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Mathematics

MAT223H1

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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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