MTH 210: Intro To Linear Algebra Fall 2019 Course Notes .
MTH 210: Intro to Linear AlgebraCourse NotesFall 2019Drew ArmstrongLinear algebra is the common denominator of modern mathematics. From the most pure tothe most applied, if you use mathematics then you will use linear algebra. It is also a relativelynew subject. Linear algebra as we know it was first systematized in the 1920s.1 Since then ithas slowly made its way into the school curriculum and is still increasing in importance everyyear. In the future I expect that introductory linear algebra will be taught as a two-semestersequence, similar to the way that we treat calculus. Until then we have to make do with justone semester. I will do my best to show you the most important ideas.Contents1 Vector Arithmetic1.1 Cartesian Coordinates . . . . . . .1.2 Points vs. Directed Line Segments1.3 Subtraction of Vectors . . . . . . .1.4 The Pythagorean Theorem . . . .1.5 The Dot Product . . . . . . . . . .1.6 The Concept of a Vector Space . .1.7 Exercises with Solutions . . . . . .2279111420252 Systems of Linear Equations2.1 Simultaneous of Equations . . . .2.2 What is a Hyperplane? . . . . . .2.3 What is a Line? . . . . . . . . . .2.4 What is a d-Plane? . . . . . . . .2.5 Systems of Linear Equations . . .2.6 The Idea of Elimination . . . . .2.7 Gaussian Elimination and RREF.35354046474849493 Matrix Arithmetic3.1 Harmless Formalism . . . . . . . .3.2 Concept of a Linear Function . . .3.3 Matrix Multiplication . . . . . . .3.4 Some Examples . . . . . . . . . . .3.5 Matrix Inversion . . . . . . . . . .3.6 Secret: The Fundamental Theorem.494950505151521The ingredients of linear algebra were developed in the 1800s but they didn’t come together as a coherentsubject until the development of quantum physics.1
4 Least Squares Regression534.1 Projection onto a line or plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Projection onto a subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.3 The normal equation AT Ax “ AT b . . . . . . . . . . . . . . . . . . . . . . . . . 535 Spectral Analysis153Vector ArithmeticBefore discussing linear algebra proper, we must first develop the correct language, which isbased on vectors. These, in turn, are based on the concepts of coordinate geometry, so that’swhere we’ll begin.1.1Cartesian CoordinatesThe subject of geometry is fundamentally about points in space. But what is “space,” andwhat is a “point”? Our modern understanding is based on the revolutionary ideas of RenéDescartes and Pierre de Fermat from the early 1600s.The Idea of Coordinate GeometryA point is an ordered list of numbers.Apparently Descartes was lying in bed one morning when he saw a fly buzzing in the corner:2
He realized that the position of the fly could be uniquely specified by measuring the (perpendicular) distance to the two walls and the floor. We can visualize this as a rectangular boxwith dimensions a, b, c. If we arrange these numbers in some fixed order (say a is the distanceto wall 1, b is the distance to wall 2 and c is the distance to the floor), then we can say thatpa, b, cq “ “the (Des)Cartesian coordinates of the fly.”In this course we prefer to express Cartesian coordinates as a vertical column:2 av “ b‚cSuch a column v is called a vector, and we say that a, b, c are the coordinates of the vector.By allowing negative coordinates we can specify the position of any point in space. The pointwith all coordinates equal to zero is called the origin of the coordinate system: 00 “ 0‚.0///But this is more than just a notation because it suggests new things that we can do withpoints. For example, we can “add” them.2Remark: In the notes I will use a boldface font for vectors. On the whiteboard I will use the notationv “ v , since I am not good at drawing boldface letters.3
