We define a Euclidean space as a space in which the axioms and conclusions of classical Euclidean Geometry are valid. Thus, a Euclidean space is meant to describe the physical space of our everyday experience.

Admittedly, this definition is not beyond reproach. First of all, mathematicians have for centuries expressed doubt with respect to the internal consistency of Euclid's framework and continue to do so to this day. Second, it has never been a foregone conclusion that Euclidean Geometry accurately describes the three-dimensional space of our experience. This profound question took on even greater significance since Albert Einstein's General Theory of Relativity was put forth in 1915. Nevertheless, we will not let these eternal uncertainties impede our progress. One must find a reasonable starting point and use it as a fulcrum for moving forward. Ours is the Euclidean space as most of us intuitively understand it. From it, we will proceed to develop an analytical framework that will later compel us to return to our starting point and reexamine its assumptions.

In addition to the three-dimensional Euclidean space, we may also consider two- and one-dimensional Euclidean spaces. A two-dimensional Euclidean space is a plane. A one-dimensional Euclidean space is a straight line. The plane will provide us with many of our concrete examples since planar objects are often easier to visualize than their three-dimensional counterparts. The straight line, on the other hand, is not a sufficiently rich object for many illustrations and will not figure frequently in our discussions. Nevertheless, it is an important concept that we must keep in mind when making general statements about Euclidean spaces.

Our geometric approach does not allow for four- and higher-dimensional Euclidean spaces. Unlike two- and one-dimensional spaces, four- and higher-dimensional spaces cannot be visualized and, therefore, our intuitive understanding of a Euclidean space cannot be extended to dimensions greater than three. While this point is more-or-less uncontroversial, it is often met with surprise. The reason behind the surprise is the ingrained association, or even equivalence, between Euclidean spaces and $\mathbb{R} ^{3}$. And as $\mathbb{R}^{3}$ extends effortlessly to $\mathbb{R}^{4}$ and beyond, Euclidean spaces are often brought along for the ride, as if it were the very same thing. However, it is not the very same thing, it is a very different thing, and it is very important to break the ill-conceived equivalence between Euclidean space and $\mathbb{R}^{n}$, so that a well-conceived association can be properly constructed.

It is worth reiterating the utmost importance of accepting the Euclidean space and objects within it as pure geometric objects. The tradition of imagining a coordinate system imposed upon the space has become so deeply entrenched, that for some of us it is difficult to imagine the geometric space without also imagining a coordinate system. Relearning to do so may take some time as well as a certain degree of introspection. But it is essential. The goal of Tensor Calculus is effective use of coordinate systems which is not possible without realizing that coordinate systems are meant to serve geometric ideas and not to replace them.

The primary object in a Euclidean space is a directed segment known as a geometric vector: We will tend to drop the word geometric with the understanding that the term vector refers to a directed segment. We will denote vectors by bold capital letters, such as

Despite its apparent simplicity, the geometric vector is an object of great richness as it combines geometric and algebraic characteristics. Its geometric nature is self-evident. Its algebraic nature is found in the fact that vectors are subject to the operations of addition, multiplication by numbers, and the dot product.

We should note that the very ability to accommodate vectors is an important characteristic of a Euclidean space that is consistent with the fact that we intuitively understand the space of our experience to be fundamentally straight. Yet, we have made no attempt to define either space or straight. For the sake of forward progress, we will leave these terms undefined and simply agree that we all intuitively understand these concepts in the same way and, furthermore, that we recognize that the physical space of our everyday experience can accommodate straight lines. We may continue to worry that some of our assumptions are vague or, worse, intrinsically inconsistent. Should that be the case, we hope that our future investigations will help clarify some of the vagueness, shed light on the existing inconsistencies, and present us a way of removing some of the more glaring inconsistencies in favor of more subtle and therefore more profound contradictions to be addressed at a later date.

One of the intrinsic concepts available in a Euclidean space is that of length. To highlight its primary, axiomatic nature, it is referred to as Euclidean length. We will use the symbol to denote the length of the vector $\mathbf{U}$.

In order to define a unit of length, an arbitrary segment is singled out as a reference. Then the length of any other segment is expressed by a multiple relative to the reference segment.

The length of a geometric vector is also known as its magnitude or absolute value. The term length is used more commonly for vectors that represent geometric quantities such as displacements. Meanwhile, the terms magnitude and absolute value are used for vectors corresponding to physical quantities such as velocity, acceleration, and force.

