the no-normie zone - 2017-08-06

Balarka Sen

13:49

@Blue Let's do it here.

Anonymous

Hi

Anonymous

Yep

Anonymous

Let's decide on the sequence in which we will cover first?

Anonymous

Let's start with these topics first?

Balarka Sen

I think we should just start off with the basics first. I have Artin in front of me, where there is a nice sequence. It's where I learnt lin. alg. actually

Anonymous

13:51

Ok. Sure. Go ahead

0celóñe7

a vector space is a module over a field @Blue

now prove linear algebra

Balarka Sen

Shit, I don't want other people here.

Let's make this room gallery lol

Anonymous

ya...lol :P

user685252

@0celóñe7 shhh!!! do you see your name on this room?

Feeds

Balarka Sen has added Blue to the list of this room's owners.

0celóñe7

13:53

yes

Balarka Sen

room mode changed to Gallery: anyone may enter, but only approved users can talk

Done.

Anonymous

Ok =P

Anonymous

I've downloaded the Artin pdf

Balarka Sen

Ok, a "real vector space" $V$ is a set with a notion of addition, that is, a map $+ : V \times V \to V$ and scalar multiplication, which is another map $\cdot : \Bbb R \times V \to V$

The first map is written as $(w, v) \mapsto w + v$ and the second as $(c, v) \mapsto cv$

Such that + and scalar multiplication satisfies all the rules it should.

That is, addition is associative, has inverses, commutes, has identity $0$, and scalar multiplication is associative, has identity $1$, and distributes over addition.

Anonymous

@BalarkaSen What does $\times$ imply?

Anonymous

13:58

In $V \times V$

Balarka Sen

It's the Cartesian product of the sets.

You remember that, right?

Anonymous

@BalarkaSen Yes

Anonymous

One sec. Lemme read through what you wrote

Anonymous

$V$ is a "set" with a "notion of addition". What does "notion of addition" mean?

Balarka Sen

I just defined it. It's a map $V \times V \to V$ satisfying the axioms I mentioned.

Anonymous

14:01

It would be helpful if you could give me an example of a real vector space

Balarka Sen

This is a synthetic way to define binary operations. Didn't you have this in 12th grade?

@Blue Ok. Take $V = \Bbb R^n$, and the usual notion of addition and scalar multiplication.

The addition is a map $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$ such that $(v, w) \mapsto v + w$.

Here addition means component-wise addition of the vectors (or, if you prefer, matrix addition - vectors are row/column matrices after all).

And scalar multiplication is also componentwise scalar multiplication

Does that sound ok?

Anonymous

What's an example of a map like $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$ ?

Anonymous

Could you give a concrete example with numbers?

Anonymous

I'm not acquainted with these symbols

Balarka Sen

It I tell you, define that map as $(v, w) \mapsto v + w$, why is that unclear?

You can ask questions, I just need to know exactly where that notation does not make sense.

By vector addition I mean, like, $[1, 2] + [2, 3] = [3, 4]$

Anonymous

14:06

I don't even know what that means. What are $v$ and $w$ ?

Anonymous

Vectors in n-D space?

Balarka Sen

$(v, w)$ is an element of $\Bbb R^n \times \Bbb R^n$. So $v$ and $w$ are vectors in $\Bbb R^n$, yes.

Anonymous

Like say (1 i + 2 j + 3 k) and (5 i + 6 j + 7k) in 3-D space?

Anonymous

We can add then by adding individual components to get 6 i + 8 j + 10 k

Balarka Sen

Yes. Ah, I see now, you're used to that physics notation for vectors.

I will always write 1i + 2j + 3k as [1, 2, 3]

Anonymous

14:08

@BalarkaSen Ok!

Anonymous

One more thing

Balarka Sen

Simply because in high enough dimension there are not enough letters :P

Ok?

Anonymous

What does $(v,w)$ is an element of $\Bbb R^n \times \Bbb R^n$ mean?

Balarka Sen

Say $A, B$ are sets. You know what $A \times B$ means, right?

Anonymous

((1 i + 2 j + 3 k),(5 i + 6 j + 7k)) is an element of $R^3 \times R^3$ ?