Addition of Points in the PlaneGiven vectors u “ pu1 , u2 q and v “ pv1 , v2 q we define them sum as follows:3 ˆˆ ˆ u1 v1v1u1:“. u v “u2 v2v2u2That is, we add two vectors by adding their respective components.This definition is completely natural in terms of numbers, but what does it mean in terms ofgeometry? First let us draw the two points in the Cartesian plane (note that I have labeledthe two axes arbitrarily):If we draw the point u v then we observe that the four points 0, u, v and u v lie at thefour vertices of a parallelogram:3Since column vectors take up lots of vertical space I will sometimes write them as a horizontal list separatedby commas.4
We will formalize this observation by calling it the “parallelogram law.”The Parallellogram LawLet u and v be any two vectors.4 Then the four points 0, u, v and u v are the verticesof a parallelogram.2D Example: Let u “ p 2, 2q and v “ p2, 0q. We compute the sum as follows:ˆ ˆ ˆ ˆ 22 2 20u v “ ““.202 02And here is the picture:4These vectors could have two or three coordinates. Later we will allow them to have n coordinates andpretend the the parallelogram lives in “n-dimensional space.”5
3D Example: The parallelogram law also works in 3D. Let u “ p0, 0, 2q and v “ p2, 3, 0q, sothat u v “ p2, 3, 2q. Here is the picture:Do you see the 2D parallelogram living in 3D space? (Actually, it is a rectangle.)We will assume that the parallelogram law holds for vectors with any number n of coordinates,even though we can’t draw a picture of n-dimensional space when n ě 4.6
1.2Points vs. Directed Line SegmentsThe parallelogram law suggests that there is more to a vector than just the point that itrepresents. Sometimes we choose to view the vector v “ pv1 , v2 q as a directed line segment.The subtle thing about this point of view is that we are allowed to pick up the directedline segment and move it, as long as we don’t change its length or direction:It will take a while for you to internalize this idea, so I will repeat myself frequently. Thereason that we want to move vectors is because it allows us to give a purely geometric definitionof vector addition.Addition of Directed Line SegmentsLet u and v be directed line segments. To compute u v we first move v so that its tailcoincides with the head of u. Then u v is defined as the directed line segment fromthe tail of u to the head of v. In other words:directed line segments add head-to-tail.Here is a picture:7
It turns out that the geometric definition of vector addition (head-to-tail) agrees with thealgebraic definition (componentwise addition) because of the parallelogram law:This picture also demonstrates that u v “ v u. In other words, the addition of directedline segments is commutative. The advantage of this idea is that we can add many vectorstogether without worrying about the coordinate system. For example:Because of commutativity, adding the vectors a, b, c, d, e in any order (there are 5! “ 120different ways to do this) always gives the same vector f . [Remark: One must also checkthat addition of vectors is associative: u pv wq “ pu vq w. You will do this on thehomework.] This picture would be a mess if we had to keep track of all the coordinates.In fact, we can draw the same kind of picture for vectors in any number of dimensions. Ididn’t mention it at the time, but the vectors a, b, c, d, e above live in four-dimensional space.8
1.3Subtraction of VectorsWhat about subtraction of vectors? Let u and v be vectors with the same number of coordinates. Which vector w deserves to be called “u v”? Whatever this vector is, it shoulddefinitely satisfy the equationv w “ u.Therefore we have the following picture:Here is the official definition.Subtraction of VectorsLet u “ pu1 , u2 , . . . , un q and v “ pv1 , v2 , . . . , vn q be vectors of the same dimension. Interms of algebra we define u1v1u1 v1 u2 ‹ v2 ‹ u2 v2 ‹ ‹ ‹ ‹u v “ . ‹ . ‹ :“ ‹. . ‚ . ‚ ‚.unvnun vnIn terms of geometry, we place the directed line segments u and v “tail-to-tail.” Thenu v is the directed line segment from the head of v to the head of u:These two definitions agree because of the parallelogram law.9