Vectors can be added together and multiplied by numbers. Addition is carried out according to the well-known tip-to-tail rule: given two vectors $\mathbf{U}$ and $\mathbf{V}$, their sum $\mathbf{W=U}+\mathbf{V}$ is constructed by shifting the tail of $\mathbf{V}$ to the tip of $\mathbf{U}$ and connecting the tail of $\mathbf{U}$ to the tip of $\mathbf{V}$. This construction is illustrated in the following figure. This definition makes intuitive sense since it agrees with our understanding of movements.

An equivalent way to add to vectors is by the equally-well-known parallelogram rule. According to the parallelogram rule, the tails of $\mathbf{U}$ and $\mathbf{V}$ are placed at the same point and their sum is the vector connecting their tails to the fourth vertex of a completed parallelogram, as illustrated in the following figure. The advantage of the parallelogram rule is that it assigns equal roles to the two vectors and does not require shifting one of the vectors. Its disadvantage, however, is that it does not work when the two vectors are collinear or when one or both of the vectors are zero. We must also remark that both definitions rely on the much-debated Fifth Postulate, which reminds us that mathematical origins are always accompanied by doubt.

Next, let us describe multiplication of geometric vectors by numbers. If $\alpha$ is a positive number, then the product $\mathbf{W}=\alpha\mathbf{U}$ is the vector that points in the same direction as $\mathbf{U}$ and has length that equals the product of $\alpha$ and the length of $\mathbf{U}$. If $\alpha$ is negative, the direction of the resulting vector is reversed, i.e. $\alpha\mathbf{U}$ points in the direction opposite of $\mathbf{U}$, and its length is the product of $\left\vert \alpha\right\vert$ and the length of $\mathbf{U}$. Since multiplication by a number has the effect of scaling, i.e. changing the length of the vector, the word scalar is commonly used to mean number.

An expression that combines multiplication by numbers with addition, such as is known as a linear combination.

Vectors can also be subtracted. One way to define the difference is in terms of addition, i.e. is a vector such that Alternatively, $\mathbf{V}-\mathbf{U}$ can be defined as the linear combination It is a straightforward task to show that the two definitions are equivalent.

The result of a subtraction is easier to construct geometrically than that of addition. If the tails of the vectors $\mathbf{U\ }$and $\mathbf{V}$ are at the same point then $\mathbf{V}-\mathbf{U}$ is the vector from the tip of $\mathbf{U}$ to the tip of $\mathbf{V}$. This construction is illustrated in the following figure.

Note that our entire discussion has been, and will continue to be, conducted without introducing a coordinate system. Furthermore, vector operations have been described without a reference to their components with respect to some basis. Hopefully, this observation helps convince the reader that geometric vectors in a Euclidean space represent a logically consistent and algebraically complete mathematical entity.

As Tensor Calculus is a close descendant of Linear Algebra in the sense of merging the worlds of Algebra and Geometry, it is not surprising that our narrative will rely heavily on various fundamental concepts from Linear Algebra such as linear independence, basis, linear decomposition, linear transformation, eigenvalues, and others. In particular, we will frequently describe geometric vectors as forming a linear space. In this context, it is important to think of all vectors that may be potentially added or subtracted as emanating from a single point. When vectors appear in a different arrangement -- for instance, when implementing the tip-to-toe construction -- it should be considered a transient configuration created for some temporary convenience.

Of course, we will later consider vector fields, i.e. collections of vectors defined at different points in space. In the context of a vector field, vectors naturally emanate from different points. However, for the purposes of performing algebraic analysis on those vectors, we should once again treat them as emanating from a single point.

Note that this point of view will become particularly relevant in the near future when we introduce curvilinear coordinate systems. In the context of a curvilinear coordinate system, the concept of a linear space becomes highly localized. Namely, at each point there will be its own unique linear space of geometric vectors that is completely separate from its neighbors.

Without a doubt, the dot product is one of the most beautiful concepts in Mathematics. Before we present its well-known definition, we should note one of its most important and, at the same time, surprising properties: with the help of the dot product, it is possible to express virtually all geometric quantities by algebraic expressions. It is this particular universality of the dot product that ultimately enables coordinate space analysis to become a logically complete self-contained framework.

