Chapter 1 Euclidean Space - Rice University

1Euclidean spaceChapter 1Euclidean spaceA. The basic vector spaceWe shall denote by R the field of real numbers. Then we shall use the Cartesian productRn R R . . . R of ordered n-tuples of real numbers (n factors). Typical notation forx Rn will bex (x1 , x2 , . . . , xn ).Here x is called a point or a vector , and x1 , x2 , . . . , xn are called the coordinates of x. Thenatural number n is called the dimension of the space. Often when speaking about Rn andits vectors, real numbers are called scalars.Special notations:R1R2R3xx (x1 , x2 ) or p (x, y)x (x1 , x2 , x3 ) or p (x, y, z).We like to draw pictures when n 1, 2, 3; e.g. the point ( 1, 3, 2) might be depicted as

2Chapter 1We define algebraic operations as follows: for x, y Rn and a R,x yax xx y (x1 y1 , x2 y2 , . . . , xn yn );(ax1 , ax2 , . . . , axn );( 1)x ( x1 , x2 , . . . , xn );x ( y) (x1 y1 , x2 y2 , . . . , xn yn ).We also define the origin (a/k/a the point zero)0 (0, 0, . . . , 0).(Notice that 0 on the left side is a vector, though we use the same notation as for the scalar0.)Then we have the easy facts:x y(x y) z0 xx x1x(ab)xa(x y)(a b)x0xa0 y x;x (y z);x;0;x;a(bx);ax ay;ax bx;0;0.in other words all the“usual” algebraic rulesare valid if they makesenseSchematic pictures can be very helpful. One nice example is concerned with the linedetermined by x and y (distinct points in Rn ). This line by definition is the set of all pointsof the form(1 t)x ty,Here’s the picture:where t R.

3Euclidean spaceThis picture really is more than just schematic, as the line is basically a 1-dimensionalobject, even though it is located as a subset of n-dimensional space. In addition, the closedline segment with end points x and y consists of all points as above, but with 0 t 1. Thissegment is shown above in heavier ink. We denote this segment by [x, y].We now see right away the wonderful interplay between algebra and geometry, somethingthat will occur frequently in this book. Namely, the points on the above line can be describedcompletely in terms of the algebraic formula given for the line. On the other hand, the line isof course a geometric object.It is very important to get comfortable with this sort of interplay. For instance, if wehappen to be discussing points in R5 , we probably have very little in our background thatgives us geometric insight to the nature of R5 . However, the algebra for a line in R5 is verysimple, and the geometry of a line is just like the geometry of R1 .Similarly, it is helpful to represent triangles with a picture in the plane of the page. Thusif we have three noncollinear points x, y, z in Rn , there is a unique plane which contains them.This plane lies in Rn of course, but restricting attention to it gives a picture that looks likean ordinary plane. The plane is the set of all points of the formp t1 x t2 y t3 z,where t1 t2 t3 1. Sometimes the scalars t1 , t2 , t3 are called barycentric coordinates ofthe point p.

4Chapter 1The point displayed inside this triangle is 13 (x y z), and is called the centroid of the triangle.PROBLEM 1–1. We need to examine the word collinear we have just used. In fact,prove that three points x, y, z in Rn lie on a line there exist scalars t1 , t2 , t3 , not allzero, such thatt1 t2 t3 0,t1 x t2 y t3 z 0.PROBLEM 1–2. Prove that x, y, 0 are collinear x is a scalar multiple of y or yis a scalar multiple of x.PROBLEM 1–3. Prove that if x, y, z in Rn are not collinear and if p belongs to theplane they determine, then the real numbers t1 , t2 , t3 such thatt1 t2 t3 1,t1 x t2 y t3 z p,are unique.

