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Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Practice midterm 3 Name: Solutions • This is a practice exam. The real exam will consist of 4 problems. • In the real exam, no calculators, electronic devices, books, or notes may be used. • Show your work. No credit for answers without justification. • Good luck! 1. /15 2. /15 3. /15 4. /10 5. /10 6. /10 Total: /75 1 Section 4.2 Notation: Given an n × n matrix A , let us denote by μ alg ( λ ) and μ geo ( λ ) the algebraic and geometric multiplicities, respectively, of an eigenvalue λ of A . In fact, both notions make sense for any scalar λ : both are zero when λ is not an eigenvalue of A . Problem 1a. (5 pts) Let A be an n × n matrix. Prove the inequality rank A ≥ X λ 6 =0 μ alg ( λ ) . (1) rank A = n dim Null A by the ranknullity theorem = n μ geo (0) = X λ μ alg ( λ ) μ geo (0) = X λ 6 =0 μ alg ( λ ) + μ alg (0) μ geo (0) ≥ X λ 6 =0 μ alg ( λ ) since μ alg (0) ≥ μ geo (0) . b. (5 pts) When A is diagonalizable, prove that (1) is in fact an equality. If A is diagonalizable, then we have μ alg (0) = μ geo (0), which gives rank A = X λ 6 =0 μ alg ( λ ) + μ alg (0) μ geo (0) = X λ 6 =0 μ alg ( λ ) . 2 c. (5 pts) When A is nondiagonalizable, prove that we cannot conclude (in general) whether (1) is an equality or a strict inequality. In other words, provide an example of nondiagonalizable matrix A such that (1) is an equality and an example of nondiagonalizable matrix B such that (1) is a strict inequality. Consider the nondiagonalizable matrix A = 7 1 0 7 . It satisfies rank A = 2 = μ alg (7) = X λ 6 =0 μ alg ( λ ) . Consider the nondiagonalizable matrix B = 0 1 0 0 . It satisfies rank B = 1 > 0 = X λ 6 =0 μ alg ( λ ) . 3 Section 5.1 Problem 2. Consider the complex vector space C n × n of complex n × n matrices. One of the following two formulas defines a complex inner product on C n × n : 1. ( A,B ) = tr( A B ) 2. (2....
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 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra

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