## Transcript — Introduction

In the last lecture, we defined vectors in Rn and saw how to find a linear combination of vectors in Rn. In this lecture, we will look at ten important properties of addition and scalar multiplication of vectors in Rn. We will then start looking at a fundamental concept in linear algebra, called spanning. Geometric interpretation of sets of vectors in Rn will be used to help explain the concept of spanning.

## Properties of Vectors in Rn

We begin with an important theorem. Theorem 1.1.1: For all vectors x, y, and w in Rn, and scalars c and d in R, we have

- Property V1: x + y is in Rn.
- Property V2:(x + y) + w = x + (y + w)
- Property V3: x + y = y + x
- Property V4: There exists a vector, called the zero vector, such that x + 0 = x for all vectors x in Rn.
- Property V5: For each vector x in Rn, there exists a vector –x in Rn such that x + –x = 0.
- Property V6: cx is also a vector in Rn.
- Property V7: c(dx) = (cd)x
- Property V8: (c + d)x = cx + dx
- Property V9: c(x + y) = cx + cy

And finally,

- Property V10: 1x = x

Property V1 is the closure under addition property. More generally, a set being closed under addition means that if we add any two vectors in the set, then the sum must be another vector, which also satisfies the condition of the set.

Property V2 is the usual associative property of addition.

Property V3 is the commutative property of addition.

Property V4 shows the existence of an additive identity in Rn, which we call the zero vector.

Property V5 shows the existence of additive inverses in Rn.

Observe that property V6 looks a little like property V1. In this case, it is the closure of scalar multiplication property. It says that if x is any vector in the set, then any scalar multiple of x is also in the set.

Property V7 is the associative property of scalar multiplication.

Properties V8 and V9 are both distributive properties.

And property V10 might seem rather obvious, but it is important. It shows that 1 is the scalar multiplicative identity.

Shortly, the program will pause to let you look over the properties more closely. As you read over them, there are a couple of things you should think about. First, make sure that all of the properties make sense. If there is a property you don’t quite understand, try doing a numerical example. That is, actually pick some vectors in some simple Euclidean space, say R2 or R3, and make sure the property holds. The other thing to look for is the two different types of properties. Some properties are about the relationship between the operations of addition and scalar multiplication and the set, while the other properties are just about addition and scalar multiplication. Finally, I will note that we are going to see these same ten properties many times during the course. It is really important that you learn them as soon as possible. Take a minute now to read them over.

We will now prove properties V1, V3, and V4. Proving the rest of the properties is left as an exercise.

To prove property V1, we need to show that if we pick any two vectors in Rn, then their sum is also a vector in Rn. So we let x and y be vectors in Rn. Then by definition, there exists real numbers x1 to xn, and y1 to yn, such that x = [x1 to xn] and y = [y1 to yn]. By definition of addition of vectors in Rn, we get that x + y is the vector [x1 + y1 up to xn + yn]. Now observe that x + y satisfies the conditions of the set Rn—namely, all of its n components are real since the sum of two real numbers xi and yi is a real number. Thus, x + y is in Rn as required. Take a minute to read over the proof again. Make sure that you can justify and understand each step of the proof.

To prove property V3, we need to show that addition in Rn is commutative. Once again, we pick any two vectors x and y in Rn. Observe that by definition of addition, and the fact that the components of x and y are real numbers, we get that x + y = [x1 + y1 to xn + yn] = [y1 + x1 to yn + xn], which by definition of addition is y + x as required. Once again, take a minute to reread the proof.

Property V4 requires us to prove an existence statement. Thus, we need to construct the required zero vector and prove that it has the desired properties. Thinking about the key steps in the proofs of properties V1 and V3, we think to take the zero vector to be the vector in Rn whose n components are all zeroes. This vector satisfies the conditions of the set Rn—all of its components are real numbers—and hence is in Rn. Now, if we pick any vector x in Rn, then we get that x + 0 = [x1 + 0 up to xn + 0], which equals the vector [x1 to xn], which is just the vector x as required. Take a minute to look over the proof.

Although the proofs above are fairly short and straightforward, there are a couple of key (but perhaps subtle) points that are important. These will be stressed with a short exercise.

