why it is false.** (d) If \(\lambda + 2\) is a factor of the characteristic polynomial of \(A\), then 2 is an eigenvalue of \(A\). (e) In order for an \(n \times n\) matrix \(A\) to be diagonalizable, \(A\) must have \(n\) distinct eigenvalues. --- - **Explanation**: - For statement (d), if \(\lambda + 2\) is a factor of the characteristic polynomial of a matrix \(A\), it implies that \(\lambda = -2\) is an eigenvalue of \(A\), not \(\lambda = 2\). Thus, the statement is false. - For statement (e), while having \(n\) distinct eigenvalues guarantees that an \(n \times n\) matrix is diagonalizable, it is not necessary. A matrix can still be diagonalizable with repeated eigenvalues, provided there are enough linearly independent eigenvectors. Therefore, this statement is also false.
Transcribed Image Text:**True/False: If the statement is false, justify why it is false.** (d) If \(\lambda + 2\) is a factor of the characteristic polynomial of \(A\), then 2 is an eigenvalue of \(A\). (e) In order for an \(n \times n\) matrix \(A\) to be diagonalizable, \(A\) must have \(n\) distinct eigenvalues. --- - **Explanation**: - For statement (d), if \(\lambda + 2\) is a factor of the characteristic polynomial of a matrix \(A\), it implies that \(\lambda = -2\) is an eigenvalue of \(A\), not \(\lambda = 2\). Thus, the statement is false. - For statement (e), while having \(n\) distinct eigenvalues guarantees that an \(n \times n\) matrix is diagonalizable, it is not necessary. A matrix can still be diagonalizable with repeated eigenvalues, provided there are enough linearly independent eigenvectors. Therefore, this statement is also false.
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