5Euclidean spacePROBLEM 1–4. In the triangle depicted above let L1 be the line determined by xand the midpoint 21 (y z), and L2 the line determined by y and the midpoint 21 (x z).Show that the intersection L1 L2 of these lines is the centroid. (This proves the theoremwhich states that the medians of a triangle are concurrent.)PROBLEM 1–5. Prove that the interior (excluding the sides) of the above triangle isdescribed by the conditions on the barycentric coordinatest1 t2 t3 1,t1 0, t2 0, and t3 0.As an example of our method of viewing triangles, think about an equilateral triangle. Ifwe imagine it conveniently placed in R2 , the coordinates of the vertices are bound to be rathercomplicated; for instance, here are two ways:x2x2(0, 3 )(0, 1)( 1, 0)(1, 0)x1x1( 3 /2, 1/2)( 3 /2, 1/2)But a really elegant positioning is available in R3 , if we simply place the vertices at (1, 0, 0),(0, 1, 0) and (0, 0, 1):

6Chapter 1(0, 1, 0)(1, 0, 0)(0, 0, 1)Now this looks much better if we view this triangle as it lies in the plane x1 x2 x3 1:We can’t draw coordinate axes in this plane, or even the origin, though we could imagine theorigin as sitting “behind” the centroid:origin

7Euclidean spacePROBLEM 1–6. Prove that the four points w, x, y, z in Rn are coplanar thereexist real numbers t1 , t2 , t3 , t4 , not all zero, such thatt1 t2 t3 t4 0,t1 w t2 x t3 y t4 z 0.As further evidence of the power of vector algebra in solving simple problems in geometry,we offerPROBLEM 1–7.Let us say that four distinct points w, x, y, z in Rn define aquadrilateral , whose sides are the segments [w, x], [x, y], [y, z], [z, w] in that order.zyzyxxwwProve that the four midpoints of the sides of a quadrilateral, taken in order, form thevertices of a parallelogram (which might be degenerate). In particular, these four pointsare coplanar. (Note that the original quadrilateral need not lie in a plane.) Express thecenter of this parallelogram in terms of the points w, x , y , and z.

8Chapter 1PROBLEM 1–8. We have seen in Problem 1–1 the idea of three points being collinear,and in Problem 1–6 the idea of four points being coplanar. These definitions can of coursebe generalized to an arbitrary number of points. In particular, give the correct analogousdefinition for two points to be (say) “copunctual” and then prove the easy result that twopoints are “copunctual” if and only if they are equal.PROBLEM 1–9. Give a careful proof that any three points in R1 are collinear; andalso that any four points in R2 are coplanar.It is certainly worth a comment that we expect three “random” points in R2 to be noncollinear. We would then say that the three points are in general position. Likewise, fourpoints in R3 are in general position if they are not coplanar.B. DistanceNow we are going to discuss the all-important notion of distance in Rn . We start with R2 ,where we have the advantage of really understanding and liking the Pythagorean theorem. Weshall freely accept and use facts we have learned from standard plane geometry. Thus we saythat the distance between x and y in R2 is#2axis x 2 y2 #1 axis x 1 y1 d(x, y) p(x1 y1 )2 (x2 y2 )2 .We can even proceed easily to R3 by applying Pythagoras to the right triangle with legs given

9Euclidean spaceby the two segments: (1) from (x1 , x2 , x3 ) to (y1 , y2 , x3 ) and (2) from (y1 , y2 , x3 ) to (y1 , y2 , y3 ).y( y 1, y2, x3 )xThe segment (1) determines the distancep(x1 y1 )2 (x2 y2 )2 ,and the segment (2) determines the distance x3 y3 .If we square these distances, add the results, and then take the square root, the distance wefind isp(x1 y1 )2 (x2 y2 )2 (x3 y3 )2 .Though we are hard pressed to draw a similar picture for R4 etc., we can easily imagine thesame procedure. For R4 we would consider the “horizontal” line segment from (x1 , x2 , x3 , x4 )to (y1 , y2 , y3 , x4 ), and then the “vertical” segment from (y1 , y2 , y3 , x4 ) to (y1 , y2 , y3 , y4 ). Thetwo corresponding distances arep(x1 y1 )2 (x2 y2 )2 (x3 y3 )2and x4 y4 ,respectively. (We have used our formula from the previous case of R3 .) We certainly want tothink of these two segments as the legs of a right triangle, so that the distance between x and

