Bu ders doğrusal cebir ikili dizinin birincisidir. Doğrusal uzaylar kavramı, doğrusal işlemciler, matris gösterimleri ve denklem sistemlerinin hesaplanabilmesi için temel araçlar vb. konuları içermektedir. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler: Bölüm 1: Doğrusal Cebirin Matematikdeki Yeri ve Kapsamı Bölüm 2: Düzlemdeki Vektörlerin Öğrettikleri Bölüm 3: İki Bilinmeyenli Denklemlerin Öğrettikleri Bölüm 4: Doğrusal Uzaylar Bölüm 5: Fonksiyon Uzayları ve Fourier Serileri Bölüm 6: Doğrusal İşlemciler ve Dönüşümler Bölüm 7: Doğrusal İşlemcilerden Matrislere Geçiş Bölüm 8: Matris İşlemleri ----------- This is the first of the sequence of two courses. It develops the fundamental concepts in linear spaces, linear operators, matrix representations and basic tools for calculations with systems of equations. The course is designed with a “content based” emphasis, answering the “why” and “where“ of the topics, as much as the traditional “what” and “how” leading to “definitions” and “proofs”. Chapters: Chapter 1: Place and Contents of Linear Algebra Cebirin Chapter 2: Learning From Vectors in the Plane Chapter 3: Learning From Equations For Two Unknowns Chapter 4: Linear Spaces Chapter 5: Function Spaces and Fourier Series Chapter 6: Linear Operators and Transformations Chapter 7: From Linear Operators to Matrices Chapter 8: Matrix Operations ----------- Kaynak: Attila Aşkar, “Doğrusal cebir”. Bu kitap dört ciltlik dizinin üçüncü cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 2: "Çok değişkenli fonksiyonlarda türev ve entegral" ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Linear Algebra, Volume 3 of the set of Vol1: Calculus of Single Variable Functions, Volume 2: Calculus of Multivariable Functions and Volume 4: Differential Equations.
İç Çarpım ve Öklit Uzayı / Inner Product and Euclidean Spaces
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來自 Koç University 的課程
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
26 個評分
Koç University
26 個評分
從本節課中
Doğrusal Uzaylar / Linear Spaces
與講師見面
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
Starting from a set of our previous session, cluster
a community of only certain components which are features common features.
We define two operations on it.
We have identified four of the proposition with each multiplication and collection and a number of them.
Common side of these propositions meet in two groups.
In a group, the result obtained being independent of the sequence of the transaction.
The second group in an impartial, neutral item availability,
The collection can be described as an opposite to it.
Let us remember that the number of the reverse shock.
Because h is born, as we know, we know, it has always been based on the number of collisions,
for example, there is no inverse of zero.
There are five but not all of the numbers, but collecting every item has an opposite.
The opposite would be equal to zero in itself, but has a feature defined.
Now here we describe another process.
This inner product.
We already see this in the vector in the plane.
We can calculate the angle of the product of two vectors was possible.
We will move to a higher-dimensional inner product space.
Of course, then we do not have the same results, but that the possibility of drawing,
giving domestic product in the state in private plane
We will start with a proposition team.
These statements superiority of the team,
not limited to the plane because they were too abstract.
We can not draw the N-dimensional space and even infinite-dimensional
function provides the possibility to expand the space.
The vector of those propositions, wherein the plane in the same direction,
giving them a special status propositions.
The first proposition when we hit with it, plus get valuable.
This is nothing but the Pythagorean theorem.
We indicate the length of a vector.
And that if and only if it is zero
We have to be a size zero to be given time.
Length plus valued magnitude of a vector.
However, the length of zero verktör equal to zero.
When we hit the plane u v Again we know,
if u equal to the product and because of the length of the vector cosine theta again.
u v theta angle while on the go,
u going to have minus is the angle theta.
But minus cosine of theta in the plane equation is equal to the cosine of theta
feature does not change the result in the vector sequence.
Yet here the sequence of different processes left and right, but it turns out the same results.
We stood before a vector by a number on the left.
Then we take the inner product, inner product before we take the right,
Then we stood by a number.
