linear algebra.pdf - University of Colorado at Denver...

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University of Colorado at Denver — Mathematics Department Applied Linear Algebra Preliminary Exam With Solutions 16 January 2009, 10:00 am – 2:00 pm Name: The proctor will let you read the following conditions before the exam begins, and you will have time for questions. Once the exam begins, you will have 4 hours to do your best. This is a closed book exam. Please put your name on each sheet of paper that you turn in. Exam conditions : Submit as many solutions as you can. All solutions will be graded and your final grade will be based on your six best solutions . Each problem is worth 20 points; parts of problems have equal value. Justify your solutions: cite theorems that you use, provide counter-examples for dis- proof, give explanations, and show calculations for numerical problems. If you are asked to prove a theorem, do not merely quote that theorem as your proof; instead, produce an independent proof. Begin each solution on a new page and use additional paper, if necessary. Write legibly using a dark pencil or pen. Notation: C denotes the field of complex numbers , R denotes the field of real numbers, and F denotes a field which may be either C or R . C n and R n denote the vector spaces of n -tuples of complex and real scalars, respectively. T * is the adjoint of the operator T and λ * is the complex conjugate of the scalar λ . v T and A T denote vector and matrix transposes, respectively. Ask the proctor if you have any questions. Good luck! 1. 5. 2. 6. 3. 7. 4. 8. Total
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On this exam V is a finite dimensional vector space over the field F , where either F = C , the field of complex numbers, or F = R , the field of real numbers. Also, F n denotes the vector space of column vectors with n entries from F , as usual. For T ∈ L ( V ) , the image (sometimes called the range ) of T is denoted Im ( T ) . 1. Suppose that P ∈ L ( V ) (the vector space of linear maps from V to itself) and that P 2 = P . (a) (6 points) Determine all possible eigenvalues of P . (b) (10 points) Prove that V = null ( P ) Im ( P ) . (c) (4 points) Is it necessary that all possible eigenvalues found in part (a) actually must occur? Prove that your answer is correct. Solution: P 2 - P = 0 implies that the minimal polynomial p ( x ) of P divides x 2 - x = x ( x - 1) . Hence p ( x ) = x , or ( x - 1) , or x ( x - 1) . So in general the eigenvalues are each equal to either 0 or 1. But p ( x ) = x if and only if P = 0 , in which case V = null( P ) and { 0 } = Im( P ) . And p ( x ) = x - 1 if and only if P = I . In this case V = Im( P ) and null( P ) = { 0 } . In these two cases the condition in part (b) clearly holds, and we see that part (c) is also answered. Finally, suppose p ( x ) = x ( x - 1) , so that both 0 and 1 are eigenvalues of P . If v null( P ) Im( P ) , then P ( v ) = 0 on the one hand, and on the other hand there is some w V for which v = P ( w ) = P 2 ( w ) = P ( v ) = 0 . Hence null( P ) Im( P ) = { 0 } . But also for any v V we have v = ( v - P ( v ))+ P ( v ) , where P ( v - P ( v )) = p ( v ) - P ( v ) = 0 . So v - P ( v ) null( P ) and clearly P ( v ) Im( P ) . Hence V = null( P ) Im( P ) .
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