University:
Whitman College
Course:
MATH-240 | Linear Algebra
Academic year:

2024
Views:

360

Pages:

1
Author:

Efrain Villa

HW Solutions, 2.1 2.1, 22 Show that if the columns of B are linearly dependent, then so are the columns of AB. SOLUTION 1: Let B = [b1 , . . . bk ]. If the columns of B are linearly dependent, then there is a set of constants c1 , . . . , ck , not all zero, so that c1 b1 + c2 b2 + · · · + ck bk = 0 We note that AB is formed as the matrix: AB = A[b1 , . . . bk ] = [Ab1 , Ab2 , . . . Abk ] so that these are the columns of AB. Now, if we multiply the previous equation by A, we must have constants c1 , . . . , ck , not all zero, so that: c1 Ab1 + c2 Ab2 + · · · + ck Abk = 0 Therefore, the columns of AB are linearly dependent as well. And, as we noted in class, this implies that the dependence relation in the columns of B are exactly the same as in AB. Alternative Solution: We could use a theorem about linear dependence instead of the definition. In that case, if the columns of B are linearly dependent, there is a nontrivial solution x so that Bx = 0. Multiply both side by A, and we see that x is a nontrivial solution to: (AB)x = 0 so the columns of AB are linearly dependent. 2.1, 24 Suppose that AD = Im . Show that for any b ∈ IRm , the equation Ax = b has a solution. SOLUTION: We show that x = Db: Ax = A(Db) = (AD)b = Ib = b 1

Linear Algebra Solutions 2.1

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