This account does not change the result.
In the same context, the process still left in the fourth proposition,
We're here before the collection process, we get a vector inner product,
The individual right of this vector we take the inner product of u and v, w.
Then we collect to bring the two.
Here are providing these features and you think, see how plain and simple propositions.
The power of a little math here.
You get a very strong structure begins with a very simple premise.
We call processing domestic product enabling these features.
Just watching this definition.
We started with a cluster, we describe two processes and procedures
When we get this set is closed under linear space.
Now we go straight to the alien.
We define a new process in a linear space.
Inner product operation.
In the space of this kind it is called Euclidean space.
You know Euclid, I guess, BC 3.
an important mathematician who lived in the century.
The founder of the geometry, in which the father,
there is said to make interesting business.
He still lived in Alexandria, not far away from us
and there is a huge library of Alexandria in that period.
There is also a very important movement around this library science.
Until the philosophy of science, as an outstanding center for art.
Then the library burning someone crazy.
If ye follow current events,
new library in Alexandria a few years ago with the help of the United Nations
and the architecture of the library was also reflected in the way the so-called old.
He who knows a little suspicious, but the old architecture,
According to sources they could find something like that here.
Euclid Archimedes again 50 years from the great names of mathematics
a man who lived long ago.
I hope you interested in this historical information.
We knew the plane for domestic products,
X. The length of y multiplied by x times y length
Starting from intermediate cosine theta x component,
multiply the y component with the first component,
hit the second components, we were getting gathering.
Likewise, only two Rn
not the component, the one-component vectors which he described.
So the generalized vector of the size of the plane.
In the same plane as the inner product of component here
It is derived from the sum of the products.
Now we do it here, but defining the inner product,
also it provides all of these four propositions.
We define a space so important.
This is a Euclidean space, because R is a linear space,
When we define an inner product here, a new space,
superior virtue, we arrive at a space type of properties.
We show it with the most.
M, coming from Euclid software in Western languages.
We show that the N-dimensional case.
Of course we are not able to draw these vectors.
But we know cosine theta sense, between two vector
angle geometry is not as we know it now.
But still conceptual, abstract gives us an opportunity for identification.
We identify gene functions in a similar domestic product definitions in space.
Function spaces may be infinite dimensional spaces.
They can also be finite-dimensional subspace, but genes
The most common function space becomes continuous partial functions.
These may include but discontinuous leap
With this combination of functions not only in space, anywhere definition,
In a single leap from the left and right limits are different.
Making a continuous function of x indicator task.
Wherein i determines the index number of components,
indicators, these discrete as one, two, three, four, five he goes.
But x is going continuously.
Where?
In the EU range.
This is how we stand out here components,
for the inner product of the function g in the FX GX fomksiyon
function integration in the EU range of means.
They also define a Euclidean space.
Because partial continuous functions were characterized by a linear space.
Define a process that also here, the Define the inner product in this way by
and that we already have all the four propositions.
Able to demonstrate easily that provide them all.
It come across as two important space.
Just as the vector of the plane
If we take for example i and j, these were the vector perpendicular to each other.
Here the definition of steepness possible.
If length times the cosine of the angle in the plane of the intermediate product of two vectors,
We see that in the inner product as a result of zero.
Because we plead here for the same definition it has zero cosine of 90 degrees.
In all finite n-dimensional Euclidean space as rn
size in number to the vector component,
which functions as elements in infinite dimensions
For these domestic product zero call them to each other.
Now and in the future multiple pages
this being perpendicular vector will show the importance of being upright.
The first theorem is linearly independent vectors that are perpendicular to each other.
This is an extremely important thing because vectors i and j independently from each other.
No projection on the other.
This same thing can show all generalized in space.
As proof of this: EN was up to us from a e1
You are given a set of vector and each element,
Take each item the couple could get to each other.
So e1 e2 to e1 ej'y to e1 Or, EN was in any e will ej'y upright.
Of course, he also has one pair between them own the multiplication of vectors.
We know from his proposition could also be zero.
These vectors are hit with numbers ca following linear equation
My rule: This will give us the condition of linear equations independence.