But there is one more way to think about subtraction. For any directed line segment v we let“ v” denote the directed line segment with the same length, but opposite direction:Then we can think of “u minus v” as “u plus the opposite of v”:u v “ u p vq.Of course we are allowed to pick up any of these vectors and move them around. We observethat everything fits together very nicely:At first it seems confusing that there are so many different ways to think about vectors. Butthis is actually the reason why linear algebra is so useful. When in doubt, here is a usefulmnemonic.Mnemonic: Head Minus Tail(any vector) “ (its head) (its tail)10
Indeed, the following picture demonstrates that for any three points u, v, x, the vector fromv to u is equal to the vector from x v to x u:1.4The Pythagorean TheoremThe subject of geometry deals with distances and angles. For any vector v in the plane wedenote its length with double5 absolute value signs:}v} :“ the length of the directed line segment v.In the plane we can easily compute this length using the Pythagorean Theorem. Suppose thatv “ pv1 , v2 q in Cartesian coordinates. By definition this means that we have a right triangle:5I guess these are supposed to be “blackboard boldface” absolute value symbols.11
Therefore the Pythagorean Theorem tells us that}v}2 “ v12 v22and henceb}v} “ v12 v22 .Using vector subtraction, this same formula allows us to compute the distance between anytwo points in the plane. Consider any points u “ pu1 , u2 q and v “ pv1 , v2 q. Then we have(distance between u and v) “ (length of the line segment from u to v)“ }v u}a“ pv1 u1 q2 pv2 u2 q2 .What about in three dimensions? We can think of the vector v “ pv1 , v2 , v3 q as the directedline segment between two opposite corners of a rectangular box:In order to compute the length }v} we will also draw the line segment from p0, 0, 0q to pv1 , v2 , 0q.Then we obtain two right triangles:12
Applying the Pythagorean Theorem to each triangle separately givesd2 “ v12 v2 ,}v}2 “ d2 v32 ,and hence}v}2 “ d2 v3“ pv12 v22 q v32“ v12 v22 v32 .We can view this as a three dimensional analogue of the Pythagorean Theorem. What abouthigher dimensions? If v “ pv1 , v2 , v3 , v4 q is a directed line segment in “four dimensional space,”is it true thatb}v} “ v12 v22 v32 v42 ?Sure, why not? Whether “four dimensional space” really exists or not, the mathematicalconcept of distance is easy to work with, so we just go ahead and define it.Length and Distance in GeneralFor any vector v “ pv1 , v2 , . . . , vn q we define its length as follows:b}v} :“ v12 v22 vn2 .Then for any points u “ pu1 , u2 , . . . , un q and v “ pv1 , v2 , . . . , vn q we define the distancebetween them as the length of the line segment v u:a(distance between u and v) :“ }v u} “ pv1 u1 q2 pvn un q2 .13
Since the orientation of the line segment doesn’t matter, we also havea}v u} “ }u v} “ pu1 v1 q2 pun vn q2 .We will see later that points in higher dimensional space can be used to represent data setsor sequences of observations. The distance between two data sets represents the amount ofdifference between them. The technique of least squares regression in statistics is based onminimizing the distance between points in higher dimensional space.1.5The Dot ProductAnd what about angles? Let u and v be any two vectors in n-dimensional space. By placingthem tail-to-tail these two vectors determine a 2-dimensional slice of n-dimensional space.Here is a picture:Inside this 2-dimensional slice we can measure the angle between the vectors. Actually, thereare two angles θ and ψ with θ ψ “ 360 . It doesn’t really matter which angle we choosebecause the cosines are the same: cos θ “ cos ψ.In order to compute the angle θ, let us draw the triangle of vectors with sides u, v and u v:14
