Linear Algebra with Applications (3rd Edition)

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

110.201 Quiz 3 Solutions: Friday March 28, 2005 Problem 1 1. Yes, the map is linear. It is not an isomorphism; its kernel contains, for example, f ( x ) = x , which means the map is not one-to-one 2. The map is linear; this much is obvious. It is also clearly injective, and since an injective linear map T : U V between linear spaces of the same finite dimension is an isomorphism, the map is an isomorphism. 3. The map is not linear. T ( kA ) = k 2 T ( A ) for all real k . Problem 2 1. There are several ways to solve this problem. One is the following: Let B 2 = { x 4 , 2 x 3 - 1 , 1 - x 2 , 3 x - 1 , 2 x } ≡ { f 1 , f 2 , f 3 , f 4 , f 5 } . Observe that 1 = (3 / 2) f 5 - f 4 (1) x = f 5 / 2 (2) x 2 = (3 / 2) f 5 - f 4 - f 3 (3) x 3 = f 2 / 2 (4) x 4 = f 1 (5) Therefore span { f 1 , f 2 , . . . , f 5 } = span { 1 , x, x 2 , x 3 , x 4 } = P 4 . It follows immediately that the f i are a basis of P 4 , since there are only five of them and dim P 4 = 5. The change-of-basis matrix from B 1 to B 2 can be found immediately by writing the elements of B 2 in terms of the standard basis: 1
Image of page 1

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

f 1 f 2 f 3 f 4 f 5
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