{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math110s-hw1sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 1 SOLUTIONS 1. Prove that the following are vector spaces over R : (a) polynomials of degree not more than d , P d = { a 0 + a 1 x + · · · + a d x d | a i R for all i } , (b) m -by- n matrices R m × n = { [ a ij ] m,n i,j =1 | a ij R for all i, j } . The addition and scalar multiplication operations for polynomials and matrices are as defined in the lectures. Solution. Routine. 2. Let V be a vector space over R with addition and scalar multiplication denoted by + and · respectively. Let W = V × V = { ( v 1 , v 2 ) | v 1 , v 2 V } . Prove that W is a vector space over C with addition defined by ( u 1 , u 2 ) ( v 1 , v 2 ) = ( u 1 + v 1 , u 2 + v 2 ) for all ( u 1 , u 2 ) , ( v 1 , v 2 ) W and scalar multiplication defined by ( a + bi ) ( v 1 , v 2 ) = ( a · v 1 - b · v 2 , b · v 1 + a · v 2 ) for all a + bi C and ( v 1 , v 2 ) W . Here i = - 1 and a, b R . Solution. Routine. 3. Which of the following are subspaces of R 2 ? Justify your answers. (a) U a = { ( x, y ) R 2 | x 2 + y 2 = 0 , x, y R } , (b) U b = { ( x, y ) R 2 | x 2 - y 2 = 0 , x, y R } , (c) U c = { ( x, y ) R 2 | x 2 - y = 0 , x, y R } , (d) U d = { ( x, y ) R 2 | x - y = 0 , x, y R } , (e) U e = { ( x, y ) R 2 | x - y = 1 , x, y R } . If we replace R by C and R 2 by C 2 above, will any of your answers change? Solution. Note that U a = { (0 , 0) } and so is a subspace. Let α, β R . U d is a subspace by Theorem 1.8 : if x 1 - y 1 = 0 and x 2 - y 2 = 0, then ( αx 1 + βx 2 ) - ( αy 1 + βy 2 ) = α ( x 1 - y 1 ) + β ( x 2 - y 2 ) = 0. (0 ,
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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