Given two vectors $\mathbf{U}$ and $\mathbf{V}$, the dot product $\mathbf{U}\cdot\mathbf{V}$, is the product of their lengths and the cosine of the angle $\gamma$ between them, i.e. Once again, note the geometric nature of this definition: it is stated in terms of directly measurable geometric quantities and does not make any references to the components of the vectors.

It is clear that the dot product is symmetric, i.e. Furthermore, it is linear with respect to vector addition, i.e. and multiplication by numbers, i.e. These two forms of linearity are usually combined into the single distributive property While commutativity and linearity with respect to multiplication by numbers are rather obvious, linearity with respect to vector addition needs a careful demonstration. This is left as an exercise for the reader.

For a fixed vector $\mathbf{G}$, think of the dot product $\mathbf{G} \cdot\mathbf{X}$ as a function $F$ of $\mathbf{X}$, i.e. Then, thanks to the distributive property of the dot product, $F\left( \mathbf{X}\right)$ is a linear function, i.e. Importantly, the converse is also true: every linear function $F\left( \mathbf{X}\right)$ can be represented as a dot product with a unique vector $\mathbf{G}$. We will make use of this crucial insight when discussing directional derivatives in Chapter 4.

Let us now discuss a few examples that illustrate how the dot product can be used to express various geometric quantities.

First, consider the dot product $\mathbf{U}\cdot\mathbf{U}$ of a vector $\mathbf{U}$ with itself. The angle $\gamma$ between a vector and itself is understood to be $0$ and therefore Consequently, the length of a vector equals the square root of the dot product with itself, i.e.

Of course, one may be skeptical with regard to the utility of this identity. Clearly, this is not a practical way to determine the length of a vector. After all, in order to evaluate a dot product in the first place, one must know the lengths of the vectors involved. However, from the conceptual of view, this equation is indispensable. First, as we will see in Section 2.6, the dot product can be effectively evaluated in the component space by cataloging the values of the dot product for the elements of a basis. In other words, having evaluated the dot product for a handful of vectors by the geometric definition, the dot products of any two vectors can be evaluated by working with their components. Second, as we will discuss in Section 2.7, in more general vector spaces, the dot product is a primary concept that precedes those of length and angle. Consequently, in more general vector spaces, the identity acts as the definition of length.

It is a similar story for the dot product as a test of orthogonality. If two vectors $\mathbf{U}$ and $\mathbf{V}$ are orthogonal then their dot product is zero since The converse is also true for nonzero vectors: if the dot product of two nonzero vectors is zero then the vectors are orthogonal. More generally, the dot product can be used to measure the angle $\gamma$ between two nonzero vectors according to the formula that follows directly from the definition of the dot product, i.e. Importantly, we can write the expression on the right strictly in terms of the dot product which helps illustrate the general idea that all geometric quantities can be expressed in terms of the dot product.

As a further illustration of the great utility of the dot product, let us show how it can be used to construct an algebraic proof of the Pythagorean theorem for a right triangle with legs of lengths $a$ and $b$ and a hypotenuse of length $c$: The following figure illustrates two ways of representing a right triangle by associating its legs with orthogonal vectors $\mathbf{U}$ and $\mathbf{V}$ and the hypotenuse with a third vector $\mathbf{W}$. In the first configuration while in the second Both configurations work equally well for our purposes. Continuing with the first configuration, dot each side of the identity with itself By the combination of the distributive and commutative properties, Since $\mathbf{U}\cdot\mathbf{U}=\operatorname{len}^{2}\mathbf{U}$, $\mathbf{V}\cdot\mathbf{V}=\operatorname{len}^{2}\mathbf{V}$, and $\mathbf{U}\cdot\mathbf{V}=0$, the above identity becomes which is precisely the statement of the Pythagorean theorem. This proof is so simple that it often leaves one with the feeling that some circular logic must have been used somewhere along the way.

The Pythagorean theorem is essentially a planar identity. However, there also exists a three-dimensional version for the diagonal of a rectangular parallelepiped illustrated in the following figure. If the length of the diagonal is $d$ and the lengths of the sides are $a$, $b$, and $c$, then the theorem states that This theorem can be proven in similar fashion by representing the sides of the parallelepiped with orthogonal vectors $\mathbf{U}$, $\mathbf{V}$, and $\mathbf{W}$ and the diagonal by a vector $\mathbf{X}$ as shown in the following figure: Then $\mathbf{X}$ is the sum of $\mathbf{U}$, $\mathbf{V}$, and $\mathbf{W}$, i.e. and it is left as an exercise to show that

We now turn our attention to the first of three other crucial applications of the dot product that include linear decomposition by the dot product, orthogonal projection, and the evaluation of the dot product in the component space. Note that these discussions will not make a single reference to the explicit definition of the dot product Instead, they will rely exclusively on its commutative and distributive properties This important insight shows that the dot product has a great deal of algebraic autonomy.