Balarka Sen

14:10

What does an element of $A \times B$ look like?

Anonymous

@BalarkaSen Say the sets are {1,2} and {3,4} then the product will be {(1,3),(1,4),(2,3),(2,4)}

Balarka Sen

Yes. So an element of $A \times B$, by definition, is an ordered pair $(a, b)$ where $a \in A$ and $b \in B$.

So an element of $\Bbb R^n \times \Bbb R^n$ is by definition an ordered pair $(v, w)$ where $v \in \Bbb R^n$ and $w \in \Bbb R^n$.

Which is to say, $v, w$ are vectors in the n dimensional Euclidean space.

Anonymous

@BalarkaSen Oh. Now I got it

Anonymous

My major problem is I don't know most notations

Anonymous

Okay, lemme go back to your initial message

Balarka Sen

14:13

Yeah it's all formal nonsense. Important to know, but trivialities once you understand.

Anonymous

15 mins ago, by Balarka Sen

That is, addition is associative, has inverses, commutes, has identity $0$, and scalar multiplication is associative, has identity $1$, and distributes over addition.

Anonymous

What does "has inverse" mean here? How does vector addition have inverse?

Anonymous

Say we are adding [1,2,3] and [4,5,6]

Anonymous

We can add to get [5,7,9]

Balarka Sen

It means, for every $v \in V$, there is a $w \in V$ such that $v + w = 0$.

$0$ being the "zero vector" (synthetically defined to be the unique element $0 \in V$ such that $v + 0 = v$ for all $v \in V$).

Anonymous

14:16

Oh. Like the inverse of [1,2,3] is [-1,-2,-3] ?

Balarka Sen

Exactly.

Anonymous

Okay. :P Move ahead

Balarka Sen

Have no fear, the axioms mean no more than "everything behaves like the standard honest to god Euclidean space"

Anonymous

Euclidean space

In geometry, Euclidean space encompasses the two-dimensional Euclidean plane, the three-dimensional space of Euclidean geometry, and certain other spaces. It is named after the Ancient Greek mathematician Euclid of Alexandria. The term "Euclidean" distinguishes these spaces from other types of spaces considered in modern geometry. Euclidean spaces also generalize to higher dimensions. Classical Greek geometry defined the Euclidean plane and Euclidean three-dimensional space using certain postulates, while the other properties of these spaces were deduced as theorems. Geometric constructions are...

Anonymous

Okay :)

Balarka Sen

14:18

Euclidean space = $\Bbb R^n$ = n dimensional space.

For $n = 2$ it's the xy coordinate plane.

For $n = 3$ it's the xyz coordinate plane

Anonymous

Yes. Got it till there. !

Balarka Sen

You know, I should tell you this if it's not clear: the exponent above R is no coincidence.

$\Bbb R^n = \Bbb R \times \Bbb R \times \cdots \times \Bbb R$ (n times Cartesian product), by construction.

So for example the xy coordinate plane is $\Bbb R \times \Bbb R$. And yeah it is because a point in the XY plane looks like $(a, b)$ for $a \in \Bbb R$ and $b \in \Bbb R$ (both the x axis and the y axis are real lines, after all).

So $\Bbb R^2 = \Bbb R \times \Bbb R$.

Ok?

Anonymous

@BalarkaSen Right. It's an ordered ~~pair~~ collection such that $(a,b,c,....)$ where $a \in \Bbb R$, $b \in \Bbb R$ and so on. Yes. Got it

Balarka Sen

Exactly. Nice!

Anonymous

Yup :)

Balarka Sen

14:23

So $V = \Bbb R^n$ is a real vector space. We are using the adjective "real" because the scalar multiplication rule is for real constants. There are also complex vector spaces (think $\Bbb C^n$ where $\Bbb C$ is the set of complex numbers).

But let's not get into that.

Anonymous

@BalarkaSen OKhay

Balarka Sen

So henceforth I will drop the word "real" until we need it (If I am not wrong the space of wavefunctions - whatever that means - is a complex vector space; remember how the wavefunctions are complex-valued functions. e to the power of i times stuffy thingy)

In any case, so let's take another example of a vector space, other than the Euclidean space.

