Chapter 3. More about Vector Spaces Linear Independence, Basis and Dimension. Contents. 1 Linear Combinations, Span

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1 Chapter 3 More about Vector Spaces Linear Independence, Basis and Dimension Vincent Astier, School of Mathematical Sciences, University College Dublin 3. Contents Linear Combinations, Span Linear Independence, Basis 3 3 Dimension of a vector space 8 4 Solution spaces of systems of linear equations 9 3. Linear Combinations, Span Linear combinations Definition. A vector v is called a linear combination of the vectors v, v,..., v n if v can be expressed as where c, c,..., c n are scalars. v = c v + c v + + c n v n 3.3 Example Consider u =, v = 3 3, w =, z = 4 then w = 3u v, so w is a linear combination of u and v. On the other hand z is not a linear combination of u and v. Why? 3.4 Another example Example. Every vector v R 3 is expressible as a linear combination of the vectors i =, j =, k =, since a b = ai + bj + ck. c 3.5

2 Example Example 3. Let p (x = x x and p (x = x in R [x]. Describe the set S of all linear combinations of polynomials p and p. Is it equal to R [x]? What is a common property of the polynomials in S? 3.6 Span of a subset Definition 4. Let V be a vector space and let S = {v, v,..., v k } be a subset of vectors. The set of all possible linear combinations of vectors in the subset is called the linear span of that subset, denoted by span{v, v,..., v k }, so span{v, v,..., v k } = {c v + c v c k v k c i R}. Span is a vector space The set W = span{v, v,..., v k } is a subset of V and is in fact a subspace of V (as an exercise, check this. Definition 5. We say that the set W is the space spanned by the vectors v, v,... v k. In turn, the set S = {v, v,... v k } is called a spanning set of W. 3.7 Example Example 6. Consider the vectors i =, j =, w =, z =. and Then span{i,j} = x y x, y R span{i,j} = span{i,j,w,z}, span{i,j} = span{w,z}. Express j in two different ways as a linear combination of the other vectors. Show that there is only one way to express j as a linear combination of w and z. 3.8

3 Example 7. Consider the vectors P (x = + x x and Q(x = x in the vector space R[x] (it is an R-vector space. Determine if + x is in the span of P (x and Q(x. Remark. If w span{v, w,..., w k }, then span{w, w,..., w k } span{v, w,..., w k }. In other words, and intuitively: If w can be expressed using v, w,..., w k (by a linear combination, then everything that can be expressed using w, w,..., w k can also be expressed using v, w,..., w k (how? Just replace w by its expression in terms of v, w,..., w k. Linear Independence, Basis Linear independence, motivation We have just seen that a set S of vectors spans a vector space V if every vector in V can be expressed as a linear combination of the vectors in S. As in the last examples, it can happen that there are sometimes many different ways to express an arbitrary vector v V as a linear combination of vectors in a spanning set. Under certain conditions, however, we can have that each vector is expressible as a linear combination of the spanning vectors in exactly one way. 3.9 Linear independence, definition This leads to the following Definition 8. Let S = {v, v,..., v k } be a non-empty set of vectors in the vector space V. S is said to be linearly independent when the following equation (in the indeterminates c,..., c k K has exactly one solution, namely, c v + c v + + c k v k = c =, c =,..., c k =. If there are other solutions, S is called a linearly dependent set. 3. Some examples Example 9. Consider the vectors i =, j =, k = in R 3. Show that the set S = {i, j, k} is linearly independent. 3

4 Example. Consider the polynomials p (x = x, p (x = 5 + 3x x, p 3 (x = + 3x x. Show that these form a linear dependent set in R [x]. 3. Another example Example. Show that the set of vectors 7 v =, v = 5, v 3 = 5, 3 8 in R 4 is linearly dependent. 3. 4

