Let A be an nxn matrix. Determine whether the statement below is true or false. Justify the answer. If A is diagonalizable, then A is invertible. ..... Choose the correct answer below. O A. The statement is false. If A is diagonalizable, then det(A – Al) = 0 has a solution. Thus, A is not invertible. O B. The statement is true. If A is diagonalizable, then det(A) does not equal 0. Thus, A is invertible. O C. The statement is true. If A is invertible, then it has n distinct eigenvectors that form a basis of R". D. The statement is false. Invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.
Let A be an nxn matrix. Determine whether the statement below is true or false. Justify the answer. If A is diagonalizable, then A is invertible. ..... Choose the correct answer below. O A. The statement is false. If A is diagonalizable, then det(A – Al) = 0 has a solution. Thus, A is not invertible. O B. The statement is true. If A is diagonalizable, then det(A) does not equal 0. Thus, A is invertible. O C. The statement is true. If A is invertible, then it has n distinct eigenvectors that form a basis of R". D. The statement is false. Invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Let A be an n×n matrix. Determine whether the statement below is true or false. Justify the answer.
If A is diagonalizable, then A is invertible.
Choose the correct answer below.
A. The statement is false. If A is diagonalizable, then det(A - Al) = 0 has a solution. Thus, A is not invertible.
B. The statement is true. If A is diagonalizable, then det(A) does not equal 0. Thus, A is invertible.
C. The statement is true. If A is invertible, then it has n distinct eigenvectors that form a basis of Rn.
D. The statement is false. Invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.
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