In a classical application, the dot product is used to construct an algebraic algorithm for linear decomposition. Since the goal of Tensor Calculus is to provide an analytical framework, this approach will be used almost exclusively when facing a linear decomposition task.

Given a basis $\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}$ and a vector $\mathbf{U}$, linear decomposition is the task of finding the coefficients of the linear combination that represents $\mathbf{U}$ with respect to the basis $\mathbf{b} _{1},\mathbf{b}_{2},\mathbf{b}_{3}$. The resulting coefficients $U_{1}$, $U_{2}$, and $U_{3}$ are known as the components of the vector $\mathbf{U}$ with respect to the basis $\mathbf{b}_{1},\mathbf{b} _{2},\mathbf{b}_{3}$. We will now show that the components $U_{1}$, $U_{2}$, and $U_{3}$ can be determined by evaluating dot products without making any additional references to the relative arrangement of the vectors $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, $\mathbf{b}_{3}$, and $\mathbf{U}$.

First, assume that the basis is orthogonal. In other words, the vectors $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, and $\mathbf{b}_{3}$ are orthogonal to each other which, in terms of the dot product, reads In order to determine the coefficient $U_{1}$, dot both sides of the identity with $\mathbf{b}_{1}$ to obtain Apply the distributive property on the right, i.e. and notice that all terms on the right but the first one vanish due to the orthogonality of the basis. Thus, and, solving for $U_{1}$, we find Obviously, the same argument works for $U_{2}$ and $U_{3}$ so, collectively, we have Thus, for an orthogonal basis, all coefficients $U_{i}$ can be determined one at a time by evaluating two dot products, $\mathbf{b}_{i}\cdot\mathbf{U}$ and $\mathbf{b}_{i}\cdot\mathbf{b}_{i}$.

This rule simplifies even further for an orthonormal (a.k.a. Cartesian) basis, where the vectors $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, and $\mathbf{b}_{3}$ are not only orthogonal to each other but each one is also unit length, i.e. Thus, for an orthonormal basis, the coefficient $U_{i}$ is given by and can therefore be determined by evaluating a single dot product.

Does this method continue to work when the basis is not orthogonal? The answer is yes, although the expansion coefficients can no longer be determined one at a time or in as few as two dot products. Instead, the coefficients need to be determined simultaneously as a set by solving a linear system of equations. The system is formed by treating the result of multiplying the identity $\mathbf{U}=U_{1}\mathbf{b}_{1}+U_{2}\mathbf{b} _{2}+U_{3}\mathbf{b}_{3}$ with each of the basis vectors as an equation for $U_{1}$, $U_{2}$, and $U_{3}$. We have Rewrite this system in the matrix form, i.e. The resulting matrix consists of the pairwise dot products of the basis vectors. In Linear Algebra, it is known as the Gram matrix or the inner product matrix. We will eventually refer to it as the covariant metric tensor.

The three coefficients $U_{1}$, $U_{2}$ and $U_{3}$ can be determined by solving the system above. While this can be time consuming, it nevertheless remains true that linear decomposition can be accomplished by evaluating dot products in combination with other elementary arithmetic operations. In terms of the matrix $M$ -- or, more specifically, in terms of its inverse, which we will later call the contravariant metric tensor -- the solution can be written in the form While it is well known that calculating the inverse is an inefficient way of solving linear systems, the concept of the inverse offers an effective way of capturing the solution of a linear system by an algebraic expression.

Finally, we note one important corollary of the above identity: a vector is uniquely determined by the values of the dot products with the elements of a basis.

Let us now turn our attention to the question of projection onto a plane which will yield an answer surprisingly similar to that of the question of decomposition. Consider a plane spanned by the vectors $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$ and a vector $\mathbf{U}$ that lies out of the plane. Let the vector $\mathbf{P}$ be the orthogonal projection of $\mathbf{U}$ onto the plane.

