MA
Exam1SampleASolution.pdf

# 5 5 let t x 1 x 2 x 3 x 4 x 1 x 2 x 3 x 4 x 1 x 2 x 3

• Test Prep
• 6

This preview shows pages 5–6. Sign up to view the full content.

5

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

5. Let T ( x 1 , x 2 , x 3 , x 4 ) = ( x 1 + x 2 - x 3 - x 4 , x 1 - x 2 - x 3 + x 4 ). Show that T is an onto linear transformation from R 4 to R 2 . Is T one to one? Solution. Observe that T is given by the matrix transformation T ( ~ X ) = A ~ X , where A = 1 1 - 1 - 1 1 - 1 - 1 1 Thus T is obviously a linear transformation. We compute the row reduced echelon from of A . 1 1 - 1 - 1 1 - 1 - 1 1 - 1 1 1 1 - 1 - 1 0 - 2 0 2 - 1 - 2 - 2 0 2 0 0 - 2 0 2 - 1 2 - 1 2 1 0 - 1 0 0 1 0 - 1 The last entry of the above tableau is the row reduced echelon form of A . The first two columns of the row reduced echelon form from the 2 by 2 identity matrix and are thus obviously linearly independent. Therefor the first two columns of A are linearly independent. Consequently T is onto. We also see that T ( x 1 , x 2 , x 3 , x 4 ) = (0 , 0) if and only if x 1 - x 3 = 0 and x 2 - x 4 = 0. There are several solutions to this system of equations for example ( x 1 , x 2 , x 3 , x 4 ) = (0 , 0 , 0 , 0) and ( x 1 , x 2 , x 3 , x 4 ) = (1 , 1 , 1 , 1). Consequently T is not one to one. 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern