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While going through a chapter of linear algebra, if some student ask me what is vector here, what should be my answer?

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While going through a chapter of linear algebra, if some student ask me what is vector here, what should be my answer?

- Mar 10, 2012

- 834

Saying that a vector is a quantity having both a magnitude and a direction is, according to me, vague.

While going through a chapter of linear algebra, if some student ask me what is vector here, what should be my answer?

A vector is an element of $\mathbb R^n$.

More generally, a vector is an element of $F^n$, where $F$ is a field (which can $\mathbb R$ or $\mathbb C$ or even a finite field or some other field).

It should be noted that the use of the word 'vector' makes sense only in the right context.

Suppose we only want to study the topology of $\mathbb R^n$. Then we don't call the points of $\mathbb R^n$ as vectors.

When we want to study the consequences of the properties of the elements of $\mathbb R^n$ of the type

$(\mathbf a+\mathbf b)+\mathbf c=\mathbf a+(\mathbf b+\mathbf c)$

and

$\alpha(\mathbf a+\mathbf b)=\alpha\mathbf a+\alpha\mathbf b$ etc,

where $\mathbf a,\mathbf b,\mathbf c\in \mathbb R^n$ and $\alpha\in\mathbb R$, then we refer to $\mathbf a,\mathbf b,\mathbf c$ as vectors and $\alpha$ as a scalar.

- Jan 26, 2012

- 236

When I teach multivariable calculus I just tell my students that a vector is an arrow. This is the easiest way of thinking about it when you first come across it.

While going through a chapter of linear algebra, if some student ask me what is vector here, what should be my answer?

You can also simply think of a vector as anything in $\mathbb{R}^n$. So it is just a $n$-tuple of numbers $(a_1,a_2,...,a_n)$. But sometimes we distinguish between saying

- Feb 15, 2012

- 1,967

Furthermore, it essentially appeals to a "polar" form of view, which becomes unwieldy in dimensions higher than 3.

Formally, a vector is an element of a vector space. This is, itself, not that informative. One needs to know what a vector space is.

That, however (the question: what is a vector space), is the question worth asking.

Informally: a vector space is a set which has elements (called vectors) you can add, and scale (by some elements of ANOTHER set, the scalars).

Formally: a vector space is an abelian group ($V$,+) together with a ring homomphism:

$F \to \text{End}_{\Bbb Z}(V)$ where $F$ is a field.

(this just characterizes in "abstract terms" what the usual vector space axioms say).

Most students undertaking vector calculus (for example), though, will never have heard of abelian groups, rings, or homomorphisms or endomorphisms, and perhaps never formally encountered fields, either.

This is often side-stepped in such courses, by taking $F = \Bbb R$ (or in early presentations, even $F = \Bbb Q$) as most students tackling vectors will have some proficency in "high-school algebra" which deals with properties of rational numbers in some detail, and real numbers in somewhat lesser detail (most students are willing to accept without proof that real numbers from a field with the rational numbers as a sub-field).

The statement that ($V$,+) forms an abelian group is often stated purely in terms of the commutativity, associativity of + (vector addition), existence of a 0-vector, and existence of additive inverses of vectors (the typical first few axioms of a vector space). Abelian groups are "user-friendly" animals, and the intuition we develop from using integers (under addition) translates well to the more general setting ("+" always acts like we expect it to). I personally feel abelian groups should be introduced much earlier in mathematical education than they are, they aren't "hard to grasp" (unlike their non-commutative brethren).

The statement:

There is a ring-homomorphism:

$F \to \text{End}_{\Bbb Z}(V)$

is a bit more complicated to explain (which is why the term "scalar multiplication" is often used instead in a first presentation).

One of the main objects of study in abelian groups is the study of "additive maps" (abelian group homomorphisms), that is, functions for which:

$f(v+v') = f(v) + f(v')$.

The notation $\text{End}_{\Bbb Z}(V)$ indicates the set of all such additive maps:

$f:V \to V$

(this is known as "the ring of endomorphsims of (the abelian group) V").

The reason for the subscript $\Bbb Z$ is a bit complicated, but it just means (in this context) that we are considering the maps that preserve +, and not other operations which may be possible on $V$.

Of course, to understand what this means, one needs to know what a ring is. While this is itself not all THAT involved, the "short" explanation is that a ring is like a field, but we might not be able to divide (find multiplicative inverses) and that it may not be commutative (the first example of such a thing encountered by many students is a ring of nxn matrices).

The way we MULTIPLY endomorphisms is to COMPOSE them:

$(f\ast g)(v) = f(g(v))$

The way we ADD endomorphisms is "point-wise":

$(f+g)(v) = f(v) + g(v)$

The particular endomorphsim $F \to \text{End}_{\Bbb Z}(V)$ we are thinking of, is, of course:

$\alpha \mapsto \alpha(\_)$,

where $\alpha(\_)$ is the function $V \to V$ that takes a given vector $v$ and returns the scalar multiple by $\alpha$, $\alpha v$.

Saying that the mapping $\alpha(\_) \in \text{End}_{\Bbb Z}(V)$ means that:

$\alpha(u + v) = \alpha u + \alpha v$

which is one of the "usual vector space axioms".

Saying it is a ring-homomorphism means it preserves the addition and multiplication of $F$:

$(\alpha + \beta)v = \alpha v + \beta v$

(that is: $(\alpha + \beta)(\_) = \alpha(\_) + \beta(\_)$)

and:

$\alpha(\beta v) = (\alpha\beta)v$

(that is: $(\alpha\beta)(\_) = (\alpha(\_))\ast(\beta(\_))$)

which are two more of the "usual axioms".

In particular, we require that:

$1(\_) = \text{id}_V$ that is:

$1v = v$

for the multiplicative identity 1 of the field $F$ (which is another axiom).

****************

That said, most people if you tried to say that to them, would just give you a blank state. If asked "what is a vector"? the correct answer is: anything which is an element of a vector space. If asked, "what is a vector space", the correct answer is: any set possessing the operations satisfying the axioms of a vector space.

Often if one then trots out the 9 or 10 axioms that define this, a typical response is then: "can you give me an example?"

And this is where the previous answers come in: $\Bbb R^n$ is a GOOD example of a vector space, and the one used the most frequently. But I want to point out that a handful of examples is NOT a definition. To become proficient in linear algebra, one must eventually fall back on the RULES (axioms) that govern vector spaces. If one wants to think of vectors as points in the plane, or points in space, or as "arrows" with "heads" and "tails" those are good AS AN ANALOGY, but any analogy has LIMITS, and when one needs more than just what help the analogy supplies, one HAS to use DEFINITIONS.

I do apologize for this, as my definition is complicated. I assure you however, that it is "what a vector space is" and any EASIER answer is incorrect.

- Jan 30, 2012

- 2,502

I would not put it like this. A vector is an element of a linear vector space, i.e., a set with two operations satisfying 8 axioms. A vector space should probably be the first definition in a linear algebra course. Of course, this definition should be followed by several examples.A vector is an element of $\mathbb R^n$.

Now, linear algebra deals primarily with finite-dimensional vector spaces, which are provably isomorphic to $F^n$, where $F$ is the underlying field and $n$ is a nonnegative integer. But this is a theorem, not a definition. Besides, I think it is better to say: "In this course, you are going to consider the special case of finite-dimensional vector spaces, but in the following courses, such as functional analysis, you will also study vector spaces of infinite dimensions, such as the space of continuous functions" instead of creating the impression that vectors are intrinsically finite-dimensional.

I am not a specialist in linear programming, but I believe it deals mostly, if not exclusively, with $\mathbb R^n$ and it does not serve as a springboard to as many other topics as linear algebra. So I see no harm in saying that in the context of linear programming vectors are finite sequences of real numbers.

- Feb 15, 2012

- 1,967

In the context of linear programming, one can simply view vectors as a one-dimensional ARRAYS, which is perhaps the form most conducive to that particular subject. In this view, morphisms of arrays (linear functions) are two-dimensional arrays, or matrices. The array values must (for these to actually be vector spaces, and not some more abstract object, like a module) come from a field, although it is not uncommon to have example problems where array values are integers.Started this reply before reading the previous one...

