11_11_concept_questions - v 1 6 = v 2 that share the same...

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Concept Questions: 11-11-2011 MA2940: Linear Algebra for Engineers Diagonalization 1) (T/F) If λ = 0 is an eigenvalue of a square matrix A , then A is not diagonalizable. 2) (T/F) If 16 is an eigenvalue of A 4 , then 2 is an eigenvalue of A . 3) (T/F) Suppose n × n matrices A and B can be diagonalized with the same eigenvector matrix S . Then AB = BA . 4) (T/F) Suppose n × n matrices A and B can be diagonalized, and AB = BA . Then they can be diagonalized with the same matrix S of eigenvectors. Applications to Differential Equations 5) (T/F) It is possible to have eigenvalues λ 1 6 = λ 2 that share the same eigenvector. 6) (T/F) It is possible to have eigenvectors
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Unformatted text preview: v 1 6 = v 2 that share the same eigenvalue. 7) (T/F) Any matrix system d dt u ( t ) = Au ( t ) an be written as a single higher-order dieren-tial equation. 8) (T/F) If w ( t ) and v ( t ) are solutions to d dt u ( t ) = Au ( t ) + b , then w + v is a solution too. 9) (T/F) If a square matrix A is unstable, then tr ( A ) > 0 and det ( A ) < 0. 10) When solving d dt u ( t ) = Au ( t ), does it matter if A isnt diagonalizable? (Yes.) Why doesnt our method work? What do we do? TA: Kyle Wilson...
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