Since $\mathbf{P}$ is the closest vector to $\mathbf{U}$ within the plane we will reuse the symbols $U_{1}$ and $U_{2}$ to denote its components with respect to the basis $\mathbf{b}_{1},\mathbf{b}_{2}$, i.e. The components of the vector $\mathbf{U}$ itself with respect to some basis in the surrounding Euclidean space will not figure in our discussion. Our goal is to find the expressions for $U_{1}$ and $U_{2}$ in terms of the dot products of the vectors $\mathbf{U}$, $\mathbf{b}_{1}$, and $\mathbf{b}_{2}$. To this end, note that, by definition, the difference $\mathbf{U}-\mathbf{P}$, in other words, the vector is orthogonal to the plane and is, therefore, orthogonal to the vectors $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$. Thus, its dot products with $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$ must vanish, i.e. Multiplying out the expressions on the left yields Organizing these equation into matrix form yields the linear system for the coefficients $U_{1}$ and $U_{2}$. Thus, $U_{1}$ and $U_{2}$ are given by The similarity between this formula and the decomposition formula is evident. In fact, the coefficients produced by the expression can be interpreted in two different ways. For a vector $\mathbf{U}$ within the plane spanned by $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$, the two coefficients are the components of $\mathbf{U}$. For a vector $\mathbf{U}$ outside of the plane, the two coefficients are the components of the projection of $\mathbf{U}$ onto the plane, i.e. the vector in the plane closest to $\mathbf{U}$.

The matrix $M$ also has another crucial application which may, perhaps, be described as more important than linear decomposition. As we are about to show, the matrix $M$ represents the dot product in the component space. Suppose that the components of the vectors $\mathbf{U}$ and $\mathbf{V}$ are $U_{1}$, $U_{2}$, $U_{3}$ and $V_{1}$, $V_{2}$, $V_{3}$, i.e. Let us introduce the vectors (in the Linear Algebra sense) in $\mathbb{R}^{3}$ whose entries consist of the components of $\mathbf{U}$ and $\mathbf{V}$. The question is: how can the dot product $\mathbf{U} \cdot\mathbf{V}$ be calculated in terms of the matrix $M$ and the vectors $U$ and $V$?

This question can be answered in straightforward fashion by substituting the linear expansions of $\mathbf{U}$ and $\mathbf{V}$ into the dot product $\mathbf{U}\cdot\mathbf{V}$, i.e. A repeated application of the distributive law yields nine terms, i.e. The resulting expression can be written concisely in two ways. The first approach uses the summation symbol $\sum$ and captures the fact that the above sum consists of all possible terms of the form $U_{i}V_{j}\mathbf{b}_{i} \cdot\mathbf{b}_{j}$. We have This is the form that we will ultimately favor once we introduce the tensor notation. At this point, however, a matrix expression will provide greater insight. In matrix terms, the sum of the nine terms can be captured by the matrix product With the help of the symbols $U$, $V$, and $M$, the same identity can be written more compactly as

The above identity represents the rule for carrying out the dot product in the component space. It clearly shows that the matrix $M$ encapsulates all the required information to calculate the dot product in the component space.

Finally, we note one important special case. If the basis $\mathbf{b} _{1},\mathbf{b}_{2},\mathbf{b}_{3}$ is orthonormal, i.e. Cartesian, then the matrix $M$ is the identity, i.e. Therefore, the dot product between $\mathbf{U}$ and $\mathbf{V}$ is given by the classical expression which is also captured by the matrix product i.e. Thus, an orthonormal basis gives the dot product its simplest possible component space representation. Looking ahead, note that with the help of the tensor notation, we will be able to achieve the same level of simplicity for any basis. You can find the equivalent expression in tensor terms in equation (10.28).

Vector spaces in Linear Algebra have all of the same elements as geometric vectors in Euclidean spaces: vectors, bases, linear combinations, dot product, and so forth. This is not surprising since the original impetus for the development of Linear Algebra was to extend, by means of Algebra, the ideas of Euclidean geometry to a broader range of objects. The subject of Linear Algebra is most often laid out from one of two distinct starting points, both of which are different from ours . The first approach is exemplified by Gilbert Strang's Introduction to Linear Algebra while the second -- by Israel Gelfand's Lectures on Linear Algebra. We will now describe how each of these approaches define vectors and the dot product. However, once the dot product is defined in either approach, the subsequent topics are treated similarly.

