Linear Combinations and Linear Transformations As you’ve probably already noticed, the two fundamental operations we’re interested in in linear algebra are addition and scalar multiplication. In particular, linear transformations are exactly functions that preserve addition and scalar multiplication of vectors. We often want to combine addition and scalar multiplication, so we’ll define a new term that describes this: ~ n n Definition 1. A vector b in R is called a linear combination of the vectors ~v1,...,~vk in R if there exist scalars ~ c1,..., ck such that b = c1~v1 + ··· + ck~vk. (This is Definition 1.3.9 in Bretscher, and you may want to read §1.3 for a different presentation of the material. For now, ignore the discussion of “rank” in §1.3; we’ll come back to that in a few weeks.) ~ A definition like this may initially look daunting because there are so many variables in it: b,~v1,...,~vk, c1,..., ck. But if we do an example, we see it’s really much less complicated than it may first seem. 3 1 0 Example 2. Let’s decide whether is a linear combination of and . In the notation of Definition 5 0 1 3 1 0 1, we are trying to decide whether ~b = is a linear combination of ~v = and ~v = . According 5 1 0 2 1 3 1 0 to the definition, that means deciding whether there are scalars c , c such that = c + c . In 1 2 5 1 0 2 1 3 1 0 this case, you can probably just look at this and see that the answer is yes: = 3 + 5 . v 5 0 1 In the previous example, we were able to look at the question and just “see” the answer; mathematicians call this solving by inspection. Let’s see an example where it’s not so easy to solve by inspection. 2 13 213 223 Example 3. Let’s decide whether 4 145 is a linear combination of 445 and 435. −9 1 7 By Definition1, we are trying to decide whether there are scalars c1, c2 such that 213 223 2 13 c1 445 + c2 435 = 4 145 (4) 1 7 −9 We can rewrite this equation in matrix form: 21 23 2 13 c 44 35 1 = 4 145 (5) c 1 7 2 −9 Of course, this is the sort of linear system that we are very familiar with by now, and we can solve it using 1 Gauss-Jordan: 21 2 13 21 2 13 44 3 145 −4(I) ! 40 −5 105 ÷ − 5 1 7 −9 −(I) 0 5 −10 21 2 13 −2(II) ! 40 1 −25 0 5 −10 −5(II) 21 0 53 ! 40 1 −25 0 0 0 Remember that we were solving for c1, c2 in equation (5), and now we see that c1 = 5, c2 = −2. Therefore, equation (4) says that 213 223 2 13 5 445 − 2 435 = 4 145 , (6) 1 7 −9 2 13 213 223 which shows that 4 145 is indeed a linear combination of 445 and 435. (Of course, it’s easy to check that −9 1 7 equation (6) is true.) v As we’ve mentioned before, a really important part of linear algebra is generalizing and thinking about all the possible things that could happen in a given situation. If you look back at the previous example, a good question to ask yourself is: what would have had to happen in order for us to conclude that a vector ~ b was not a linear combination of vectors ~v1,...,~vk? Using linear combinations to understand linear transformations In problems like Problem Set 3, #3 and Problem Set 3, #4, you have already seen how linear combinations are useful in the study of linear transformations. Let’s do one more example. Example 7. Suppose we have a linear transformation T : R3 ! R3, and all we know initially is that 02131 213 02231 243 2 13 T @445A = 425 and T @435A = 455. Since we found in Example3 that 4 145 is a linear combination 1 3 7 6 −9 213 223 02 131 of 445 and 435, we have enough information to find T @4 145A. More specifically, if we use equation (6), 1 7 −9 2 we have 02 131 0 213 2231 T @4 145A = T @5 445 + (−2) 435A −9 1 7 0 2131 0 2231 = T @5 445A + T @(−2) 435A since T preserves addition 1 7 02131 02231 = 5 T @445A + (−2)T @435A since T preserves scalar multiplication 1 7 213 243 = 5 425 − 2 455 3 6 2−33 = 4 05 3 v We can summarize