Observe that properties V2, V3, V7, V8, V9, and V10 only refer to the operations of addition and scalar multiplication, while the other properties—V1, V4, V5, and V6—are about the relationship between the operations and the set Rn. These facts should be clear from the proofs of the properties. We will make use of this soon.

One thing I will mention frequently throughout this course is the importance of understanding proofs. Not only does understanding proofs help one in creating their own proofs, but proofs often teach us more than just the fact that the statement of the theorem is true. For example, in the proof of V4, we showed more than just that the zero vector exists. The proof shows us that the zero vector is defined to be the vector with all components 0. Similarly, a proof of V5 demonstrates the additive inverse of a vector x in Rn is the vector –x, which equals -1x.

Recall that a set with property V1 is said to be closed under addition, while a set with property V6 is said to be closed under scalar multiplication. Theorem 1.1.1 shows that Rn is closed under linear combinations. That is, if v1 to vk are any vectors in Rn, then any linear combination c1v1 + up to ckvk is also a vector in Rn for any real scalars c1 to ck. This fact might seem rather obvious in Rn. However, we will soon see that there are sets which are not closed under linear combinations. In linear algebra, it will be important and useful to identify which sets are closed under linear combinations, and in fact, which sets satisfy all of the properties V1 to V10.

## Spanning

In linear algebra, we will frequently look at the set of all possible linear combinations of a set of vectors in Rn. So we make the following definition. Let B be a set of vectors {v1 to vk} in Rn. We denote the span, S, of the set B by

S = Span B, or S = Span{v1 to vk}

and define it to equal the set of all vectors of the form t1v1 + up to tkvk for all real values of t1 up to tk. We also use the terminology that S is spanned by B, and that B is a spanning set for S.

Note that since Rn is closed under linear combinations, we know that the span of a set of vectors in Rn will be a subset of Rn.

Let’s consider a couple of examples. Which of the following sets contain the vector x = [1; 2; 1]?

- The set S1 spanned by [1; 0; 0], [0; 1; 0], and [0; 0; 1]. By definition of spanning, the vector [1; 2; 1] is in the set S1 if and only if there exists real coefficients t1, t2, and t3 such that t1[1; 0; 0] + t2[0; 1; 0] + t3[0; 0; 1] = [1; 2; 1]. In this case, we can easily identify by inspection that taking t1 = 1, t2 = 2, and t3 = 1, the equation is satisfied. Hence, [1; 2; 1] is in S1.
- Now consider the set S2 = Span{[2; 0; 1], [1; 0; 1]}. The vector [1; 2; 1] is in S2 if and only if we can find real numbers t1 and t2 such that t1[2; 0; 1] + t2[1; 0; 1] = [1; 2; 1]. Simplifying the left-hand side using operations on vectors, we get the vector [2t1 + t2; 0; t1 + t2] = [1; 2; 1]. This cannot hold since the middle entries are not equal. Thus, [1; 2; 1] is not in S2.

## Viewing a Span Geometrically

We now look at a couple of examples to help explain the concept of spanning. Describe geometrically the set S spanned by [1; -1]. Applying the definition of span, we get that S is the set of all possible scalar multiples of [1; -1]—that is, all vectors of the form t1[1; -1] which equals [t1; -t1]. To interpret S geometrically, we can view it as the set of all points in R2 of the form (t1, -t1). Plotting these points, we see that S is a line through the origin.

## Vector Equation of a Set

It can sometimes be a little cumbersome to have to use set notation whenever we are dealing with spanning sets. For this reason, we invent some additional notation to write spanned sets more compactly. If S is spanned by a set of vectors {v1 to vk} in Rn, then we define a vector equation for S to be an equation of the form x = t1v1 + up to tkvk, where t1 up to tk are real numbers. Observe, this is giving us an equation for every vector x in S. For example, a vector equation for S = Span{[1; 2; 1], [1; 0; 1]} is x = t1[1; 2; 1] + t2[1; 0; 1] for any t1 and t2 in R.