I would not put it like this. A vector is an element of a linear vector space, i.e., a set with two operations satisfying 8 axioms. A vector space should probably be the first definition in a linear algebra course. Of course, this definition should be followed by several examples.

Now, linear algebra deals primarily with finite-dimensional vector spaces, which are provably isomorphic to $F^n$, where $F$ is the underlying field and $n$ is a nonnegative integer. But this is a theorem, not a definition. Besides, I think it is better to say: "In this course, you are going to consider the special case of finite-dimensional vector spaces, but in the following courses, such as functional analysis, you will also study vector spaces of infinite dimensions, such as the space of continuous functions" instead of creating the impression that vectors are intrinsically finite-dimensional.

I am not a specialist in linear programming, but I believe it deals mostly, if not exclusively, with $\mathbb R^n$ and it does not serve as a springboard to as many other topics as linear algebra. So I see no harm in saying that in the context of linear programming vectors are finite sequences of real numbers.

There is a well-known (abstract) theorem which states that for FINITE-DIMENSIONAL vector spaces these two "points of view" (the abstract characterization, and the specific interpretation as "arrays") are equivalent, once you choose a set of LINEARLY INDEPENDENT values for each array position. Of course, for this to make sense, one has to define "linear independence/depedence" or equivalently: bases, or more abstractly "minimal generating sets".

Using different "scaling units" for the array positions (such as "pennies" instead of "dollars") means we get a different correspondence:

abstract $\leftrightarrow$ concrete

so "turning numbers into arrays" also has to encode "what the numbers stand for".

- Jan 26, 2012

- 236

Not when it is a calculus course. The incorrect and informal definition is better because it develops a better intuition for the concept.I would not put it like this. A vector is an element of a linear vector space, i.e., a set with two operations satisfying 8 axioms. A vector space should probably be the first definition in a linear algebra course. Of course, this definition should be followed by several examples.

- Feb 15, 2012

- 1,967

The "advanced calculus" course I took used the text:Not when it is a calculus course. The incorrect and informal definition is better because it develops a better intuition for the concept.

Calculus of Vector Functions: Richard E. Williamson, Richard H. Crowell, Hale F. Trotter: 9780131123670: Amazon.com: Books

The vector space axioms are given on page 3. An abstract discussion in earnest begins on p. 57. Derivatives do not make their appearance until p. 115. All of Chapter 1 is devoted to Linear Algebra, and the first part of Chapter 2 is given over to topological considerations. The "honors" course (for which I was not eligible for as a freshman) used a "more advanced text" (Baby Rudin, I think, but I might be mistaken).

In my opinion there is absolutely NOTHING to be gained by "skimping" on vector spaces with multi-variate calculus. There is nothing wrong with using $\Bbb R^n$ to ILLUSTRATE, and the intuition gained *is* valuable, but students who learn ONLY THAT are being short-changed, and will accordingly suffer when they come face-to-face with functions of two complex variables, for example.

To give an example, there are some students who will say that the matrix:

$\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$

has no eigenvalues. This is not TRUE, and thinking like this can lead to wrong conclusions. Leading students to believe all vector spaces are finite-dimensional does not prepare them at ALL for Fourier analysis, or even that well for Taylor series (and if that is not mastered fully at the "easy calculus level", it augers poorly for studying analytic complex functions and Laurent series).

Mathematics continues to accelerate at a rapid pace. The abstract point of view is not "going to go away" (a passing fad like "new math") but will continue to be even more tightly integrating into the standard curriculum.

I will readily agree that a DETAILED study of $\Bbb R^n$ is the most profitable in calculus (and pays dividends enough, to be sure), but ignorance of what is true is scant justification. It's better to be "over-prepared" than under-prepared, no matter what the circumstances.

My educational experience is decades old, at this point. If it is still better than what is "commonly taught" nowadays, something is terribly wrong.

- Jan 29, 2012

- 1,151

And a "vector space over field F" is defined in Linear Algebra as an algebraic structure containing objects called (of course) vectors with an operation (called "addition") allowing you to combine two vectors and get a vector as the result and an operation (called "scalar multiplication") allowing you to combine member of the field (a "scalar") and a vector and get a vector as a result, satisfying specific properties.