5 A result Theorem. Let V be a vector space, and let S = {v, v,..., v n } be a set of vectors of V. The following two statements are equivalent.. S is linearly independent.. Every vector in span S can only be expressed in one way as a linear combination of the elements of S: If v = c v + + c n v n for some c,..., c n K and v = d v + + d n v n for some d,..., d n K, then c = d,..., c n = d n. 3.3 Basis Definition 3. Let V be a vector space and S = {v, v,..., v n, } a set of vectors in V then S is called a basis for V if the following conditions hold. S is linearly independent. S spans V. Condition says that every element of V can be written as a linear combination of v, v,...,v n and condition says that this can only be done in one way. So S = {v, v,..., v n, } is a basis for V if and only if for every vector v V there exists a set of unique scalars c,..., c n such that v = c v + + c n v n. Definition 4. The tuple S. We write c. is called the coordinates of the vector v in the basis c n coord S (v = c.. c n 3.4 An example Example 5. We have seen in previous examples that the set {i, j, k} is linearly independent and also spans R 3. Therefore this set forms a basis for R 3. This basis is called the standard basis for R 3 and arises from a geometric point of view as unit mutually orthogonal vectors in the principle directions E/W, N/S, and Up/Down. 3.5 Another example Example 6. Show that the set of vectors {v, v, v 3 } forms a basis for R 3, where ( ( 9 ( 33 v =, v = and v 3 =. Given an arbitrary x R 3, we seek unique scalars a, b, c such that av + bv + cv 3 = x. 4 5

6 If such unique numbers a, b, c exist, then the given set of vectors forms a basis for R 3. Finding these numbers is equivalent to solving the corresponding system of equations derived from ( a ( 9 ( 33 + b + c 4 ( xy =. z That is 3.6 a +b +3c = x R a +9b +3c = y R a +4c = z R3 We can try to solve this set of equations (the solution will depend on x, y and z. If the solution is unique {v, v, v 3 } is a basis for R We solve the system in any way we like... And we eventually get a = 36x +8y +z b = 5x y 3z c = 9x y 5z It follows that there exists unique values a, b, c such that ( ( 9 ( 33 ( xy a + b + c =. 4 z thus showing that {v, v, v 3 } is a basis for R We could also have written our system in matrix form: 3 a x 9 3 b = y. 4 c z Rembember what you know about the linear systems and invertible matrices. 3.9 We compute 3 det 9 3 = 4 This tells us that our system has a unique solution and vectors {v, v, v 3 } form a basis for R Remark. We thus see that R 3 has several different bases. Depending on the problem under consideration, it will be useful to use a well-chosen basis for the computations. Example 7. Compute the coordinates of the vector in the bases i, j, k and 4 v, v, v 3 of R 3. 6

7 Some more Example 8. Which of the following sets of vectors form a basis for R 3? {( ( ( },, {( 3 ( ( },, {( ( ( ( 3 },,, Example 9. Which of the following sets of vectors form a basis for R? {(, 3 } { ( 3 9, ( } 4 Example. Does the set {p, p, p } form a basis for R [x], when p (x =, p (x = x and p (x = x? p (x = + x, p (x = x, p (x = x + x? In both cases, if {p, p, p } is a basis for R [x], compute the coordinates of x 4 in this basis. 3. 7

