Assignment 3 - Solutions

Assignment 3 - Solutions - MATH 223, Linear Algebra Winter,...

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MATH 223, Linear Algebra Winter, 2010 Solutions to Assignment 3 1. Let V = M 2 ( R ) be the real vector space of 2 × 2 matrices with real entries. For each of the following subsets of V , decide if it is independent, if it is a spanning set for V , and/or if it is a basis for V . Justify your answers. (a) ‰± 1 3 2 - 1 , ± 2 0 5 0 , ± 4 1 0 - 1 ¶² . Solution: This set is independent. To see this, suppose the matrices in it are A , B and C and that α 1 A + α 2 B + α 3 C = 0. Then α 1 + 2 α 2 + 4 α 3 = 0, 3 α 1 + 0 α 2 + 1 α 3 = 0, 2 α 1 + 5 α 2 + 0 α 3 = 0 and - 1 α 1 + 0 α 2 - 1 α 3 = 0. We could easily enough set this set of four equations in three unknowns as some kind of matrix (4 × 3, in fact) but here it’s easier to note that the last equation forces α 3 = - α 1 and the second forces α 3 = - 3 α 1 ; together these force α 1 = α 3 = 0 and then α 2 = 0 follows from any of the other two. This is the point — only the trivial linear combination comes out to be zero. M 2 ( R ) is 4-d.; this set has only three vectors (i.e. 2 × 2 matrices), so it cannot be a spanning set for V . It is therefore not a basis for V . (b) ‰± 1 1 1 1 , ± 1 1 - 1 - 1 , ± 1 - 1 1 - 1 , ± 1 2 3 5 ¶² . Solution: This has just the right number of elements to be a basis for V , so we need only check whether it is independent. Suppose that the matrices are A , B , C and D (in order); if aA + bB + cC + dD = 0, this implies that a + b + c + d = 0, a + b - c +2 d = 0, a - b + c +3 d = 0 and a - b - c + 5 d = 0. It is not hard to see that this system has only the trivial solution a = b = c = d = 0; if necessary row-reduce the matrix 1 1 1 1 1 1 - 1 - 1 1 - 1 1 - 1 1 2 3 5 . The set is independent and thus a basis for M 2 ( R ). (c) ‰± 1 0 0 0 , ± 1 1 0 0 , ± 1 1 1 0 , ± 1 1 1 1 , ± 0 1 1 1 ¶² . Solution: Because there are five matrices here, the set cannot be independent, and so it isn’t a basis for M 2 ( R ). It is, however, a spanning set for V . Suppose that A = ± a b c d is any vector in V ; it is not hard to see that (for instance) ( a - b ) ± 1 0 0 0 +( b - c ) ± 1 1 0 0 +( c - d ) ± 1 1 1 0 + d ± 1 1 1 1 = 1
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A . (Because of the redundancy, there are infinitely many other ways to express A as a linear combo of the five given matrices.) 2. Find a basis for each of the null space, row space and column space of the
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This note was uploaded on 02/10/2010 for the course MATH math 223 taught by Professor Loveys during the Winter '10 term at McGill.

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Assignment 3 - Solutions - MATH 223, Linear Algebra Winter,...

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