According to the approach in Strang's Introduction to Linear Algebra, a vector is an element of $\mathbb{R}^{n}$, i.e. a set of $n$ numbers, typically organized into a column and surrounded by square brackets. Thus, a typical element of $\mathbb{R}^{3}$ is Vectors are denoted by plain lowercase letters. This notation enables us to also treat vectors as $n\times1$ matrices to participate in matrix products.

Vectors, as elements of $\mathbb{R}^{n}$, can be added together and multiplied by numbers in the natural entry-by-entry fashion, i.e. and The dot product $x\cdot y$ is defined by the matrix product where $P$ is a symmetric positive definite matrix. Perhaps it is more appropriate to use the article a, as in a dot product, since any symmetric positive definite $P$ can represent a valid dot product. Nevertheless, the article the is used since once a dot product is chosen, it becomes the dot product.

The symmetry of $P$ assures that the dot product is commutative while the distributive property of matrix multiplication assures distributivity Finally, the positive definiteness of $P$ assures the positive definiteness of the dot product, i.e. provided that $x$ is not zero. This property enables us to define the length of a vector as the square root of the dot product with itself, i.e. It can be shown that the dot product satisfies the Cauchy-Schwarz inequality Note that the Cauchy-Schwarz inequality is a trivial statement with respect to geometric vectors, but it is anything but trivial for the dot product $x\cdot y=x^{T}Py$.

The Cauchy-Schwarz inequality enables us to define the angle $\gamma$ between $x$ and $y$ by the equation although this concept is rarely used and is interesting largely from the point of view of drawing a closer parallel between geometric and algebraic vector spaces. The concept of orthogonality, on the other hand, is one that used very commonly. Two vectors $x$ and $y$ are said to be orthogonal if

By contrast with vectors as elements of $\mathbb{R}^{n}$, Gelfand's Lectures on Linear Algebra aims to bring the widest possible range of objects under the umbrella of the subject. According to this approach, vectors are defined as objects of any kind that can be added together and multiplied by numbers to produce another object of the same kind. Addition and multiplication by numbers are introduced axiomatically simply by requiring that those operations satisfy a number of natural properties such as associativity, commutativity, and distributivity. Thus, the properties that had previously been corollaries are adopted as definitions in this approach. The totality of vectors of a particular type is referred to as a vector space or linear space. The two instances of vectors introduced above, i.e. directed segments and elements of $\mathbb{R}^{n}$, satisfy all of the required properties and are therefore considered vectors in the axiomatic sense.

In the context of axiomatic vector spaces, the dot product is known as inner product and, instead of writing we now write Naturally, an axiomatic approach to vectors requires an axiomatic approach to the inner product since an explicit expression is not possible given the diverse nature of objects now considered vectors. As usual, properties that had previously been corollaries are now adopted as definitions. Thus, an inner product is defined as a real-valued function of two vectors $\mathbf{x}$ and $\mathbf{y}$ that is symmetric, i.e. distributive, i.e. and positive definite, i.e. The geometric dot product and the dot product for vectors as elements of $\mathbb{R}^{n}$ are examples of inner products. For functions, an example of an inner product is where $w\left( x\right) \gt 0$.

The Cauchy-Schwarz inequality follows from the axiomatic definition. Its elegant proof is as follows. For any real number $\theta$, the inner product of the vector $\theta \mathbf{x}+\mathbf{y}$ with itself is nonnegative, i.e. thanks to the positive definite property of the inner product. A repeated application of the distributive property, transforms the above inequality into The expression on the left is a quadratic polynomial where Since $f\left( \theta\right)$ has either one root or no roots, its discriminant $b^{2}-ac$ must be either zero or negative, i.e. In other words, which is equivalent to the Cauchy-Schwarz inequality.

The length of a vector is, of course, defined by the identity and the angle $\gamma$ between the vectors $\mathbf{x}$ and $\mathbf{y}$ is although, as we mentioned previously, this quantity is rarely used.

Once the inner product is defined, whether it is the dot product for geometric vectors, the dot product in $\mathbb{R}^{n}$, or a general inner product, the rest of the discussion can proceed as before. In particular, we can decompose by the inner product and we can carry out the inner product in the component space by an appropriate multiplication. In the context of Linear Algebra, a Euclidean space is defined to be any vector space combined with an inner product.

Exercise 2.1Demonstrate that the dot product satisfies the distributive property Note that your demonstration must be geometric: you should not introduce a basis in order to refer to the components of the vectors.

Exercise 2.2Show that the distributive property of the dot product implies that products of sums can be distributed in familiar fashion, e.g.

