# F10HW10 - M341 Fall 2010 Homework 10 Section 4.4 Problem 10...

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M341, Fall 2010, Homework 10. Section 4.4 Problem 10. Since det A =det A T , we may consider the determinant of the matrix with columns u , v , w . These 3 vectors are linearly independent if and only if the three columns of this matrix are pivot columns which happens if and only if there are 3 pivots in the matrix or if and only if the matrix is invertible or if and only if the determinant of the matrix is ̸ = 0. Problem 17. The premise means that in f ( x ) there are two monomials, say a k x k and a x , with k ̸ = and a k ̸ = 0, a ̸ = 0. Then if we had a dependence relation between f ( x ) and x f ( x ), say u f ( x ) + vx f ( x ) = 0 , we should have u ( a k x k ) + v ( xka k x k 1 ) = 0 and u ( a x ) + v ( xℓa x 1 ) = 0 . The homogeneous system: { ua k + vka k = 0 ua + vℓa = 0 should have non trivial solutions in u and v . Since a k ̸ = 0, a ̸ = 0, this system is equivalent to the system: { u + vk = 0 u + vℓ = 0 the determinant of which is k ̸ = 0 and there are no non trivial solutions. Problem 19. a) T = { A v 1 , . . . , A v k } is linearly independent S = { v 1 , . . . , v k } is linearly inde- pendent. (otherwise:Premise: T = { A v 1 , . . . , A v k } is linearly independent. Conclusion: S = { v 1 , . . . , v k } is linearly independent.) Assume T is linearly independent. By contradiction. If S was not linearly independent there would be real numbers a 1 , . . . , a k not all zeroes, such that a 1 v 1 + . . . + a k v k = 0 . When A acts on each member of this equality, by the properties of the matrix multi- plication (distributivity, A ( c v ) = cA ( v ), A 0 = 0 ), we get: a 1 A v 1 + . . . + a k A v k = 0 , and this would be a dependence relation in T . But this is not possible since T is assumed to be linearly independent. b) Converse: S is linearly independent T is linearly independent. For a counter example don’t try to be smart, just make it simple and take A = O , the zero matrix. (Actually any non invertible matrix would give a counter example.) c) If A is square and non singular (invertible), the converse is true. If A is invertible, A 1 exists and if we put T = { w 1 , . . . , w k } , we have S = { A 1 w 1 , . . . , A 1 w k } and we may use a) in this new context. Section 4.5 Preliminary. In an n -dimensional vector space, for a set S of vectors any 2 of the 3 following properties yield the third one: - S # = n ( S contains n vectors). - S is linearly independent. 1

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- S is a spanning set. Nevertheless, in Problems 1 and 2, the wording of the question brings me to think that we better answer the question by showing the 2nd and 3rd properties (even though the 1st one is clearly satisfied).
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