The law of cosines gives the following relationship between the side lengths and the angle θ:}u v}2 “ }u}2 }v}2 2}u}}v} cos θ.On the other hand, we can compute the length }u v} in terms of Cartesian coordinates.For example, suppose we are working in 2D space with u “ pu1 , u2 q and v “ pv1 , v2 q. Thenapplying our formula for the distance between points gives}u v}2 “ pu1 v1 q2 pu2 v2 q2“ pu21 2u1 v1 v12 q pu22 2u2 v2 v22 q“ pu21 u22 q pv12 v22 q 2pu1 v1 u2 v2 q.We can simplify this formula a bit by substituting u21 u22 “ }u}2 and v12 v22 “ }v}2 to obtain}u v}2 “ }u}2 }v}2 2pu1 v1 u2 v2 q.By comparing the two boxed equations we conclude that}u}}v} cos θ “ u1 v1 u2 v2 .This is a rather strange formula. It relates the geometric concepts of length and angle onthe left to a surprising computation on the right involving the Cartesian coordinates. Thisformula was discovered in the 1840s by the Irish mathematician and physicist William RowanHamilton, but his notation was a bit metaphysical. In the 1880s, the scientists Josiah WillardGibbs and Oliver Heaviside distilled the most useful ideas from Hamilton and they definedthe modern language of vectors. Here is their most important definition.Definition of the Dot ProductFor any vectors u “ pu1 , u2 , . . . , un q and v “ pv1 , v2 , . . . , vn q we define the dot product:u ‚ v :“ u1 v1 u2 v2 un vn .Note that this operations takes a pair of vectors u and v to a single number u ‚ v (calleda scalar). For this reason the dot product is sometimes also called the scalar product.6 Itis worth noting that the length of a vector can be expressed in terms of the dot product:a?}u} “ u21 u22 u2n “ u ‚ u}u}2 “ u21 u22 u2n“ u ‚ u.This definition allows us to state the following theorem.6Sometimes it is also called the inner product of vectors. We will see why later when we discuss the outerproduct of vectors.15
Theorem (The Angle Between Two Vectors)If θ is the angle between the vectors u and v (measured in the two-dimensional planethat they share), then the same argument as above can be used to show that7u ‚ v “ }u}}v} cos θ.It follows from this that the vectors u and v are perndicular (written u K v) if and onlyif the dot product is zero:8uKvðñcos θ “ 0ðñu ‚ v “ 0.Finally, we can use this formula to solve for (the cosine of) the angle θ in terms of the Cartesiancoordinates of u and v:cos θ “u‚vu1 v1 u2 v2 un vna“a 2.}u}}v}u1 u2n v12 vn2Let’s see some examples.2D Example: Compute the angle θ between the vectors u “ p2, 3q and v “ p5, 2q:We apply the dot product formula to obtaincos θ “u‚v2 5 3 216? “ 0.824“?“?2222}u}}v}13 292 3 5 27Occasionally you will see a textbook that uses the formula u ‚ v “ }u}}v} cos θ as the definition of thedot product. I don’t like that.8Here we assume that u ‰ 0 and v ‰ 0 so that }u}}v} ‰ 0. I guess we could say that the zero vector isperpendicular to every vector.16
and henceθ “ arccosp0.824q “ 34.5 or 325.5 .Based on our picture we choose the smaller angle.///1D Example: It might seem a bit silly, but we should make sure that we understand theformula u ‚ v “ }u}}v} cos θ in very simple case when n “ 1.Let u “ puq be any “1-dimensional vector,” also known as a “real number.” Observe that thelength of u is the same as the absolute value of u:}u}2 “ u2?}u} “ u2 “ u .Indeed, this is why we use the absolute value sign for the length of a vector. If v “ pvq is anyother number, observe that the dot product of u and v is just the product of numbers:u ‚ v “ uv.If θ is the angle between u “ puq and v “ pvq then we conclude thatuv “ u v cos θ.Does this make any sense? Sure. The only possible values for θ are 0 (when u and v have thesame sign) and 180 (when u and v have opposite signs). Since cosp0 q “ 1 and cosp180 q “ 1we have the following rule for the product of real numbers:# u v if u and v have the same sign,uv “ u v if u and v have opposite signs.This is correct, and interesting.///3D Example: The Tetrahedral Angle. Every chemistry textbook quotes θ “ 109.5 asthe angle between two hydrogen atoms in a molecule of methane (or any other molecule withtetrahedral symmetry):17
Why is this the correct answer?In order to minimize energy, the four hydrogen atoms sit at the vertices of a regular tetrahedroncentered on the carbon atom. Let’s place the carbon atom at the origin 0 “ p0, 0, 0q. Thenwhat are the coordinates of the hydrogen atoms? First let me observe that the eight pointsp 1, 1, 1q form the vertices of a cube centered at the origin:By choosing four alternate vertices, we obtain a regular tetrahedron centered at 0. There aretwo choices and it doesn’t matter which one we pick. For example, let’s place the hydrogenatoms at p 1, 1, 1q, p1, 1, 1q, p1, 1, 1q and p 1, 1, 1q:18