If all c are difficult or if another solution and to zero,
We say this to the linearly independent vectors.
They are very easy to show that individual.
This is easy using the steepness.
Let this equation.
This linear relation Let.
e1'le Carpalin.
When we hit e1'l, the product comes e1'l e1.
e1'l to E2, e1'l to ej'n and EN e1'l comes to multiplication.
By definition, all initially as the assumptions that
E We agree that to each other.
He selected a team.
Therefore, "e1, e2, e1 EJ e1 at all be zero.
One of them will be zero.
It is also the vector e1 is multiplied by itself to where we used to e1.
Consequently back here c1, e1 of time multiplied by its own pros
zero zero zero zero remains.
So here,
We see that c1, e1 e1 times zero is not zero.
The same procedure when applying for the others,
all car we see is zero.
There is no other solution.
Therefore, the perpendicular up to this EN was e1
vector without at the same time that no other account
Through them, we see that the theorem independently.
Indeed, we saw in the plane and space, j and k
important because it provides this feature.
Let's do a couple of examples.
Let's take one of the four dimensional space.
Here of course we do not draw these vectors.
To see the steepness of these
The only tools we have for this inner product.
This means that the inner product of vectors,
writing the first component of vectors side by side, the second components,
The third and fourth components to collect bounced between them.
Once one, minus two times, three times three,
four times minus two, giving these numbers as you can see.
These are the times that we collect plus 10 minus 10 equals zero,
This means that two orthogonal vectors.
For example if we take the first of these two components, we also see whether they are upright.
One, two and one and a negative one, another well
not steep but overall we take these two orthogonal vectors.
Meanwhile we know it is not open to intermediate angles can not draw them,
In a more abstract sense, a higher level terms.
But that is perpendicular vector of this we are treated this way.
Let the function.
the sine function f x g cosine function and let x
Let's reset functions identified between the two pi.
This intermediate inner product of the cosine of f and g sünüsl the integration of the product.
Where?
Given the range.
Reset between the two pi.
We see this easily is zero.
Sinus x u to say, other cos x dx happens.
Wherein the term means becomes due once.
other times one half of the frame.
we know that a split of the two sine sinus x is x squared happens.
Zeros will take two pie.
As you can see, the two pita zero sine fell.
Zero zero again, he fell.
That gave us zero minus zero zero here.
Now we provide here is an important initiative.
The steepness of the function definition in the function space
We see can provide.
Now we have to emphasize the importance of not more than steepness but it
We also know from the plane.
and otherwise work with i j, with base functions,
much easier to work with the base vector.
Why do I emphasize that it is easy to be a bit more open.
Here before,
cosine sine X. X is not only,
the cosine of the variable x
n times and when m and n to be a multiple m of each other
When we find that there are different each other of these functions.
Bunda cosine of a times b multiplied by the cosine formula
If we use the cosine of a times b is a split two times cosine
Dividing a minus b cosine minus cosine
We know that a plus b is divided into two.
It turns out that the integral right sinus of check accounts.
We can see that the sine zero.
We not only do the same process in the sinus cosine.
Here's the product of sine, cosine here again, this time plus similar
We use the identity checked.
Be integral to the sinuses again we see that out of this zero and zero.
Of course, when n and m are different from each other.
Because interest is equal to n and m refer here cosine square.
Not zero integral value of the cosine square and a plus because everywhere
function.
There is both a plus and minus for taking these functions,
hence for taking a zero pros cons.
n to m is importantly different.
And for the same n and m are also important to observe that there will be zero.
Similarly, the cosine we hit the sinuses
When we use this time in a times the cosine sine b formula
here also it consists of half of the difference and the sum of the sinuses.
This integral values taken by these interests.
Here as important, we are revealed a very important result.
Reset both the cosine and sine m in the range of pi x,
We see those functions they are to each other.
Therefore, a steep, we are the team that has achieved a base of infinite elements.
This is actually not a very new thing conceptually,
j k is the opposite of the infinite-dimensional function space.
How is that j k are perpendicular to each other, here with not infinite
An important advantage, you can move to the infinite dimension of function space,
we are determined that they are upright.