Exercise 2.3Show that

Exercise 2.4On the basis of the above equation conclude that there cannot be another definition of the dot product that is symmetric, distributive, and has the property that the dot product of a vector with itself produces the square of its Euclidean length.

Exercise 2.5Show that every linear function $F\left( \mathbf{X}\right)$, i.e. a function that satisifies the identity for all $\alpha$, $\beta$, $\mathbf{X}$, and $\mathbf{Y}$, can be represented as a dot product with a unique vector $\mathbf{G}$, i.e. Hint: Select a basis $\left\{ \mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b} _{3}\right\}$, let $\mathbf{G}=\alpha_{1}\mathbf{b}_{1}+\alpha_{2} \mathbf{b}_{2}+\alpha_{3}\mathbf{b}_{3}$, and then determine $\alpha_{1}$, $\alpha_{2}$, and $\alpha_{3}$.

Exercise 2.6Use the dot product to show that if $\mathbf{U}$ and $\mathbf{V}$ are equal length, then $\mathbf{U}+\mathbf{V}$ and $\mathbf{U}-\mathbf{V}$ are orthogonal.

Exercise 2.7Conversely, use the dot product to show that if $\mathbf{U}+\mathbf{V}$ and $\mathbf{U}-\mathbf{V}$ are orthogonal then $\mathbf{U}$ and $\mathbf{V}$ are equal length.

In the following exercises, use vector algebra and the dot product to accomplish the stated task.

Exercise 2.8Prove the Pythagorean theorem in three dimensions

Exercise 2.9 Prove the Law of Cosines, i.e. for a triangle with sides $a$, $b$, and $c$ and the angle $\gamma$ between the first two sides,

Exercise 2.10Prove the three-dimensional Law of Cosines for the parallelepiped in the following figure. Namely, for a parallelepiped with sides of lengths $a$, $b$, and $c$ and angles $\alpha$, $\beta$, and $\gamma$, the diagonal $d$ is given by

Exercise 2.11 Prove the parallelogram law, i.e. in a parallelogram with sides of length $a$ and $b$ and diagonals of length $c$ and $d$,

Exercise 2.12Suppose that $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$ are unit-length vectors that form an angle of $\pi/3$. Describe, by geometric means, the vector $\mathbf{U}$, such that $\mathbf{U\cdot b}_{1}=1$ and $\mathbf{U\cdot b} _{2}=0$.

Exercise 2.13Suppose that $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$ are unit-length vectors that form an angle of $\pi/3$. Find the vector $\mathbf{U}$, in terms of $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$, such that $\mathbf{U\cdot b}_{1}=1$ and $\mathbf{U\cdot b}_{2}=1$.

Exercise 2.14Demonstrate that the matrix is symmetric and, provided that $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, and $\mathbf{b}_{3}$ are linearly independent, positive definite.

Problem 2.1Use vector algebra to show that the medians of a triangle intersect at the same point. Hint: If $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ denote vectors pointing to the vertices of the triangle from an arbitrary common point, show that three medians intersect at the point, known as the centroid of the triangle, located at the tip of the vector

Problem 2.2This problem requires proficiency in a number of Linear Algebra concepts that exceeds the required level for the rest of the narrative. The goal of this problem is to determine the extent to which the basis $\mathbf{b} _{1},\mathbf{b}_{2},\mathbf{b}_{3}$ can be reconstructed from the matrix Note that the matrix $M$ depends only on the lengths of the vectors $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, and $\mathbf{b}_{3}$ and their relative arrangement. Thus, if the basis $\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}$ is subjected to an arbitrary orthogonal transformation, i.e. rotation and/or reflection (in other words, transformations that do not change the relative arrangement of vectors or their lengths), the new basis will result in the exact same dot product matrix $M$. Therefore, any attempt at reconstructing the basis $\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}$ from $M$ can be accomplished, at best, to within an orthogonal transformation. Prove that this is also the worst case scenario. Namely, show that if two bases $\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}$ and $\mathbf{c} _{1},\mathbf{c}_{2},\mathbf{c}_{3}$ produce the same dot product matrix $M$, then they must be related by an orthogonal transformation.

Problem 2.3Devise an algorithm that reconstructs a basis $\mathbf{b} _{1},\mathbf{b}_{2},\mathbf{b}_{3}$ that corresponds to a given dot product matrix $M$.