Math 51 Tutorial { August 10
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1 SSEA Summer 7 Math 5 Tutorial { August. The span of a set of vectors {v, v,..., v } is given by span (v, v,..., v k ) {c v + c v + + c k v k c, c,..., c k R}. This set can be viewed algebraically as an infinite collection of vectors or geometrically as a set of points (identified by those vectors). These points, typically, lie on a line, on a plane, or comprise the entire space. For each set of vectors below, determine whether its span is a line, a plane, or the entire space. (a). (b) Solution: The span is a line in R 3 since every vector of the form c for some real constant c will identify a point on the line through the origin and the point (,, ). {[ ] [ ]},. 5 Solution: The span is a plane in R. How do we know that the span is not a line in R? Because, the vectors are not in the same direction and therefore span the entire space R. This means, any point (a, b) in the R space can be identified by a suitable linear combination of the two vectors as follows: [ ] a b a }{{} c [ ] + b [ ]. }{{} 5 5 c (c) {[ ] [ ]},. 5
2 SSEA Math Module Math 5 Tutorial, Page of 6 August, 7 Solution: Since span [ ] [ ] 5, we have: 5 ([ ] [ ]) { [ ], c 5 { [ ] c { { c [ ] + c 5 } c, c R ] [ } + c 5 c, c R [ ] } (c + 5c ) c, c R [ ] } ([ ]) c R span, (d) where we have set c c + 5c to be a single constant. This we know is a line in R that passes the through the origin and the points (, ) and (, 5). 5, 8, 6. Solution: By inspection, we see that and that the first and the last vectors are not multiples of each other as in the previous problem - that is, they are not in the same direction. Therefore, 5 5 span, 8, 6 c + c 8 + c 3 6 c, c, c 3 R c + c c 3 6 c, c, c 3 R (c + c ) + (c + c 3 ) 6 c, c, c 3 R d + d 6 d, d R, where we have set c + c d and c + c 3 d. Thus, the span is a plane in R 3 that contains the vectors and 6.
3 SSEA Math Module Math 5 Tutorial, Page 3 of 6 August, 7 (e) 4, 8. 3 Solution: The span is a line in R 3 that passes through the origin and the points (,, 3) and (4, 8, ) because the vectors are in the same direction: (f),,. Solution: The span is the entire R 3 space since any point (a, b, c) can be represented as a linear combination of the three vectors: a b a + b + c.. Solve Example.8 from the text. 3. Find a parametric representation of the hyperplane in R 4 which contains the points P (, 3, 4, 5), Q(,,, ), R(,,, ) and S(,, 3, ). Solution: In the previous problem we saw that we needed two parameters to parametrize a plane in R 3. To define a hyperplane in R 4, we need three parameters: which is one less than the 4 in R 4. Each of these parameters will multiply a direction vector, each of which has its tail, say, at P (, 3, 4, 5). Thus we have: P Q OQ OP P R OR OS P S OS OP 3 4 3, ,
4 SSEA Math Module Math 5 Tutorial, Page 4 of 6 August, 7 Let us represent x OP, a P Q, b P R, and c P S. We can now write a parametrization of the hyperplane passing through the point P and parallel to span (a, b, c) as follows: 3 {x + sa + tb + uc s, t, u R} s 3 + t 4 + u 5 s, t, u R Let u, v, and w be linearly independent vectors in R 3. Now consider the following set for some real constant a: {v u, aw v, u w}. For what values of a is this set linearly dependent and linearly independent? Solution: If there are nonzero values of c, c, and c 3 (at least one of them ) such that: c (v u) + c (aw v) + c 3 (u w), then the set of vectors {v u, aw v, u w} is linearly dependent. If the equation above is possible only for c, c, and c 3, then the set of vectors is linearly dependent. Let us rewrite the equation above in the following form: ( c + c 3 )u + (c c )v + (ac c 3 )w. Since the set of vectors {u, v, w} is linearly independent, the equation above is possible only if c + c 3 c c ac c 3. From the first equation, c c 3, and from the second equation c c. Therefore, c c c 3 from the first and second equations. From the third equation, c 3 ac. Now if a, then this results in c 3 c. This means, for a, we can pick arbitrary equal values of c, c, and c 3 to satisfy the given equations. Since there exist nonzero values of c, c, and c 3 such that: c (v u) + c (aw v) + c 3 (u w),
5 SSEA Math Module Math 5 Tutorial, Page 5 of 6 August, 7 the set of vectors {v u, aw v, u v} is linearly dependent. For the set of vectors to be linearly independent, we require c c c 3. This is possible only if a, which forces c, c, and c 3 to be equal to in order to satisfy the given equations. Therefore, the set {v u, aw v, u v} is linearly independent for all values of a. 5. Optional Challenge Problem. Let vectors u, u,..., u r be linearly independent. If vector v is a linear combination of u, u,..., u r, while vector w is not, show that the vectors: {tv + w, u, u,..., u r } form a linearly independent set for any real constant t. Solution: The vectors u, u,..., u r form a linearly independent set implies that if the following statement is true for some real constants c, c,..., c r : c u + c u + c r u r, then all these constants are equal to. Now let us consider the following linear combination: (tv + w) + c u + c u + c r u r. We will now prove that if (tv + w) + c u + c u + r u r, then. Suppose, if is not equal to, then we can divide by on both sides to get the following: This implies: (tv + w) + c u + c u + cr u r. w tv + c u + c u + cr u r. Since v can be written as a linear combination of u, u,..., u r, we have: v d u + d u + d r u r, for some real constants d, d,..., d r. Using this result in the previous equation w t [d u + d u + d r u r ] + c u + c u + cr u r c ( w td + c ) ( u + td + c ) ( u + td r + c ) r u r,
6 SSEA Math Module Math 5 Tutorial, Page 6 of 6 August, 7 which means the vector w can be written as a linear combination of u, u,..., u r. However, this is contradicts the fact that w is linearly independent of u, u,..., u r. Therefore, our initial assumption that resulted in this contradiction must be false; that is should indeed equal to. That is: (tv + w) + c u + c u + c r u r c u + c u + c r u r c, c..., c r, since the vectors u, u,..., u r are given to be linearly independent. Therefore, the vectors: {tv + w, u, u,..., u r } form a linearly independent set.
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