Since the angle between any two hydrogen atoms is the same, we can can pick two at random,say u “ p 1, 1, 1q and v “ p1, 1, 1q. Then the dot product formula givescos θ “u‚v 1 1p 1q 1 1 p 1q 1 1a“? ? ““a,222122}u}}v}33 3p 1q 1 1 1 p 1q 1and it follows thatˆθ “ arccos 13 “ 109.4712206 degrees.///4D Example: Compute the angle in the following picture:9In order to compute the dot product u ‚ v we must first translate the vectors u and v intostandard position. For this purpose, we subtract p0, 1, 1, 0q from all three points to obtain9Note that the three points of the triangle live in a 2D slice of 4D space. The angle is defined inside this 2Dslice.19
Then we compute 11 1‹ 1 ‹‹ ‹u‚v “ 2‚‚ 0 ‚ “ 1 p 1q 1 1 2 0 2 1 “ 2,12 11‹ ‹ 12222‹ 1‹}u}2 “ 2‚‚ 2‚ “ 1 1 2 2 “ 10,22 1 1 ‹ ‹1‹ ‚ 1 ‹ “ p 1q2 12 02 12 “ 3.}v}2 “ 0‚ 0‚11Plugging these into the theorem givescos θ “u‚v2? “ 0.365“?}u}}v}10 3and henceθ “ arccosp0.365q “ 68.6 .1.6The Concept of a Vector SpaceIn the previous sections we developed the basic properties of geometry in terms of Cartersiancoordinates. From the modern point of view, we take these formulas as the definition of(Euclidean)10 geometry.Definition of Euclidean SpaceRecall that R denotes the set of real numbers. We define Rn as the set of ordered n-tuplesof real numbers, thought of as column vectors: , x1’/&. . ‹nR :“ x “ . ‚ : x1 , . . . , xn P R ./’%xn10There exist non-Euclidean geometries, such as the Lorentzian geometry which is used in Einstein’s theoryof relativity.20
For any two column vectors x, y P R2 we define their dot product as follows: y1x1 . ‹ . ‹x ‚ y “ . ‚‚ . ‚ :“ x1 y1 xn yn .xnynWe define the length }x} P R so that the generalized Pythagorean Theorem holds:}x}2 “ x ‚ x “ x21 x2n .Then the angle θ between x and y satisfies the following equation, which is analogous tothe law of cosines:x‚yx‚ycos θ ““?}x}}y}x‚x y‚yThe set Rn together with the dot product function ‚ : Rn ˆRn Ñ R is called n-dimensionalEuclidean space.There is one further algebraic operation that we can define on vectors. For any vector v “pv1 , . . . , vn q P Rn we have defined v1 v1 ‹ ‹ v “ . ‚ “ . ‚vn vnand we have observed that v is the vector with the same length as v, but with oppositedirection. More generally, we make the following definition.Definition of Scalar MultiplicationLet v “ pv1 , . . . , vn q P Rn be any n-dimensional vector and let t P R be any real number,called a scalar. Then we define the vector tv P Rn as follows: v1tv1 . ‹ . ‹tv “ t . ‚ :“ . ‚.vntvnWhat is the geometric meaning of this operation? We observe that the length of tv satisfiesbb?a}tv} “ ptv1 q2 ptvn q2 “ t2 pv12 vn2 q “ t2 v12 vn “ t }v .21
In other words, multiplying a vector v P Rn by a number t P R “scales its length” by the factor t . This is why we call numbers scalars. The direction of tv is the same as v when t ą 0 andit is opposite to v when t ă 0. The concept of scalar multiplication is easier to understand ifyou visualize the set of points ttv : t P Ru as the infinite line generated by the vector v:We say that this line is “parametrized by t.” I would even call it the “t-axis.” Now we havethree different algebraic operations related to vectors:11 Addition : Rn ˆ Rn Ñ Rn takes a (vector,vector) pair to a vector. Scalar multiplication : R ˆ Rn Ñ Rn takes (scalar,vector) pair to a vector. The dot product ‚ : Rn ˆ Rn Ñ R takes a (vector,vector) pair to a scalar.It is worth recording all of the abstract properties that these operations must satisfy.Properties of Vector ArithmeticFor any vectors u, v, w P Rn and for any scalars a, b P R we have u v “v u u pv wq “ pu vq w u 0“u 0u “ 0 1u “ u apbuq “ pabqu11For any sets S and T , the notation S ˆ T denotes the set of ordered pairs ps, tq with s P S and t P T . Ithas nothing to do with vector multiplication.22
