Conversion of BCD (8421) numbers 101 0101 to a decimal numbers.

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Chapter 86, Problem 17A

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Conversion of BCD (8421) numbers 101 0101 to a decimal numbers.

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Each answer must be justified and all your work should appear. You will be marked on the quality of your explanations. You can discuss the problems with classmates, but you should write your solutions sepa- rately (meaning that you cannot copy the same solution from a joint blackboard, for exam- ple). Your work should be submitted on Moodle, before February 7 at 5 pm. 1. True or false: (a) if E is a subspace of V, then dim(E) + dim(E) = dim(V) (b) Let {i, n} be a basis of the vector space V, where v₁,..., Un are all eigen- vectors for both the matrix A and the matrix B. Then, any eigenvector of A is an eigenvector of B. Justify. 2. Apply Gram-Schmidt orthogonalization to the system of vectors {(1,2,-2), (1, −1, 4), (2, 1, 1)}. 3. Suppose P is the orthogonal projection onto a subspace E, and Q is the orthogonal projection onto the orthogonal complement E. (a) The combinations of projections P+Q and PQ correspond to well-known oper- ators. What are they? Justify your answer. (b) Show…

1. True or false: (a) if E is a subspace of V, then dim(E) + dim(E+) = dim(V) (b) Let {i, n} be a basis of the vector space V, where vi,..., are all eigen- vectors for both the matrix A and the matrix B. Then, any eigenvector of A is an eigenvector of B. Justify. 2. Apply Gram-Schmidt orthogonalization to the system of vectors {(1, 2, -2), (1, −1, 4), (2, 1, 1)}. 3. Suppose P is the orthogonal projection onto a subspace E, and Q is the orthogonal projection onto the orthogonal complement E. (a) The combinations of projections P+Q and PQ correspond to well-known oper- ators. What are they? Justify your answer. (b) Show that P - Q is its own inverse. 4. Show that the Frobenius product on n x n-matrices, (A, B) = = Tr(B*A), is an inner product, where B* denotes the Hermitian adjoint of B. 5. Show that if A and B are two n x n-matrices for which {1,..., n} is a basis of eigen- vectors (for both A and B), then AB = BA. Remark: It is also true that if AB = BA, then there exists a common…