Exam2-solution - Exam 2 — Version 2 — Solutions Math 340L Spring 2007 Purcell 1(8 points total Let A be the matrix 4 1 2 1 6 0 2 3 3 1 0 0 1 4

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Unformatted text preview: Exam 2 — Version 2 — Solutions Math 340L, Spring 2007, Purcell 1. (8 points total) Let A be the matrix: 4 1 2- 1 6 0 2 3- 3- 1 0 0- 1 4 0 0 0 0 6 (a) (2 points) What is the rank of A ? Explain. Solution: The rank is 4. This is because A has 4 pivot columns. (If this isn’t obvious, swap the last two rows to put A into echelon form. Now count: 4 pivots.) The number of pivot columns is the dimension of the column space, which is the rank. (b) (2 points) Is A diagonalizable? Why or why not? Solution: Yes A is diagonalizable. It has 5 distinct eigenvalues, which means it has 5 linearly independent eigenvectors. (c) (4 points) Find a basis for Col( A ). Solution: A basis consists of columns corresponding to pivots. These are the first, second, third, and fifth columns. 4 , 1 2 , 2 3- 1 , 6- 1 4 6 1 2. (9 points total, 3 points each) (a) If the null space of a 4 × 6 matrix A is 3–dimensional, what is the dimension of the row space of A ? Why? Solution: The row space is 3-dimensional. Dimension of the row space is the dimension of the column space is the rank of the matrix. Rank plus dimension of the null space equals the number of columns = 6. So dimension of the row space is 6- 3 = 3. (b) If A is a 5 × 7 matrix, what is the largest possible dimension of the column space of A ? Why? Solution: The column space has dimension equal to the number of pivots. Since A can’t have more pivots than it has rows, the largest dimension is 5. (c) If A is a 2 × 4 matrix, then the null space of A is a subspace of R n . What is n ?...
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This note was uploaded on 03/20/2008 for the course M 340L taught by Professor Pavlovic during the Spring '08 term at University of Texas at Austin.

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Exam2-solution - Exam 2 — Version 2 — Solutions Math 340L Spring 2007 Purcell 1(8 points total Let A be the matrix 4 1 2 1 6 0 2 3 3 1 0 0 1 4

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