apu vq “ au av pa bqu “ au bu u‚v “v‚u u ‚ pv wq “ u ‚ v u ‚ w apu ‚ vq “ pauq ‚ v “ u ‚ pavqThese properties are mostly obvious so we won’t bother proving them.12 Instead, here is anexample showing how to use the rules.?D Example: Let x and y be any two vectors satisfyingx‚x“y‚y “1and x ‚ y “ 0.(I won’t tell you the dimension of these vectors because it doesn’t matter.) Compute the anglebetween the vectors u “ x 2y and v “ 3x y.Solution 1: We can solve this problem without thinking by applying the rules of vectorarithmetic. First we compute the dot products u ‚ u, v ‚ v and u ‚ v:u ‚ u “ px 2yq ‚ px 2yq “ x ‚ x 4x ‚ y 4y ‚ y “ 1 4 0 4 1 “ 5,v ‚ v “ p3x yq ‚ p3x yq “ 9x ‚ x 6x ‚ y y ‚ y “ 9 1 6 0 1 “ 10,u ‚ v “ px 2yq ‚ p3x yq “ 3x ‚ x 7x ‚ y 2y ‚ y “ 3 1 7 0 2 1 “ 5.If θ is the angle between u and v then it follows thatcos θ “ ?51u‚v?“? ? “?u‚u v‚v5 102and henceˆθ “ arccos1?2 “ 45 .Solution 2: Or we can draw a picture. The identities x ‚ x “ y ‚ y “ 1 tell me that}x} “ }y} “ 1 and the identity x ‚ y “ 0 tells me that x and y are perpendicular. We saythat x and y are perpendicular unit vectors. Here is a picture:12You will prove the identity u pv wq “ pu vq w on the homework.23
(This picture might live in 10-dimensional space. It doesn’t matter.) Now we can add thevectors u “ x 2y and v “ 3x y to our picture:From this point of view we see that the angle between u and v is the same as the anglebetween the vectors p1, 2q and p3, 1q in the Cartesian plane. It follows thatcos θ “1p1, 2q ‚ p3, 1q1 3 2 15?“?“? ? “? ,}p1, 2q}}p3, 1q}5 10212 22 32 11and we get the same answer θ “ 45 . The key idea here is to treat the vectors x and y as the“basis for a coordinate system.” Then we might as well express any vector ax by using thefollowing notation:ˆ aax by “.bSince the basis vectors x and y are perpendicular and have length 1 (we say that they forman “ortho-normal basis”) it turns out that any computation using this language will give the24
correct answer. For example:13ˆ ˆ 31“ 1 3 2 1 “ 5.‚u ‚ v “ px 2yq ‚ p3x yq “12///The point of view in this example is rather abstract. There are two reasons that we might wantto work with an abstract coordinate system such as x and y instead of the usual Cartesiancoordinates: Some objects, such as a 1D line or a 2D plane in 3D space, do not come with a standardcoordinate system. In this case we have to introduce our own coordinate system. Wewill see many examples of this in the next chapter. In the twentieth century, mathematicians observed that there are structures other thanCartesian space that obey the same rules of vector arithmetic. (We call these abstractvector spaces.) For example, in the theory of probability we can think of random variablesas “vectors” and we can think of covariance as the “dot product.” For another example,the subject of signal processing is based on the idea of treating a signal as some kind of“vector.” If f ptq represents the amplitude of a signal at time t, then the “dot product”between two signals f ptq and gptq is defined by the integralżf ptqgptq dt.These kinds of abstract vector spaces do not come with a built-in coordinate system; ifyou want to work in coordinates then you have to define your own.14 We won’t purseexotic kinds of vector spaces in this course, but you should be aware that they exist andthat they follow the same general theory.1.7Exercises with Solutions1.A. Define the standard basis vectors e1 “ p1, 0, 0q, e2 “ p0, 1, 0q and e3 “ p0, 0, 1q.(a) Draw the cube with the following 8 vertices:0, e1 , e2 , e3 , e1 e2 , e1 e3 , e2 e3 , e1 e2 e3 .(b) Draw the triangle in 3D with corners at e1 , e2 , e3 . Compute the side lengths and theangles of this triangle by using the dot product.13Remark: If the basis vectors x and y are not ortho-normal then addition and scalar multiplication are stillcorrect, but the dot product will give the wrong answer.14In the theory of signal processing it is common to use the “pure sine waves” sinptq, sinp2tq, sinp3tq, . . . andcosptq, cosp2tq, cosp3tq, . . . as a coodinate system.25