Using this base function Fourier series,
function series in space is obtained.
And Fourier series of important technological revolution in which we live today,
digital revolution that is the most important mathematical digital revolution
There is also the matrix contributes about them.
Matrix also the importance of the topics we discussed in the second part of this course.
Any signal through the use of Fourier series, torch
speech signal, we can turn to transmit the video signal number.
That such CDs, DVDs that are made this way,
we call this concept in digital television is made in this regard.
Our voice transmission, the transmission of voice as digital,
the same communication, communication through this Fourier series.
It also has some advanced ones, but the essence of this kind of Fourier series.
Now other vectors told the importance of the team.
Let us show an example of the importance of this.
Whether a given first vector in two dimensions.
No feature of this vector, randomly selected vector.
e1 and e2 are given vectors in two.
These two are independent of each other but not to each other.
Because we always seem to hit e2 e1'l the product here, it turns out four.
These two are not perpendicular to each other is not zero.
But also independent of each other, e1, e2, not because a particular floor.
Therefore it can be used as a base.
That means any vector in the plane, we can breed from the E1 and E2.
Process that we do here.
We are writing to v.
e1, e2 times v1 plus v2 times where everything is known.
Only v1, v2 unknown.
Our goal is to find the v1 and v2.
There are two basic methods for this.
We synchronize each of the components of this vector.
Cons would be equal to twice v1 plus v2.
Three must be equal to one times v1 plus v2 two times.
V1 v2 we can solve this equation with two unknowns.
We did this several times in similar jobs.
Now let's make this process using the inner product.
So let's determine whether v1 v2.
Here v1, e1,
We stood before v2 e2 e1'l to take.
On the left side there are known.
e1, one had two, had been there a minus three.
bilmediklerimiz v1 and v2.
Bu e1 ile e1'i e1'le e2'yi
As we hit the easily see here, we get the following equation.
The first equation.
Where v is the same as given by e2 hit after the details of this time,
With a simple internal crash, we get a second equation.
v1 and v2 the first equation to the second equation them,
four to one hit here, we remove the other one hit the feeder,
one hit as the one we find these solutions, we hit such interests.
This can be considered the following question: There is always a solution, the only solution we can find?
Yup. Because we got theorem.
A theorem we have seen before.
He says that, given a vector component in a team is only as certain base.
This is the e1,
e2 base will be one of the components of v in the team is guaranteed.
So here it is impossible to have no solution or infinite solutions.
We saw two equations in two unknowns may be an infinite number of solutions,
but the number of solutions that can be zero independently E1 and E2
To theorem, we see before us this theorem
It guarantees that the components will be found to be single-valued.
Now here we spend a great effort.
Because it consists of two equations with two unknowns but two equations with two unknowns
We also had to solve.
Now the same vector, still the same
minus a third vector, perpendicular to one another, this time a
Let's get to the bottom team based components.
It seems that quite easily that two vectors.
Once minus one, minus one, once one plus one gives zero.
We work with a team of other base means.
They also base this on a previous e1 e2 of the base
to show that different.
This is the component of the vector v
Let the radius is limited to the base of e1 and e2 e1 base typing the base.
As you can see here it is calculated by multiplying the base of e1 e1 base.
But there is multiplied by the base of e1, e2 base.
But the definition of the base e1, e2 base of zero.
Therefore, although equation with two unknowns here
the second component is visible,
The second unknown coefficients turns out to be zero for an unknown in an equation.
We find here, just as a v1.
When we made a similar process with the base e2, e2 be calculated by multiplying the base by,
As you can see here e1, e2 calculated by multiplying the base of the base will be zero.
But there are guarantees that there will be zero multiplied by the base of e2 e2 base.
Because the basic premise of the domestic product also shows that it is not zero.
Similarly, the multiplication of the base e1 e1 where the base was not zero.
Indeed, we find it.
Here, the coefficient of zero base v1,
v2 v2 coefficient obtained from the bases of the two that we find.
When you come to a seemingly self again, even if two unknown
so it is a known equation.
Therefore v1 from the first equation right here,
We solve the second equation and solve the equation v2
You need to get a solution to this place without eliminating.