Anonymous

@BalarkaSen Sure.What example?

Balarka Sen

Some definition: Let $C(\Bbb R)$ denote the set of all continuous functions $\Bbb R \to \Bbb R$. Notice that for any two continuous functions $f, g : \Bbb R \to \Bbb R$, I can "add them" in a natural way; $f + g$ is defined as $(f + g)(x) := f(x) + g(x)$ for all $x \in \Bbb R$.

I can also scalar multiply a function $f : \Bbb R \to \Bbb R$; for any $c \in \Bbb R$, I can define $c \cdot f$ as $(c \cdot f)(x) := c \cdot f(x)$ for all $x \in \Bbb R$.

Anonymous

Got it

Balarka Sen

14:29

I claim that under these operations, $C(\Bbb R)$ gets promoted from a boring big set to a (real) vector space.

Take this as an exercise. It's a straightforward walk through the axioms.

https://i.sstatic.net/OiU4w.jpg
These are the axioms you need to check, just as reference.

Anonymous

As the real valued functions satisfy the axioms of a vector? Okay..I'm trying

Anonymous

Ok. Yes. The addition is associative

Anonymous

It's commutative

Anonymous

Identity element exists : f+0=f....where 0 is a constant function

Balarka Sen

Hold up.

It's not just some arbitrary constant function.

$47$ is also a constant function, defined by $f(x) := 47$ for all $x \in \Bbb R$.

But is that the identity element?

Anonymous

14:33

Okay. Let me reframe. 0 is an element of R such that f+0=f as f is a real valued function

Anonymous

$f: R \to R$

Anonymous

@BalarkaSen Nope. 0 is unique

Balarka Sen

Still not good. $0$ can't be an element of $\Bbb R$: you have to find an element $g$ of $C(\Bbb R)$ such that $f + g = f$ for all $f \in C(\Bbb R)$.

Adding an element of $\Bbb R$ with an element of $C(\Bbb R)$ does not make sense.

I know what you are thinking of, but I just want you to say it a bit rigorously. It's good to practice rigor.

Anonymous

C(R) is a set of all continuous functions. Right? And 0 is an element of that set C(R)?

Balarka Sen

To be an element of C(R) means it has to be a continuous function R --> R

0, per se, is a real number. How is that a continuous function?

Anonymous

14:36

@BalarkaSen 0 is a continuous function from R--->R, isn't it? We can just plot the 0 function on the graph. It can input any real value of x and "spit" out a real value 0!

Balarka Sen

I'm being really really pedantic at this point. Don't make me do this :P

@Blue Aha!

That's what I wanted you to say.

Anonymous

:)

Balarka Sen

So, yes. The "zero function" is defined as $f(x) := 0$ for all $x \in \Bbb R$. It's the constant function which is zero everywhere.

But yes, I agree now. So the zero function is your identity element.

what's next in the axioms?

Anonymous

Compatibility....one sec

Balarka Sen

You missed inverse.

Anonymous

14:39

Oh

Anonymous

Yeah

Anonymous

Inverse will be just -f(x) ?

Balarka Sen

Yup.

Anonymous

Since it is also an element of C(R)

Anonymous

Okay

Balarka Sen

14:39

By scalar multiplication!

Notice that you are scalar multiplying f by -1 there.

That's what -f means.

Anonymous

Yes. Agreed

Anonymous

I can't understand the next part...compatibility

Balarka Sen

So the axiomatic structure is all knitting up to a big beautiful picture.

@Blue Ah, what ails you?

Anonymous

"compatibility of scalar multiplication with field multiplication"

Anonymous

Lemme check the definition of field once

Balarka Sen

14:41

Don't be bothered by the name.

No don't bother

Anonymous

Accha. Then?

Anonymous

I forgot what field is

Balarka Sen

Scalar multiplication of $c$ and $f$ is $cf$, right?

Anonymous

@BalarkaSen Yea

Balarka Sen

Now take another constant $a$. Scalar multiplication of $a$ and $cf$ is $a(cf)$.