(a) Observe that the cube is made out of many parallelograms (actually, they are squares):(b) Now consider the dotted triangle with vertices e1 , e2 , e3 . Since each side of the triangleis the diagonal? of a unit square, we conclude from the Pythagorean Theorem that each sidehas length 2. Then since all three sides of the triangle have the same length, we concludethat the three interior angles are equal, i.e., each is 60 . Let’s check this result using the dotproduct.Let u and v be the vectors with tails at e1 and heads at e2 and e3 , respectively. In orderto compute the dot products u ‚ u, v ‚ v and u ‚ v we translate these vectors into standardposition by subtracting e1 from all three points:26
Using the fact that up 1, 1, 0q and v “ p 1, 0, 1q givesu ‚ u “ p 1q2 12 02 “ 2,v ‚ v “ p 1q2 02 12 “ 2,u ‚ v “ p 1qp 1q 1 0 0 1 “ 1.We conclude that }u} “ }v} “?2 andcos θ “ ?u‚v11?“? ? “ ,u‚u v‚v22 2which implies θ “ arccosp1{2q “ 60 . Alternatively, we can use vector arithmetic and the factthat the standard basis vectors satisfy#1 if i “ j,ei ‚ ej “0 if i ‰ j.Then since u “ e2 e1 and v “ e3 e1 we haveu ‚ u “ pe2 e1 q ‚ pe2 e1 q “ e2 ‚ e2 2e1 ‚ e2 e1 ‚ e1 “ 1 2 0 1 “ 2,v ‚ v “ pe3 e1 q ‚ pe3 e1 q “ e3 ‚ e3 2e1 ‚ e3 e1 ‚ e1 “ 1 2 0 1 “ 2,u ‚ v “ pe2 e1 q ‚ pe3 e1 q “ e2 ‚ e3 e1 ‚ e2 e1 ‚ e3 e1 ‚ e1 “ 0 0 0 1 “ 1.The computations for the other two angles are similar, so we omit them.1.B. Let u “ p1, 2q and v “ p3, 1q.(a) Draw the points u and v together with the points11u v,2213u v,4411u v,44u v.(b) Draw the infinite line ttv tuu where t is any real number. [Hint: It is enough to drawtwo points on this line and then use a ruler.](c) Draw the infinite line tsu p1 sqvu where s is any real number. [Hint: Same as (b).](d) Shade the finite region of the plane defined by tsu tv : 0 ď s ď 1, 0 ď t ď 1u.(e) Shade the infinite region of the plane defined by tsu tv : 0 ď s, 0 ď tu.(a) First we observe that the points 0, u, v and u v are the four vertices of a parallelogram.Next let me observe that11u v1u v “ pu vq “.222227
We can view this as the fourth vertex of the parallelogram with vertices 0, u{2 and v{2. Orwe can view it as half of the vector u v. Or we can view it at the midpoint (the center ofmass) of the two points u and v. More generally, the midpoint of any two points x, y P Rnis px yq{2. (Can you explain why?) It follows that 14 u 14 v “ 14 pu vq is the midpointbetween 0 and 21 pu vq, and 41 u 34 v is the midpoint between 12 u 12 v and v:ˆ pu{2 v{2q vu{2 3v{21 1313““u v “ u v.222 2244Here is the picture:(b) I have drawn the line tu tv “ tpu vq in the picture above. The line passes throughthe points 0, pu vq{4, pu vq{2 and u v, in that order.(c) I have drawn the line su p1 sqv in the picture above. This line passes through thepoints v, u{4 3v{4, u{2 v{2 and u, in that order. We can also think of this as the linev spu vq that starts at the point v when s “ 0 and then moves in the direction of thevector u v as s increases.(d) The set of points tsu tv : 0 ď s ď 1, 0 ď t ď 1u is the filled parallelogram with verticesat 0, u, v and u v:28
We call this the “parallelogram spanned by u and v.”(e) The set of points tsu tv : 0 ď s, 0 ď tu is the “cone” spanned by u and v:We can think of this as the “first quadrant” of the coordinate system defined by u and v.This coordinate system is a bit squished.1.C. Let u and v be any two vectors of length 2. Compute the following dot products:(a) u ‚ p uq(b) pu vq ‚ pu vq(c) pu 2vq ‚ pu 2vq(d) Explain why you do not have enough information to compute pu vq ‚ pu vq.(a) Since u has length 2 we know that u ‚ u “ }u}2 “ 4 and henceu ‚ p uq “ u ‚ pp 1quq “ p 1qu ‚ u “ 4.29
(b) Since v has length 2 we also know that v ‚ v “ }v}2 “ 4 and hencepu vq ‚ pu vq “ u ‚ u u ‚ v u ‚ v v ‚ v “ u ‚ u v ‚ v “ 4 4 “ 0.Evidently the vectors u v and u v are perpendicular. I do not know the angle between uand v, but here is a picture:We can regard this as a theorem that the two diagonals of a rhombus are perpendicular. [Ifu and v did not have the same length, this parallelogram would not be a rhombus and itsdiagonals would not be perpendicular.](c) We have pu 2vq ‚ pu 2vq “ u ‚ u 2u ‚ v 2v ‚ u 4v ‚ v “ u ‚ u 4v ‚ v “ 12.I cannot dr
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