If T has vector equation x = s1[1; 0] + s2[0; 1] + s3[2; -1], with s1, s2, and s3 all real numbers, then we may write T = Span{[1; 0], [0; 1], [2; -1]}. Notice, however, that the third vector in the spanning set for T is a linear combination of the first two vectors. In this case, we may simplify the vector equation, and hence the corresponding spanning set. In particular, observe that we can write the vector [2; -1] as 2 times the first vector plus -1 times the second vector. That is, we can write the vector equation of T as x = s1[1; 0] + s2[0; 1] + s3(2[1; 0] + -1[0; 1]). Using the properties of vectors in Rn, i.e. Theorem 1.1.1, we can rearrange this to get x = (s1 + 2s3)[1; 0] + (s2 – s3)[0; 1]. Replacing (s1 + 2s3) with t1, and (s2 – s3) with t2, we get that a simpler vector equation for T is x = t1[1; 0] + t2[0; 1], where t1 and t2 can take any real values since s1, s2, and s3 could take any real values. We cannot simplify the vector equation any further since neither of the remaining vectors is a scalar multiple of the other. Hence, we call this new equation a simplified vector equation for T. This leads to a simpler spanning set. In particular, we have T is spanned by [1; 0] and [0; 1].

## Simplifying a Spanning Set

In linear algebra, it will often be very important to find a spanning set that contains as few vectors as possible. First, we prove a theorem, which shows us under what condition we can remove a vector from a spanning set. Theorem 1.1.2: If vk can be written as a linear combination of v1 to vk-1, then Span{v1 to vk} = Span{v1 to vk-1}. (In addition to learning linear algebra, part of the purpose of this course is to help you learn to prove mathematical results. Thus, for a few proofs in this course, we will break down the thought process which goes into making a proof to help you learn to prove results on your own. The first thing you should do when thinking about a proof is to identify what assumptions you are making, what you are trying to prove, and how you are going to try to prove it.)

Proof: In this case, we are assuming that vk can be written as a linear combination of v1 to vk-1. Hence, by definition, we are assuming that there exists real scalars c1 to ck-1 such that c1v1 + up tp ck-1vk-1 = vk. We want to prove that Span{v1 to vk} = Span{v1 to vk-1}. So, we think, how can we show that two sets are equal? The main way of doing this is to show that the sets are subsets of each other. That is, we will need to show that Span{v1 to vk} is a subset of Span{v1 to vk-1}, and then show that Span{v1 to vk-1} is a subset of Span{v1 to vk}. Next, we ask ourselves, how do you show one set A is a subset of another set B? To do this, we need to show that every element in the set A also belongs to the set B. Now that we have figured out what we know, and what we are trying to show, we begin trying to do it. For the first part, we want to show that Span{v1 to vk} is a subset of Span{v1 to vk-1}. So, as we figured out, our first step is to pick any vector x in Span{v1 to vk}. We want to show that x also belongs to Span{v1 to vk-1}. We ask ourselves, what does that mean? This means that we want to write x as a linear combination of the vectors v1 to vk-1. Now the rest of the proof is just trying to fill in the steps from knowing that x is in Span{v1 to vk} to writing x as a linear combination of the vectors v1 to vk-1. If our goal is to write x as a linear combination of vectors, we should start by writing it as a linear combination of vectors. By definition, if x is in Span{v1 to vk}, then there exist real scalars d1 to dk such that x = d1v1 + up to dkvk. We see that we want to remove the vector vk from this linear combination. We recall that our initial assumption is that vk is a linear combination of v1 to vk-1. So making the substitution, we get x = d1v1 + up to dk-1vk-1 + dk(c1v1 + up to ck-1vk-1). Rearranging this, we find that we have written x as a linear combination of v1 to vk-1 as desired. So x is in Span{v1 to vk-1}, and hence Span{v1 to vk} is a subset of Span{v1 to vk-1}. So now, we just need to show that Span{v1 to vk-1} is a subset of Span{v1 to vk}. Observe that if y is in Span{v1 to vk-1}, then there exist real coefficients t1 to tk-1 such that y = t1v1 + up to tk-1vk-1. But we can just rewrite this as y = t1v1 + up to tk-1vk-1 + 0vk. So y is a linear combination of v1 to vk as required. Therefore, we have shown that Span{v1 to vk} = Span{v1 to vk-1}, poof.

