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# nsol2 - Solutions to Review Problems for Exam Two 1 Find...

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Solutions to Review Problems for Exam Two 1. Find the dimension and a basis for the set S of symmetric 2 × 2 matrices. The basis is { 1 0 0 0 , 0 0 0 1 , 0 1 1 0 } and tThe dimension is 3. 2. Find the dimension and a basis for the set S of polynomials in P 4 p ( x ) such that p (2) = 0. The basis is { x - 2 , x 2 - 4 , x 3 - 8 } and the dimension is 3. 3. Suppose that the m × n matrices A and B are row equivalent. Which of the following are True/False: (a) A and B have the same rank (b) A and B have the same rowspace (c) A and B have the same column space (d) A and B have the same nullspace (a), (b) and (d) are true–see (19). (c) is false, by the following counterexample: A = 1 0 1 0 and B = 1 0 0 0 . 4. Suppose the vector space V has dimension n and r < n < m . Which of the following are True/False: (a) Any set of m vectors must span V . (b) Any set of m vectors must be dependent. (c) Any set of r vectors must be independent. (a) is false, by the counterexample { (1 0) T , (2 0) T , (3 0) T } (b) is true –see (21). (c) is false, by the counterexample { (1 0 0) T , (2 0 0) T } in R 3 . 5. Suppose that a 3 × 5 matrix A has rank 3. Which of (a) -(d) are True/False: (a) The rows of A linearly independent (b) The rows of A span R 5 (c) The columns of A linearly independent (d) The columns of A span R 3 (e) Find the dimension of the nullspace of A. Recall that the rank of A is the dimension of the rowspace, which equals the dimension of the column space by (19). (a) is true, since the rows are 3 vectors in the 3-dimensional rowspace. (b) is false since 3 rows cannot span a 5-dimensional space. (c) is false since 5 vectors in a 3-dimensional space must be dependent. (d) is true, since the column space is a 3-dimensional subspace of R 3 . (e) Rk ( A ) + Null ( A ) = 5, so that dim ( nullspace ( A ) = Null ( A ) = 5 - 3 = 2. 6. Let A be an m × n matrix, so that T A : F n F m . Which of the following statements is equivalent to the statement that T A is onto? (a) rk(A) = m (b) rk(A) = n (c) the columns of A span F m (d) the columns of A are independent (e) for each -→ b F m , there is at most one solution -→ x such that A -→ x = -→ b (f) for each -→ b F m , there is at least one solution -→ x such that A -→ x = -→ b (g) A has a left inverse (h) A has a right inverse By definition, T A is onto if and only if every v F m is equal to T A ( x ) = Ax for some x R n . Since A ( x 1 , x 2 , . . . , x n ) T = x 1 A 1 + · · · + x n A n , this is equivalent to saying that the columnspace of A is R m . Recall also that the rank(A) is the dimension of the column space by (22).

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(a), (c), (f) and (h) are true. T a onto ⇐⇒ (f) and (f) ⇐⇒ (c) by the discussion above. (c) ⇐⇒ (a), since the columnspace is a subspace of F m and therefore has dimension m if and only of it equals F m , which is if and only if the columns span F m . (f) (h), since we can find a right inverse by simultaneously solving Ax = e i for all i . (h) (f), since once we have AR = I , we can solve Ax = b by letting x = Rb .
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