8 3 Dimension of a vector space Definition Definition. Let V be a nonzero vector space. If V contains a finite set of vectors {v, v,..., v n } that is a basis, then n is called the dimension of V. If no such set exists then V is said to be infinite dimensional. The only K-vector space of dimension is the vector space containing only the zero element. Theorem. If S and S are two bases of a vector space V, then S and S have the same number of elements It shows that the definition of dimension does make sense (we will get the same number of elements from any basis of V. Example 3. We have seen that {i, j, k} is a basis for R 3. So More examples dim R 3 = {i, j, k} = 3. Example 4. The space of all m n matrices M m,n (R has dimension mn with a basis consisting of m n matrices with exactly one nonzero entry equal to. Example 5. The set {, x, x,..., x n } is a basis for P n (R and so P n (R has dimension n +. We finish this section with an extremely important result about bases and one consequence Theorem 6.. Let L be a linearly independent subset of V. Then there is a basis B of V such that L B.. Let T be a subset of V such that T generates V. Then there is a basis B of V such that B T. In other words:. You can always add elements (possibly no element at all to a set of linearly independet vectors to get a basis (the elements have to be chosen carefully, of course.. You can always remove elements to a set that generates V, to get a basis of V (again, we have to be clever about what elements to remove. Proposition 7. Let U and W be subspaces of V, such that U V. Assume dim W is finite. Then. dim U dim W.. U = W if and only if dim U = dim W. Is means that the dimension is a good way to measure the size of a vector space. Example 8. Let v =, v =, v 3 =, v 4 = be vectors in R 3. What can we now say without doing many computations?. The dimension of R 3 is 3 (we know that {i, j, k} is a basis.. The set {v, v, v 3, v 4 } is not linearly independent: Because otherwise we could add elements (possibly no element at all to the set {v,..., v 4 } to get a basis of R 3. But this would give us a basis of R 3 with at least 4 elements, impossible. 8

9 3. The set {v,..., v 4 } generates R 3. To see it, it suffices to show that i, j, k span{v, v, v 3, v 4 }. Can you explain why? 4. So, by the previous theorem, we know that we can find a basis of R 3 inside {v, v, v 3, v 4 }. Which vector(s should we remove? I claim that you should remove a vector that is a linear combination of the others, and keep doing this until you have only 3 vectors left. Can you explain why? 4 Solution spaces of systems of linear equations A system of linear equations can be written in the matrix form as: where A M n,m (K, x = x. x n Ax = a, and a K n. If the righthand-side of the system is equal to zero, i.e. a =, then we say that it is a system of homogenous linear equations. Theorem 9. The set of solutions of the system of equations Ax = forms a vector space. We now consider some examples to illustrate how to determine a basis for and the dimension of the solution space of a system of homogenous linear equations. 3.6 Example 3. Determine a basis for and the dimension of the solution space in R 5 of the system x +x x 3 +x 5 = x x +x 3 3x 4 +x 5 = x +x x 3 x 5 = x 3 +x 4 +x 5 = Solution We solve the system (in any way we like, and find the solution ( x x x 3 x 4 x 5 = ( s t s t t = ( s s + ( t t t = s ( + t where s, t R. Your solution may very well be expressed in a different way, and 3.8 still be correct (there is more than one way to express the same set of solutions. This shows that if v = (, v = (. then {v, v } spans the space and since v, v are linearly independent, {v, v } is a basis of the set of solution of this system, which has dimension. 3.9 More examples ( Example 3. Determine a basis for and the dimension of the set of solutions of the following systems of linear equations (to be solved in R: { x +x x 3 = x x +x 3 = x +x 3 = 3.7 9

10 { x 3x +x 3 = x 6x +x 3 = 3x 9x +3x 3 = Example 3. Verify that the set of solutions in R 3 to the system 3.3 x +x +3x 3 = x +5x 3 = x +x 3 = has dimension zero. Example 33. In order to grow a certain crop, it is recommended that each square foot of ground be treated with phosphorus, potassium and nitrogen. Suppose that there are three brands of fertilizer on the market- say brand X, brand Y and brand Z. One kilogram of brand X contains units of phosphorous, 3 units of potassium, and 5 units of nitrogen. One pound of brand Y contains unit of phosphorous, 3 units of potassium, and 4 units of nitrogen. One pound of brand Z contains only unit of phosphorous and unit of nitrogen.. It is recommended that each square foot of ground be treated with units of phosphorus, 9 units of potassium and units of nitrogen. Is it possible to meet the recommendations by applying some combination of three brands of fertilizer.. Is it possible to meet the recommendations if the number of recommended units for nitrogen is changed to 9? 3. Explain what combinations of phosphorus, potassium and nitrogen are possible? 3.3

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