Those properties are, effectively, that the vectors together with the "addition" operation form a commutative group and that for scalars [tex]\alpha[/tex] and [tex]\beta[/tex] and vectors u and v, [tex]\alpha(\beta v)= (\alpha\beta)v[tex], [tex]\alpha(u+ v)= \alpha u+ \alpha v[/tex], and [tex](\alpha+ \beta)v= \alpha v+ \beta v[/tex].

An example of a vector space is the set of all polynomials of order less than or equal to n (for n some specific positive integer), over the field of real numbers, with addition defined by [tex](a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0)+ (b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x+ b_0)= (a_n+ b_n)x^n+ (a_{n-1}+ b_{n-1})x^{n-1}+ \cdot\cdot\cdot+ (a_1+ b_1)x+ (a_0+ b_n)[/tex]

and scalar multiplication defined by [tex]\alpha(a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0)= \alpha a_nx^n+ \alpha a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ \alpha a_1x+ \alpha a_0[/tex]

- Jan 26, 2012

- 236

Probably the most confusing thing for students is that a scalar can be a vector also. For example, when viewing the complex numbers over the real numbers. Every real number is simultaneously a vector and a scalar.

- Feb 15, 2012

- 1,967

In fact, one of the applications of linear algebra is in extension theory:

You start with a field. Extend it to a ring. Now you have a vector space, with "scalar multiplication" being the ring multiplication operation.

This is what happens in associative algebras. In the ring $F[x]$, we can regard scalars as "outside entities that act on polynomials as multipliers" OR

"constant polynomials".

The same thing happens with the ring of continuous real functions on $[a,b]$. Scalars have a dual nature, we can regard them as something we multiply by, pointwise:

$(af)(x) = a\cdot f(x)$

or we can view them as the "constant functions":

$g(x) = a$

(differentiating them this way using the product rule yields the same result).

This also happens in the ring of $nxn$ matrices over a field, where we have $F$ embedded as the matrices:

$\lambda I$, for $\lambda \in F$.

So vectors as "arrows" isn't really correct, and can lead to wrong intuitions if the notion is too firmly established.

SO yes, scalars CAN BE vectors. Every field is a vector space over itself. Every extension ring is a vector space over a subring which happens to be a field.

One proves the the expression $a + b\sqrt{2}$ for rational $a,b$ as a real number is unique, by showing linear independence of $\{1,\sqrt{2}\}$ over $\Bbb Q$. This says no more nor no less than $\sqrt{2}$ is irrational.

These are not "pathological examples" they are COMMON.

- Jan 26, 2012

- 236

- Feb 15, 2012

- 1,967

It is hard to divine what the OP is after. He may WANT an "easy answer", and as I indicated, there really isn't one, there are just "easy inaccurate answers".

As far as an axiomatic definition goes, I believe I was in 9th or 10th grade, when I was given a list of "field axioms" (although if I recall correctly, they were not called such, and the details of something like set-builder notation were nowhere in sight) listed like so:

The Law of Commutativity of Addition:

For all numbers a,b,c:

a+(b+c) = (a+b)+c

Etc.

If someone taking a calculus class does not know these already for SOME algebraic structure (presumably the rational numbers, but hopefully the real numbers) then I fail to see how they will accomplish the arithmetic required to answer even remedial questions.

There is nothing "hard" about the idea of an abelian group, per se, and most CHILDREN show faculty with symbolic manipulation of "formal sums" at a fairly early age (it is not uncommon to see polynomials, for example, taught to 8-th graders).

There is nothing hard about the idea of a field, either. Fields are well-behaved mathematically, and they "intuitively" act like we think they should (finite fields can be a bit strange, but they do not need to be examined in any detail for calculus, so one can merely mention that some exist, or neglect them for a later date).

I am all in favor of using arrows to REPRESENT vectors, but I firmly believe that one should understand when a representation is figurative, and not literal. True, this is a level of intellectual maturity that need not be required of high-school students, but encouraging it couldn't hurt.

I do not think that any educator, at any level, should tell students things that are not true. Omitting difficult proofs because they are hard to understand is one thing, mis-representing the truth is entirely another.