This is extremely important.
From important ways: We can solve the equation in two unknowns anyway
Although unknown to many, but in a way, even though hundreds of unknown,
A computer program can solve them in putting but quite a few
an energy expenditure of our time we spend, we must make an effort.
In particular, we passed to the function space,
there are infinite unknown.
They resolved to again cut somewhere to turn a finite infinite solutions
We find about solutions, but we are working with other base
It is an extremely important feature when it does not need to solve the equation.
Let's make this a little more clearly see that what I said just now.
Get a VN-dimensional space.
v n are one component.
They are obtained by multiplying of the new e representation of this.
E certain.
They also may be perpendicular to each other.
there are also provided.
Our goal is that the one you have, find and VJs.
v1, v2, so vn'y find.
It grains respectively, by multiplying the e2'yl to en'yl to e1'l,
Taking the inner product of the equation, we get one.
These çarpsak this equation with the addition of any ordinary type,
v The ek'yl the product, multiplied by the total ek'yl.
If we include ek'y because the inner product of a linear
because here the importance of a property in order
He said that the third and fourth propositions.
Here will be the product of the EJ ek'yl as we have seen.
Additional specific, we choose e1, e2 like.
EJ collection on the variables taken.
Among those products are all zero, except one.
when j is equal to k.
When we do this, we see that immediately.
A lot of multiplication zero zero.
K characteristics term stays only.
ek'yl to guarantee that there will be zero in the WWE's product.
My inner diameter so that the first proposition.
Therefore, as you can see here, the left side is no longer certain.
On the right side remained unknown.
There was therefore no need to solve the equation.
That was the equation we have reached.
I am writing here again.
We're here to say that the unknown VK divided by the addition and multiplication WWE.
One feature more accounts can facilitate some more.
If the addition of the size of the unit is configured such that when you see
ek'yl will be in addition to the ground and VK solution will be much more simple.
I do not want to say easy.
There is no difficulty here, but will work with less.
Because here you have to carry the grain product.
Whether at the beginning of the same transaction, it can not be an additional longitudinal
dividing paint can configure the unit at all times.
See how much simplification.
Anyway i, j, k widespread use of this reason.
I. first component 1, the secondary 0.
The second component of the others 0 1 j respectively.
The third component in the k 1 and others are 0.
We see immediately that they are perpendicular to each other easily.
We see easily that one of their length.
That is why i, j and k, we prefer our action.
If we came to the function space,
We can also here the equivalent of the previous N-dimensional space.
A fx given function,
to each other, for example, we saw cosine and sine of the above.
They came out to each other.
Such a base in the show, the representation of such a base, respectively fk'yl to,
Taking the inner product as a little before our creatng addition,
all inner product will be 0, except one.
It is also the place where the product of fk.
This is the product of fk, I know that it is not 0.
So, the number of unknowns ck all
We can deduce the need to solve the equation.
Again in the RN-dimensional finite-dimensional
As the space inside the fi function
If we configure the product will be 1,
As you can see in the denominator of the frame fk,
fk'n the inner product of this integration is that this fk
There needs to account to, we found direct direct CK-in.
f multiplied by fk.
This software also functions in open space
domestic product as we have seen, these two functions hit
given a, b integrally to receive income equivalent intervals.
That's observations
In Fourier series of sine and cosine,
even exponential function of the virtual combination thereof
including base, we can produce all functions.
Indeed, this feature in the next main section
We handle using Fourier series.
Now it will be useful here to take a break.
We passed a new concept.
This is really important,
Not a big change, but there will be a crowning of this space.
This is called a Hilbert space.
1900l Hilbert, a German scientist who lived in the year to the life sciences,
In mathematics, the physical sciences and mathematics,
In the very important contribution it has been used in engineering science,
even a person who does the president of the World Mathematical Society.
The transition to the year 1900, so go to the ends 1800 of 1900
The main problems of the next century, he has identified a number of problems.
Some of them were dissolved.
We know that some of them still waiting to be resolved.
Bye now.
So do we process, to define an inner product in linear space.
But a remarkable new concept that enriches applications.
