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 . O 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.
1 1 2 Example 3. Let’s decide whether 14 is a linear combination of 4 and 3. −9 1 7
By Definition1, we are trying to decide whether there are scalars c1, c2 such that
1 2 1 c1 4 + c2 3 = 14 (4) 1 7 −9
We can rewrite this equation in matrix form:
1 2 1 c 4 3 1 = 14 (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:
1 2 1 1 2 1 4 3 14 −4(I) → 0 −5 10 ÷ − 5 1 7 −9 −(I) 0 5 −10 1 2 1 −2(II) → 0 1 −2 0 5 −10 −5(II) 1 0 5 → 0 1 −2 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 1 2 1 5 4 − 2 3 = 14 , (6) 1 7 −9 1 1 2 which shows that 14 is indeed a linear combination of 4 and 3. (Of course, it’s easy to check that −9 1 7 equation (6) is true.) O
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 1 1 2 4 1 T 4 = 2 and T 3 = 5. Since we found in Example3 that 14 is a linear combination 1 3 7 6 −9 1 2 1 of 4 and 3, we have enough information to find T 14. More specifically, if we use equation (6), 1 7 −9
2 we have
1 1 2 T 14 = T 5 4 + (−2) 3 −9 1 7 1 2 = T 5 4 + T (−2) 3 since T preserves addition 1 7 1 2 = 5 T 4 + (−2)T 3 since T preserves scalar multiplication 1 7 1 4 = 5 2 − 2 5 3 6 −3 = 0 3
O
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.
1 0 Example 10. What does span 0 , 0 look like? 0 1
1 0 According to the definition of span, this is the set of all linear combinations of 0 and 0. So, it is 0 1 1 0 c1 exactly the set of all vectors of the form c1 0 + c2 0 = 0 . These vectors form the plane y = 0 in 0 1 c2 R3 (i.e., the xz-plane). O Example 11. We can now describe Problem Set 0, #W2(c) in new language: in that problem, we saw that
3 1 2 1 span 2 , 2 is the plane in R3 which goes through the origin and includes the vectors 2 and 3 1 3 2 2. O 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 1 0 0 1 0 span , = 2. Similarly, in Problem Set 3, #4, knowing T 0, T 1, and T 0 0 1 R 0 0 1 1 0 0 enabled us to find T(~x) for all ~x in R3 since span 0 , 1 , 0 = 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 ”. 1 0 0 1 0 As we’ve seen already, and span 2, while 0 , 1 , 0 span 3. We can generalize this 0 1 R R 0 0 1 pattern.
Definition 12. We’ll define the following vectors in Rm: 1 0 0 0 1 0 0 0 0 ~e1 = . , ~e2 = . ,..., ~em = . . . . . 0 0 0 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 1 1 1 0 talking about , 0, , or something else. Therefore, to communicate clearly when we use this 0 0 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. As we start to assign more reading from the textbook, you should make it a habit to look out for things like this; linear algebra requires very careful reading!
m m Example 14. The vectors ~e1,...,~em in R form a basis of R . This particular basis is called the standard basis of Rm. O
Note on Terminology
As you can already see, linear algebra is a branch of mathematics with a great deal of its own terminology. The point of introducing new terms is to give names to ideas we will use frequently, so that we build up a common language that we can use to talk about these ideas. It is absolutely crucial to learn the terminology, as it will eventually enable us to describe complicated ideas very precisely with just a few words.
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