Anonymous

14:42

Yes

Balarka Sen

The axiom is saying scalar multiplication of $ac$ with $f$, that is, $(ac)f$ is the same function as $a(cf)$.

Kinda trivial, right?

Anonymous

Yup. Trivial :D

Balarka Sen

OK. When you move ahead into math, you will know that this is what makes the scalar multiplication into a "group action".

But don't bother at all now. This is a crucial condition, but that's it.

Anonymous

@BalarkaSen Let's do this till 8:30. After that you might be having school work and I have some hw to do. :) We can do linear algebra for 30 mins each day. (BTW thanks a ton for this!)

Anonymous

@BalarkaSen Okay

Anonymous

14:44

Next one

Balarka Sen

@Blue Exactly my thought.

Ok, next

Anonymous

Identity element of scalar multiplication

Anonymous

1.f=f

Balarka Sen

Where 1 is?

Anonymous

1 is the multiplicative identity. It is a constant function belonging to C(R) again

Balarka Sen

14:46

(Same pedanticism as the 0 thing I did up there)

@Blue Right, constant function which is 1 everywhere.

Anonymous

@BalarkaSen yea

Balarka Sen

The next two axioms are trivial too. Want to stop here?

You can check it out later if you want.

Anonymous

Okay. Sure

Anonymous

Anything more you wanna discuss?

Anonymous

Or tomorrow?

Balarka Sen

14:47

Yeah, in the next few minutes.

Anonymous

13 mins to go :D

Anonymous

Go on

Balarka Sen

So $C(\Bbb R)$ is a vector space of all continuous real functions.

Anonymous

Yes. I get it

Balarka Sen

You can similarly check that $C^k(\Bbb R)$, the set of all $C^k$ functions - this means $k$ times continuously differentiable functions - is a vector space.

Under the same addition and scalar multiplication operation.

Take this as homework :P It's again very easy.

Anonymous

14:49

Should be quite similar. Yea

Balarka Sen

But here is some more definition:

$V$ be a vector space. Let $W \subset V$ be a subset. If $W$ is closed under addition and scalar multiplication operation of $V$, and those operations satisfy the vector space axioms, then $W$ is called a vector subspace of $V$.

Now, being closed under those operations means the following.

Addition is an operation $V \times V \to V$, given by $(v, w) \mapsto v + w$.

But if $v, w$ are elements of $W$, who guarantees $v + w$ would also be an element of $W$?

Anonymous

Sorry to interrupt. I got confused with this:

Balarka Sen

If it is, then $W$ is said to be closed under vector addition.

Ok?

Anonymous

We know that in a real vector space like $R^3$

Anonymous

The i,j,k etc are the unitary elements which act like building blocks of any vector in 3D space

Balarka Sen

14:53

Indeed.

Anonymous

Like that what are the unitary elements which act like the building blocks of C(R)?

Balarka Sen

Ah, what an interesting question.

Firstly one has to make the notion of "unitary elements" precise. These are called basis elements. Choice of basis are not unique, however. In $C(\Bbb R)$, the notion of basis gets murky because it's an infinite dimensional vector space.

Let me pin that question. I'll get back to you on that when we discuss basis.

An answer to that question might lead us down the path of Fourier theory. Very very juicy stuff.

Anonymous

@BalarkaSen Oh. Makes sense. It could be something like all "independent" functions which are not superpositions can be called the building blocks...but that is a bit hazy

Anonymous

Ok, let us end it here today?

Balarka Sen

We'll think about those later.

Anonymous

14:56

It's nearly 8:30 :P

Anonymous

I have some hw to do

Balarka Sen

If you want. Let me pin subspaces so we can start talking about them later

Anonymous

Meanwhile I'll read up what you wrote

Anonymous

@BalarkaSen Sure

Anonymous

So let us meet tomorrow at around 7-7:30 again if you will be free

Anonymous

14:58

I'll get back home by 5 pm

Balarka Sen

sure

Transcript for

♦ $Mathematics$ the no-normie zone

Transcript for

♦ the no-normie zone

♦ $Mathematics$ the no-normie zone