I am not so hard-hearted as to recognize that some students may have difficulty with this. Working with "undefined" basic objects that follow rules that have no "extrinsic" motivation is a bit strange, at first glance.

But this is what is done, it's the language of mathematics, not some brand of cook-book physics disguised as math. Students who will not take any math beyond calculus should at least get a taste of the "real language", and students who will take more math deserve a better preparation.

It's not like the vector space axioms don't get used (like, EVERY SINGLE TIME a vector calculation is made). Is it actually the case that some teachers out there fear that merely saying the first few rules that one ACTUALLY NEEDS to calculate anything define something called an "abelian group" will cause massive drop-out and failure? Am I to believe that college students have to be weaned from a "fear of terminology"?

As far as "preaching to the crowd" goes, yeah, I know. What endorsement I may get from other staff members or senior posters here doesn't mean as much as say, an endorsement from some young 'un who might be challenged to think about something differently, in a new way.

I don't believe "dumbing a subject down" does a service to the students or the subject. I find the current state of mathematical education in my country (the U.S.) deplorable, and it doesn't seem to be getting any better. Kind of tough luck for us, that's more scientific and technical jobs going overseas.

Abstraction is not something *bad*, it's a tool to keep from repeating similar things over and over. It's quality over quantity. It's knowledge over memory. It's a more portable tool-kit, to free up more mind-space for...I dunno, fun and games.

- Jan 26, 2012

- 236

Now for the engineers who may require notions from calculus it is best to dumb-it-down for them as long as they get the right answers and most efficiently learn how to solve problems. For them mathematics is not a subject in itself to be studied but a tool used to solve other problems.

For the good math kids who learn calculus then you are correct to say that it is best to teach them the concepts in the most accurate way possible. Since they will pursue mathematics further and really care about the finer points of the definitions. In this, and only this case, will formality be appropriate. In other cases, for those people who use mathematics as a tool only, it will be a disservice.

- Feb 15, 2012

- 1,967

I believe you have your reasons, and that these may be justifiable from your point of view. I still cannot help myself, it really *does* make me feel sad you feel that way.

- Jan 26, 2012

- 236

It is not sad. It is a recognition that people have different talents. Being able to think rationally and act on choices based on deductive reasoning and empirical evidence instead of how a person feels or would like to be true is one of those talents. Most people are not like that. Just like all other talents. Some people are good at singing, others are awful at it, other people can be artistic while others cannot even draw a straight line on paper.I find that point of view sad. personally, I prefer to believe that people will rise to what you expect of them.

To think that people are all naturally interested, curious, willing to learn, pursue knowledge and carefully reason their arguments, is not only a wrong statement about people but it is damaging if a school follows such a view. Because what starts to happen is that they try to accommodate everyone as they believe that everyone is a little scientist inside and all that does is hold back the students who would legit progress with a program that has no interest to the rest of the students. The school starts to think that maybe the reason why over 90% of all students are not interested is because the subject is too hard, or perhaps because the subject is not made interesting, they will continue to come up with reasons to what to do with these disinterested students. Now if they actually realize that these students simply have no interested in solving polynomial equations they can kick them out, leave them be what they are legit interested to do, and leave the remaining academic students to learn more stuff.

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It is one thing to recognize that people have differing talents, but quite another to declare that people are "naturally stupid." I agree withIt is not sad. It is a recognition that people have different talents...

I do agree though that it is harmful to try to educate everyone in the same exact way. More effort needs to be made to foster the talents inherent to the individual rather than trying to create a one size fits all curriculum.

- Jan 26, 2012

- 236

Maybe some people do not like it but it is accurate. It is more important to ask what is correct than how people feel about it. Whether it is pessimistic or not is irrelevant, just ask how accurate it is. It is not a moral condemnation of people. Academics is not for them. That does not make them bad people but it does not make them smart either.It is one thing to recognize that people have differing talents, but quite another to declare that people are "naturally stupid." I agree with Deveno that such a declaration is indeed a pessimistic view, and even insulting to humanity in general. I think we must resist the temptation to declare those who do not share the same general talents with ourselves to be in deficit.