In the first two sections of this chapter we looked at vectors in 2-space and 3-space.  You probably noticed that with the exception of the cross product (which is only defined in 3-space) all of the formulas that we had for vectors in 3-space were natural extensions of the 2-space formulas.  In this section we’re going to extend things out to a much more general setting.  We won’t be able to visualize things in a geometric setting as we did in the previous two sections but things will extend out nicely.  In fact, that was why we started in 2-space and 3-space.  We wanted to start out in a setting where we could visualize some of what was going on before we generalized things into a setting where visualization was a very difficult thing to do.

 

So, let’s get things started off with the following definition.

 

Definition 1  Given a positive integer n an ordered n-tuple is a sequence of n real numbers denoted by .  The complete set of all ordered n-tuples is called n-space and is denoted by

 

In the previous sections we were looking at  (what we were calling 2-space) and  (what we were calling 3-space).  Also the more standard terms for 2-tuples and 3-tuples are ordered pair and ordered triplet and that’s the terms we’ll be using from this point on.

 

Also, as we pointed out in the previous sections an ordered pair, , or an ordered triplet, , can be thought of as either a point or a vector in  or  respectively.  In general an ordered n-tuple, , can also be thought of as a “point” or a vector in .  Again, we can’t really visualize a point or a vector in , but we will think of them as points or vectors in  anyway and try not to worry too much about the fact that we can’t really visualize them.

 

Next, we need to get the standard arithmetic definitions out of the way and all of these are going to be natural extensions of the arithmetic we saw in  and .

 

Definition 2  Suppose  and  are two vectors in . (a)  We say that u and v are equal if,                                                   (b) The sum of u and v is defined to be,                                                (c) The negative (or additive inverse) of u is defined to be,                                                         (d) The difference of two vectors is defined to be,                                        (e) If c is any scalar then the scalar multiple of u is defined to be,                                                         (f) The zero vector in  is denoted by 0 and is defined to be,

 

The basic properties of arithmetic are still valid in  so let’s also give those so that we can say that we’ve done that.

 

Theorem 1  Suppose ,  and  are vectors in  and c and k are scalars then, (a)   (b)   (c)   (d)   (e)   (f)   (g)   (h)

 

The proof of all of these come directly from the definitions above and so won’t be given here.

 

We now need to extend the dot product we saw in the previous section to  and we’ll be giving it a new name as well.

 

Definition 3  Suppose  and  are two vectors in  then the Euclidean inner product denoted by  is defined to be

 

So, we can see that it’s the same notation and is a natural extension to the dot product that we looked at in the previous section, we’re just going to call it something different now.  In fact, this is probably the more correct name for it and we should instead say that we’ve renamed this to the dot product when we were working exclusively in  and .

 

Note that when we add in addition, scalar multiplication and the Euclidean inner product to  we will often call this Euclidean n-space.

 

We also have natural extensions of the properties of the dot product that we saw in the previous section.

 

Theorem 2  Suppose , , and  are vectors in  and let c be a scalar then, (a)   (b)   (c)   (d)   (e)  if and only if u=0.

 

The proof of this theorem falls directly from the definition of the Euclidean inner product and are extensions of proofs given in the previous section and so aren’t given here.

 

The final extension to the work of the previous sections that we need to do is to give the definition of the norm for vectors in  and we’ll use this to define distance in .

 

Definition 4  Suppose  is a vector in  then the Euclidean norm is,

 

Definition 5  Suppose  and  are two points in  then the Euclidean distance between them is defined to be,

 

Notice in this definition that we called u and v points and then used them as vectors in the norm.  This comes back to the idea that an n-tuple can be thought of as both a point and a vector and so will often be used interchangeably where needed.

 

Let’s take a quick look at a couple of examples.

 

Example 1  Given  and  compute (a)   (b)   (c)   (d)   (e)   Solution There really isn’t much to do here other than use the appropriate definition. (a)                                           (b)                             (c)                                               (d)                                             (e)

 

Just as we saw in the section on vectors if we have  then we will call u a unit vector and so the vector u from the previous set of examples is not a unit vector

 

Now that we’ve gotten both the inner product and the norm taken care of we can give the following theorem.

 

Theorem 3  Suppose u and v are two vectors in  and  is the angle between them.  Then,

 

Of course since we are in  it is hard to visualize just what the angle between the two vectors is, but provided we can find it we can use this theorem.  Also note that this was the definition of the dot product that we gave in the previous section and like that section this theorem is most useful for actually determining the angle between two vectors.

 

The proof of this theorem is identical to the proof of Theorem 1 in the previous section and so isn’t given here.

 

The next theorem is very important and has many uses in the study of vectors.  In fact we’ll need it in the proof of at least one theorem in these notes.  The following theorem is called the Cauchy-Schwarz Inequality.

 

Theorem 4  Suppose u and v are two vectors in  then

 

Proof : This proof is surprisingly simple.  We’ll start with the result of the previous theorem and take the absolve value of both sides.

However, we know that  and so we get our result by using this fact.