and generalize what happened in Example7: m n Fact 8. If we are studying a linear transformation T : R ! R and we know T(~v1),..., T(~vk) for some vectors ~v1,...,~vk, then we can find T(~x) for any ~x that is a linear combination of ~v1,...,~vk. We will so often be interested in all linear combinations of ~v1,...,~vk that we give this a name, too. m Definition 9. If ~v1,...,~vk are vectors in R , the set of all linear combinations c1~v1 + ··· + ck~vk of ~v1,...,~vk is called the span of ~v1,...,~vk. So, we can restate Fact8 as: m n Fact8 restated. If we are studying a linear transformation T : R ! R and we know T(~v1),..., T(~vk) for some vectors ~v1,...,~vk, then we can find T(~x) for all ~x in span(~v1,...,~vk). Notice that the span of vectors is defined to be a set of vectors; let’s try to understand this better by interpreting it geometrically in some examples. 0213 2031 Example 10. What does span @405 , 405A look like? 0 1 213 203 According to the definition of span, this is the set of all linear combinations of 405 and 405. So, it is 0 1 2 3 2 3 2 3 1 0 c1 exactly the set of all vectors of the form c1 405 + c2 405 = 4 0 5. These vectors form the plane y = 0 in 0 1 c2 R3 (i.e., the xz-plane). v Example 11. We can now describe Problem Set 0, #W2(c) in new language: in that problem, we saw that 3 0213 2231 213 span @425 , 425A is the plane in R3 which goes through the origin and includes the vectors 425 and 3 1 3 223 425. v 1 Bases of Rm You should see now that Fact8 is exactly the idea underlying Problem Set 3, #3 and #4. In Problem 1 0 Set 3, #3, knowing T and T enabled us to find T(~x) for all ~x in 2; this was because 0 1 R 02131 02031 02031 1 0 span , = 2. Similarly, in Problem Set 3, #4, knowing T @405A, T @415A, and T @405A 0 1 R 0 0 1 0213 203 2031 enabled us to find T(~x) for all ~x in R3 since span @405 , 415 , 405A = R3. Thus, we see that, if we want 0 0 1 m n to find a complete formula for a linear transformation T : R ! R , it is enough to know T(~v1),..., T(~vk) m for vectors ~v1,...,~vk whose span is all of R (i.e., all of the domain of T). This principle is an incredibly important one in linear algebra. Note on language: The term span is also often used as a verb rather than as a noun. Another way to say m m that “the span of ~v1,...,~vk is R ” is to say, “~v1,...,~vk span R ”. 213 203 203 1 0 As we’ve seen already, and span 2, while 405 , 415 , 405 span 3. We can generalize this 0 1 R R 0 0 1 pattern. Definition 12. We’ll define the following vectors in Rm: 213 203 203 607 617 607 6 7 6 7 6 7 607 607 607 6 7 6 7 6 7 ~e1 = 6 . 7 , ~e2 = 6 . 7 ,..., ~em = 6 . 7 . 6 . 7 6 . 7 6 . 7 6 7 6 7 6 7 405 405 405 0 0 1 (The j-th entry of ~ej is 1, and all other entries are 0.) Notice that the notation ~ej by itself is ambiguous: if you simply write ~e1, it’s not clear whether you’re 2 3 2 3 1 1 1 607 talking about , 405, 6 7, or something else. Therefore, to communicate clearly when we use this 0 405 0 0 3 notation, we have to say something like “~e1 in R ”. m m We’ve now defined vectors ~e1,...,~em in R with the property that span(~e1,...,~em) = R . Notice that, if we took away any one of these vectors, the remaining ones would no longer span Rm. Thus, it looks like 4 we need at least m vectors in order to span Rm; this turns out to be true, and we’ll understand exactly why in a few days. But we’ll go ahead and define one more very important term now. m m m Definition 13. We say that m vectors ~v1,...,~vm in R form a basis of R if their span is R . Notice that an important part of this definition is a little bit hidden: the definition only applies when we have m vectors in Rm – 5 vectors in R5, for example, but not 6 vectors in R5. As you can see, when you read mathematical definitions, it’s important to parse them very carefully.