10Chapter 1y should come from Pythagoras by squaring the two numbers above, adding, and then takingthe square root:pd(x, y) (x1 y1 )2 (x2 y2 )2 (x3 y3 )2 (x4 y4 )2 .We don’t really need a picture to imagine this sort of construction for any Rn , so we areled toTHE PYTHAGOREAN DEFINITION. The distance between x and y in Rn isvu nuXd(x, y) t (xi yi )2 .i 1Clearly, d(x, y) d(y, x), d(x, x) 0, and d(x, y) 0 if x 6 y. We also say that d(x, y) isthe length of the line segment [x, y].This definition of course gives the “right”answer for n 2 and n 3. (It even works forpn 1, where it decrees that d(x, y) (x y)2 x y .)It will be especially convenient to have a special notation for the distance from a point tothe origin:DEFINITION. The norm of a point x in Rn is the numbervu nuXkxk d(x, 0) tx2i .i 1Thus we have equations likeR1R2R3R4::::k 3k 3k(3, 4)k 5 k(1, 2, 3)k 14k(1, 1, 1, 1)k 2.The idea of “norm” is important in many areas of mathematics. The particular definitionwe have given is sometimes given the name Euclidean norm.We have some easy properties:d(x, y)kaxk k xkk0kkxk kx yk; a kxk for a R;kxk;0;0 if x 6 0.

11Euclidean spaceHere’s a typical picture in R3 :Oxdx1x3x2kxk2 δ 2 x23δ 2 x21 x22We close this section with another “algebra-geometry” remark. We certainly are thinkingof distance geometrically, relying heavily on our R2 intuition. Yet we can calculate distancealgebraically, thanks to the formula for d(x, y) in terms of the coordinates of the points x andy in Rn .C. Right angleNow we turn to a discussion of orthogonality. We again take our clue from the Pythagoreantheorem, the square of the hypotenuse of a right triangle equals the sum of the squares of theother two sides. The key word we want to understand is right.Thus we want to examine a triangle in Rn (with n 2). Using a translation, we maypresume that the potential right angle is located at the origin. Thus we consider from thestart a triangle with vertices 0, x, y. As we know that these points lie in a plane, it makessense to think of them in a picture such as

12Chapter 1xOyWe are thus looking at the plane containing 0, x, y, even though these three points lie in Rn .We then say that the angle at 0 is a right angle if and only if the Pythagorean identityholds:d(x, y)2 d(x, 0)2 d(y, 0)2 .I.e., in terms of the Euclidean norm on Rn ,kx yk2 kxk2 kyk2 .(P)You probably have noticed that our reasoning has been somewhat circular in nature. For weused the Pythagorean idea to motivate the definition of distance in the first place, and now weare using distance to define right angle. There is an important subtlety at work here. Namely,in defining distance we worked with right angles in coordinate directions only, whereas nowwe are defining right angles in arbitrary (noncoordinate) directions. Thus we have achievedsomething quite significant in this definition.We feel justified in this definition because of the fact that it is based on our intuition fromthe Euclidean geometry of R2 . Our planes are located in Rn , but we want them to have thesame geometric properties as R2 .

13Euclidean spaceNow we perform a calculation based upon our use of algebra in this material:kx yk2 nX(xi yi )2i 1 nX(x2i 2xi yi yi2 )i 12 kxk 2nXxi yi kyk2 .i 1Thus the Pythagorean relation (P) becomesnXxi yi 0.i 1PIn summary, we have a right angle at 0 ni 1 xi yi 0.Based upon the sudden appearance of the above number, we now introduce an extremelyuseful bit of notation:DEFINITION. For any x, y Rn , the inner product of x and y, also known as the dotproduct, is the numberx y nXxi y i .I like to make this dot huge!i 1The above calculation thus says thatkx yk2 kxk2 2x y kyk2 .Just to make sure we have the definition down, we rephrase our definition of right angle:DEFINITION. If x, y Rn , then x and y are orthogonal (or perpendicular ) if x y 0.Inner product algebra is very easy and intuitive:x y(x y) z(ax) y0 xx x y x;x z y z;a (x y) ;0;kxk2 .