problemset4solutions

# problemset4solutions - MAS 213 Linear Algebra II Problem...

This preview shows pages 1–4. Sign up to view the full content.

MAS 213: Linear Algebra II. Problem list for Week #4. Solutions . Problem 1: (Problems 2.1.4 in [FIS].) Consider the linear transformation T : M 2 × 3 ( F ) M 2 × 2 ( F ) determined by the formula T a 11 a 12 a 13 a 21 a 22 a 23 = 2 a 11 - a 12 a 13 + 2 a 12 0 0 . (i) Find a basis for N ( T ), the null space of T . (ii) Find a basis for R ( T ), the range of T . (iii) Check that the conclusions of the rank-nullity theorem holds. Solution to problem 1, part 1. The null space is the set of matrices a 11 a 12 a 13 a 21 a 22 a 23 whose entries satisfy the two equations 2 a 11 - a 12 = 0 , a 13 + 2 a 12 = 0 . This is a system of two linear equations in six variables. As an augmented matrix the system can be written as follows, and solved: 2 - 1 0 0 0 0 0 0 2 1 0 0 0 0 " 1 0 1 4 0 0 0 0 0 1 1 2 0 0 0 0 # (Recall that in this course we are assuming that F means Q , R or C . If we were allowing more general fields, we would need to find a way to avoid the two fractions appearing above. See below for an alternative approach.) So the null space has 4 independent variables, and 2 dependent variables. It is: N ( T ) = - r 4 - r 2 r s t u : r, s, t, u F . To see a basis for this space, we can rewrite it as follows: N ( T ) = r - 1 4 - 1 2 1 0 0 0 + s 0 0 0 1 0 0 + t 0 0 0 0 1 0 + u 0 0 0 0 0 1 : r, s, t, u F . It is clear at this point that a basis of N ( T ) is given by - 1 4 - 1 2 1 0 0 0 , 0 0 0 1 0 0 , 0 0 0 0 1 0 , 0 0 0 0 0 1 . 1

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

View Full Document
2 To be precise, because: (1) Any element of N ( T ) is a linear combination of these four matrices. (2) And these four matrices are obviously linearly independent. Then it follows that they are a basis for N ( T ). Next we’ll determine a basis for the range R ( T ) of this linear operator. The technique for doing this starts with a basis for the domain M 2 × 3 ( F ): 1 0 0 0 0 0 , 0 1 0 0 0 0 , 0 0 1 0 0 0 , 0 0 0 1 0 0 , 0 0 0 0 1 0 , 0 0 0 0 0 1 . By the general theory, we know that the set of images of these elements is a spanning set for the range of T . So to find a basis of R ( T ) we take the set of images of these elements and then reduce it to a basis using the standard algorithm. The images are: 2 0 0 0 , - 1 2 0 0 , 0 1 0 0 , 0 0 0 0 , 0 0 0 0 , 0 0 0 0 . Applying the standard reduction algorithm to this list leaves us with the following basis for R ( T ): 2 0 0 0 , - 1 2 0 0 . Now let’s check that the conclusion of the rank-nullity theorem holds, as expected. As expected, the dimension of the domain V (which is 6) equals the dimension of the null space of T (which is 4) plus the dimension of the range of T (which is 2). (Comment for the interested: You can solve for the null space without introducing fractions by: 2 - 1 0 0 0 0 0 0 2 1 0 0 0 0 - 2 1 0 0 0 0 0 4 0 1 0 0 0 0 And so on.)
3 Problem 2: (Problem 2.1.10 in [FIS]) Suppose that T : R 2 R 2 is a linear transformation with the properties that T (1 , 0) = (1 , 4) , T (1 , 1) = (2 , 5) . What is T (2 , 3)? Is T one-to-one? Solution to Problem 2: We are told that this function T is a linear transformation. So to determine its value on (2 , 3), we need to express (2 , 3) as a linear combination of the vectors whose values we know: (2 , 3) = ( - 1) * (1 , 0) + 3 * (1 , 1) . It follows that: T (2 , 3) = T (( - 1) * (1 , 0) + 3 * (1 , 1)) = T (( - 1) * (1 , 0)) + T ((3) * (1 , 1)) (Because T respects addition.) = ( - 1) T (1 , 0) + (3) T (1 , 1) (Because T respects scalar multiplication.) = ( - 1) * (1 , 4) + 3 * (2 , 5) (These values are given to us.) = (5 , 11) .

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

View Full Document
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