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Unformatted text preview: 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 . (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 ranknullity 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 2 1 0 0 0 → " 1 0 1 4 0 0 0 0 1 1 2 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 + 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 1 0 0 , 0 0 0 0 1 0 , 0 0 0 0 0 1 . 1 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 ranknullity 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 2 1 0 0 0 → 2 1 0 0 0 0 4 0 1 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) ....
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 Fall '10
 ANDREWKRICKER

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