A few notes about this. First, observe that figuring out the proof involved asking a lot of questions like “What does this mean?” and “How do I show this?” It also involved applying a lot of definitions and theorems. To make your own proofs, you need to know and understand all the definitions and theorems, and be good at asking yourself the right questions. Of course, learning what are the right questions to ask is the hard part, and takes practice. The extra difficulty is that, in many cases, the right questions will have multiple right answers, but only one or a couple of the answers will lead to a nice proof. Perseverance is also an important quality required for learning to do proofs. You may find at first that, quite often, your first thought of how to do a proof doesn’t work out, and so you have to figure out another way to try to do the proof. The good news is that you will learn as much from the ideas that don’t work as from the ideas that do. So when first learning to do proofs, it is important to remember the saying “If at first you don’t succeed, try try again.” Remember that professional mathematicians don’t always get it right the first time either. Andrew Wiles took about seven years to come up with the proof of Fermat’s Last Theorem, and then another year to correct a few errors in that proof. He is one of the best examples that if you work hard and don’t give up, you can accomplish great things.

Second, notice that it was really, really easy to show that Span{v1 to vk-1} was a subset of Span{v1 to vk}. In textbooks, this step would often be considered obvious, and written as “Clearly, Span{v1 to vk-1} is a subset of Span{v1 to vk}.” When reading such proofs, always make sure that you understand why it is obvious. When writing proofs, make sure that something is really obvious before writing that. There are many things in math which seem obvious, but are actually very challenging to prove. Moreover, there are actually a few things in math which seem obvious, but are, in fact, false. So, be very careful when thinking something is obvious. Essentially, unless a result follows immediately from the definition, you shouldn’t write “obvious”. For example, the whole proof of “Span{v1 to vk-1} is a subset of Span{v1 to vk}” is to see that we just take the coefficient of vk to be 0.

Take a minute now to read over the proof. As usual, make sure that you understand—and can justify—each line of the proof. Note that by rearranging the vectors in the set, we see that Theorem 1.1.2 shows that any vector that can be written as a linear combination of the other vectors can be removed from the set without changing the set it spans.

Example: Write a simplified vector equation for the set S spanned by {[1; 2; 1], [2; 4; 2], [0; 0; 0]}. By definition, we have that S = Span{[1; 2; 1], [2; 4; 2], [0; 0; 0]}. We first observe that [2; 4; 2] can be written as a linear combination of the other vectors. In particular, [2; 4; 2] = 2[1; 2; 1] + 0[0; 0; 0]. Thus, we can remove [2; 4; 2] from the spanning set by Theorem 1.1.2. Hence, S = Span{[1; 2; 1], [0; 0; 0]}. Now we notice that [0; 0; 0] is a scalar multiple of [1; 2; 1]—in particular, 0[1; 2; 1]. Hence, we can apply Theorem 1.1.2 again to get that S = Span{[1; 2; 1]}. Clearly, the spanning set of S cannot be reduced any further, so a simplified vector equation for S is x = t[1; 2; 1] for any real number t.

Example: Find a simplified spanning set for the set T with vector equation x = c1[3; 0; 1] + c2[0; 3; 2] + c3[1; 1; 1] where c1, c2, and c3 are real numbers. By definition, T is spanned by {[3; 0; 1], [0; 3; 2], [1; 1; 1]}. Observe that [3; 0; 1] = -1[0; 3; 2] + 3[1; 1; 1]. That is, [3; 0; 1] is a linear combination of the other two vectors. Hence, we can apply Theorem 1.1.2 to get that T = Span{[0; 3; 2], [1; 1; 1]}. Neither vector in this spanning set for T is a scalar multiple of the other, so this is a simplified spanning set for T.

Note that in both of these examples, we could have gotten different answers by removing different vectors. For example, another simplified spanning set for T would be {[3; 0; 1], [1; 1; 1]} since we could have written [0; 3; 2] as a linear combination of [3; 0; 1] and [1; 1; 1].

In these examples, we were able to write one vector as a linear combination of the others by inspection. However, in the real world, there could easily be thousands of vectors with thousands of entries. We need to develop a way of finding if at least one vector in a set can be written as a linear combination of the others, and if there is such a vector, what that linear combination is. We will begin looking at this in the next lecture.