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# Tutorial Problem Set #2 Solutions

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Department
Mathematics
Course
MAT223H1
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Fall

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Department of Mathematics, University of Toronto MAT223H1F - Linear Algebra I Fall 2013 Tutorial Problems 2 2 3 2 3 2 3 ( 2 0 c ) 6 1 7 6▯3 7 6c + 2 1. Find all values of c for which the set S = 7;6 7 is linearly dependent. 4▯2 5 4▯2 5 4 0 5 1 1 0 Solution: The set S is linearly dependent if and only if the system of linear equations 2 3 2 3 2 3 2 3 2 0 c 0 6 17 6▯3 7 6 c + 2 607 x16 7 + x26 7+ x36 7= 6 7 4▯25 4▯2 5 4 0 5 405 1 1 0 0 has a nontrivial solution (i.e. a solution where not all of x ;x and x are zero). This system gives rise to 1 2 3 the augmented matrix 2 3 2 0 c 0 6 1 ▯3 c + 207 6 7 : 4▯2 ▯2 0 05 1 1 0 0 Let’s row reduce it: 2 3 2 3 2 3 2 0 c 0 0 ▯2 c 0 1 1 0 0 6 1 ▯3 c + 2 07 6 0 ▯4 c + 207 60 ▯2 c 0 6 7 ▯! 6 7 ▯! 6 7 : 4▯2 ▯2 0 05 4 0 0 0 05 40 0 2 ▯ c 0 1 1 0 0 1 1 0 0 0 0 0 0 Now note that if c 6= 2 the system will have three leading entries, and therefore a unique solution (namely x1= x 2 x =30). While if c = 2 the system will have a free parameter (n3mely x ), and therefore a nontrivial solution (in▯nitely many, in fact). So we conclude that S is linearly dependent only when c = 2. 2 3 2 3 2 3 2 3 ( 0 6 ) ( 2 2 ) 6▯2 7 6 7 7 6 37 6 17 2. Let W = Span 4 5 4 5 , and S = 4 5 4 5 . Show that W = SpanS. ▯1 ▯1 0 ▯1 3 0 ▯1 2 Solution: To simplify notation, let’s1write 2 and w for the two given vectors tha1 span2W and s and s for the vectors in S. To show that W = SpanS, we have to show that W ▯ SpanS and that SpanS ▯ W. Let’s begin by showing that W ▯ SpanS. What we must do is show that every w 2 W can be written as a linear combinatio1 of s2and s . In fact, it’s enough to s1ow th2s for w and w . Indeed, suppose we know that w1= as1+ bs2 w2= cs1+ ds2; 1 of 4 where a;b;c;d 2 R. Then, given any w 2 W, we can write it 1s w =2ew +fw for some e;f 2 R, and then w = e(a1 + b2 ) + f(1s + 2s ) = (ea +1fc)s + (eb 2 fd)s : This shows that w is in SpanS. Now, to show that1w and 2 are in SpanS, we must show that the two systems x1 1+ x2 2= w 1 and x1 1+ x2 2= w 2 have solutions. These systems give rise to the augmented matrices 2 3 2 3 2 2 0 2 2 6 6 3 1 ▯27 6 3 1 7 7 6 7 and 6 7: 4 0 ▯1 ▯15 4 0 ▯1 ▯1 5 ▯1 2 3 ▯1 2 0 After row reduction (steps omitted), these matrices become 2 3 2 3 1 ▯2 ▯3 1 ▯2 0 60 1 1 7 6 0 1 17 6 7 and 6 7: 40 0 0 5 4 0 0 05 0 0 0 0 0 0 2 3 2 2 0 6 6 3 1 ▯2 77 (Note: To save some time, you can row reduce the singl4 matrix 5 :) 0 ▯1 ▯1 ▯1 ▯1 2 3 0 So both systems have solutions. This shows1th2t w ;w 2 SpanS, and therefore W ▯ SpanS. Next, we must show that SpanS ▯ W. Just as above, it will be enough to s1o2 that s ;s 2 W. This amounts to considering the systems given by 2 0 6 2 3 2 0 6 2 3 6▯2 7 3 7 and 6▯2 7 1 7: 4▯1 ▯1 0 5 4▯1 ▯1 ▯1 5 3 0 ▯1 3 0 2 After row reduction, these become 2 3 2 3 1 1 0 1 1 1 6 7 6 7 6 0 3 17 and 60 3 17: 4 0 0 05 40 0 05 0 0 0 0 0 0 So each has a solution, and consequ1nt2y s ;s 2 W, as desired. 2 of 4 n 3. Let S = fv 1v ;2::;v gkbe a set of k vectors in R . (a) Prove that if k < n then there exists a vector x 2 R such that x 62 SpanS. (b) Prove that if k > n then S must be linearly dependent. (c) Prove that if k = n then SpanS = R if and only if S is linearly independent. Solution: (a) Consider the matrix A whose columns are the vectors in S. As S has k vectors in it, the reduced echelon form of A will have at most k leading e
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