 

Here are some nice properties of the Euclidean norm.

 

Theorem 5  Suppose u and v are two vectors in  and that c is a scalar then, (a)   (b)  if and only if u=0. (c)   (d)  - Usually called the Triangle Inequality

 

The proof of the first two part is a direct consequence of the definition of the Euclidean norm and so won’t be given here.

 

 

Proof :

(c) We’ll just run through the definition of the norm on this one.

(d) The proof of this one isn’t too bad once you see the steps you need to take.  We’ll start with the following.

So, we’re starting with the definition of the norm and squaring both sides to get rid of the square root on the right side.  Next, we’ll use the properties of the Euclidean inner product to simplify this.

 

Now, notice that we can convert the first and third terms into norms so we’ll do that.  Also,  is a number and so we know that if we take the absolute value of this we’ll have .  Using this and converting the first and third terms to norms gives,

 

We can now use the Cauchy-Schwarz inequality on the second term to get,

 

We’re almost done.  Let’s notice that the left side can now be rewritten as,

 

Finally, take the square root of both sides.

 

Example 2  Given  and  verify the Cauchy-Schwarz inequality and the Triangle Inequality.   Solution Let’s first verify the Cauchy-Schwarz inequality.  To do this we need to following quantities.                       Now, verify the Cauchy-Schwarz inequality.                                       Sure enough the Cauchy-Schwarz inequality holds true.   To verify the Triangle inequality all we need is,     Now verify the Triangle Inequality.                                So, the Triangle Inequality is also verified for this problem.

 

Here are some nice properties pertaining to the Euclidean distance.

 

Theorem 6  Suppose u, v, and w are vectors in  then, (a)   (b)  if and only if u=v. (c)   (d)  - Usually called the Triangle Inequality

 

The proof of the first two parts is a direct consequence of the previous theorem and the proof of the third part is a direct consequence of the definition of distance and won’t be proven here.

 

Proof (d) : Let’s start off with the definition of distance.

 

Now, add in and subtract out w as follows,

Next use the Triangle Inequality for norms on this.

Finally, just reuse the definition of distance again.

 

We have one final topic that needs to be generalized into Euclidean n-space.

 

Definition 6  Suppose u and v are two vectors in .  We say that u and v are orthogonal if .

 

So, this definition of orthogonality is identical to the definition that we saw when we were dealing with  and .

 

Here is the Pythagorean Theorem in .

 

Theorem 7  Suppose u and v are two orthogonal vectors in  then,

 

Proof : The proof of this theorem is fairly simple.  From the proof of the triangle inequality for norms we have the following statement.

However, because u and v are orthogonal we have  and so we get,

 

Example 3  Show that  and  are orthogonal and verify that the Pythagorean Theorem holds.   Solution Showing that these two vectors is easy enough.                        So, the Pythagorean Theorem should hold, but let’s verify that.  Here’s the sum                                                         and here’s the square of the norms.                                      A quick computation then confirms that .

 

We’ve got one more theorem that gives a relationship between the Euclidean inner product and the norm.  This may seem like a silly theorem, but we’ll actually need this theorem towards the end of the next chapter.

 

Theorem 8  If u and v are two vectors in  then,

 

Proof : The proof here is surprisingly simple.  First, start with,

The first of these we’ve seen a couple of times already and the second is derived in the same manner that the first was and so you should verify that formula. 

 

Now subtract the second from the first to get,

Finally, divide by 4 and we get the result we were after.

 

 

In the previous section we saw the three standard basis vectors for , i, j, and k.  This idea can also be extended out to .  In  we will define the standard basis vectors or standard unit vectors to be,

and just as we saw in that section we can write any vector  in terms of these standard basis vectors as follows,

 

Note that in  we have ,  and .

 

Now that we’ve gotten the general vector in Euclidean n-space taken care of we need to go back and remember some of the work that we did in the first chapter.  It is often convenient to write the vector  as either a row matrix or a column matrix as follows,

 

In this notation we can use matrix addition and scalar multiplication for matrices to show that we’ll get the same results as if we’d done vector addition and scalar multiplication for vectors on the original vectors.

 

So, why do we do this?  We’ll let’s use the column matrix notation for the two vectors u and v.

 

Now compute the following matrix product.

 

So, we can think of the Euclidean inner product can be thought of as a matrix multiplication using,

provided we consider u and v as column vectors.

 

The natural question this is just why is this important?  Well let’s consider the following scenario.  Suppose that u and v are two vectors in  and that A is an  matrix.  Now consider the following inner product and write it as a matrix multiplication.

 

Now, rearrange the order of the multiplication and recall one of the properties of transposes.

 

Don’t forget that we switch the order on the matrices when we move the transpose out of the parenthesis.  Finally, this last matrix product can be rewritten as an inner product.

 

This tells us that if we’ve got an inner product and the first vector (or column matrix) is multiplied by a matrix then we can move that matrix to the second vector (or column matrix) if we simply take its transpose.

 

A similar argument can also show that,