Unformatted text preview: Answer Key to Final Examination, Version 1, Math. 33A, Spring Quarter, 2003 1. Suppose matrix A is similar to B. Prove the following facts (missing step by step description of your reasoning costs maximum a half of the allocated points): (a) There exists an isomorphism from the kernel of A onto the kernel of B (50 points). Since A is similar to B, we have an invertible matrix S such that B SAS 1 BS SA S 1B AS 1. Then if x is in the kernel of A, then BS x SA x S 0 0 . This shows S x is in the kernel of B. Thus we have a linear transformation T : Ker A Ker B given by T x S x . Since S is invertible, Ker T Ker S 0 ; so, Ker T 0 . S 1B y S 10 0 . Thus S 1 y is in the If y is in the kernel of B, AS 1 y kernel of A. In particular y SS 1 y is in the image of T . This shows that Im T Ker B . The two facts Ker T 0 and Im T Ker B combined tell us that T is an isomorphism, and T 1 is given by the multiplication by S 1 . By (a), Ker A and Ker B has the same dimension, which are the nullity of A and B, respectively. Since rank A
21 30 12 2. Compute the following numbers and vectors. a. The matrix T B (50 points).
j b. The largest eigenvalue of A (25 points). By the above computation, the largest eigenvalue is 3. c. The rank of A (25 points). Since det T B 6 0, the matrix is invertible, and the rank is 3. 1 " " By computation, A v j v j for j 1 2 3. Thus T B ! # ! 1 2 three vectors v , v and v 3 2 3 2 4 6 3 . 100 020 003 . ! Let A 10 14 6 4 6 1 1 2 0 and T x A x for x in 3. Let B be a basis of n nullity A n nullity B rank B . (c) rank A rank B (30 points). (b) nullity A nullity B (20 points). 3 made of 2 3. Label the following statements as being true or false. Statements Label The orthogonal projection to a 3 dimensional subspace of 5 has nullity 3. F If A is a n n matrix with eigenvalue 1, F then the geometric multiplicity of eigenvalue 1 is n rank A . If A and B are n n matrices, if a is an eigenvalue of A and F if b is an eigenvalue of B, then ab must be an eigenvalue of AB. There is a 3 3 matrix A such that Ker A Im A . F Even if matrix A is similar to matrix B, F rank A could be different from rank B . For any given pair of vectors v 1 v 2 and u 1 u 2 in 2 , there exists a linear transformation T F such that T v 1 u 1 and T v 2 u 2. Even if a column of a n n matrix A is F a linear combination of other columns, det A can be nonzero. The algebraic multiplicity of an eigenvalue F cannot exceeds its geometric multiplicity. 1 is not diagonalizable, Even if matrix A F the matrix A could be diagonalizable. 2 . If A is any n n matrix, then Ker A Ker A F The matrix of any orthogonal projection onto a subspace T V of n is diagonalizable. n. If an n n matrix A has rank n, we have Im A T If Ker T 0 for a linear transformation T : V V T for a finite dimensional linear space V , T is invertible. If 0 is an eigenvalue of a matrix A, det A 0. T If a triangular matrix has a diagonal entry 1, T the geometric multiplicity of 1 is equal to 1 or larger. If a matrix B is obtained from A by row operations, rank A rank B . T If a matrix B is obtained from A by row operations, T there images could be different subspaces. For any n n matrix A, nullityA5 nullityA9 . T Two similar matrices has the same trace. T If u v 0 for two nonzero vectors u and v in n , T they span two dimensional subspace of n . 4. Which of the following four sets is a field? Explain briefly your reason. a : a in with a 0 No : a and b are in No ab 0a 01 00 does not have inverse. 1 is not in the set. # a b b a : a and b are in Subset The rational numbers . yes or no reason Yes One can add, substract, multiply and divide in . By computation, the set is closed under addition, substraction and multiplication, Yes and one can do division if ab 0 " " 3 5. Answer the following questions: We need to find when the rank of the matrix is less than 3. By row operation we get From this, the matrix is invertible if either k 1 or k 2. vectors two vectors of 11 1 12 k 1 4 k2 are linearly independent. Since k 1 or 2, the rref of the above matrix is 102 k 01k 1 00 0 (c) When 11 1 1 2 k is 1 4 k2 11 1 of 1 2 k2 14k not invertible, find all linearly independent vectors among column (Each redundant vector given costs 10 points). , which shows the first By the above row operation, we find that the determinant is k 1 k (b) Compute the determinant of 11 1 12 k 1 4 k2 . 2. 2nd 1st, 3rd 2nd 11 1 12 k 1 4 k2 11 1 01 k 1 02k k 1 1st 2nd, 3rd 2 2nd 10 2 k 01 k 1 00 k 2 k 1 (a) Find all values of k when 11 1 12 k 1 4 k2 is not invertible. ...
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 Math, Linear Algebra, Algebra, Ker B, Ker A, Ker A Ker, Ker T Ker

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