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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
- 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).
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern