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.
Kapalı Küme, Doğrusal Uzay ve Alt Uzay / Closed Sets, Linear Spaces and Subspace
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En provenance du cours de Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
26 notes
À partir de la leçon
Doğrusal Uzaylar / Linear Spaces
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_KAYIT] Hello.
In the previous session,
When a process is defined and offered it as a collection
t1 to time that we accepted it as a collection,
t2, t3, and t4, we say that notice of the order
0 item can be found in a neutral
processes that are able to accept the collection.
Similarly, the operations up to C1 to C4,
It did not change the result of a sequence of three,
One still an objective of a neutral element number 1
providing that the call processing element multiplication with a number.
A cluster,
It is defined as a collection of components which are common features
but this is a very coarse community.
Without the possibility of a team making process can not go too far in that.
Therefore, with this addition and multiplication number,
When applied to the elements of a cluster creates the possibility to go further.
You need something like this alone.
This process obtained when you do not sturdy element in this cluster sturdy?
So the vector y in the plane with the vector x
When you collect the vector it here
This transaction is between vectors in the plane
You can go constantly and always make you stay in the same cluster.
If you do not have the ability to get out of this cluster Avoid longer continue operations.
Here we say that it sets off a cluster grip.
The first new element emerged at the end of the transaction set definition
If it's for this cluster
It is trying to be closed under the procedure described.
Now let's determine whether this with several examples.
If a cluster, and the number of collection
The linear spaces closed under multiplication we call this cluster.
So this sets a very rough structure, but it is being done on this process,
It can be defined, which makes propositions, but in addition to this
obtained at the end of the operations performed elements in this set
The building now belongs to the cluster at a more senior level.
We call linear space.
We call the elements of a linear vector space.
Or, as we will see this in the boyutlus the vector later
With one number, if defined by a finite number of algebraic vector is happening,
vector in this plane was defined by two points.
Seventy be defined by the number of vectors in this way very easily
be able to understand.
Our boots will no longer be possible.
There is also a subspace definition.
We know the subset definition.
We call a cluster of some elements of the subset that is not included in the set.
But at the bottom of each space
cluster can create a linear space.
If you are creating a subset of the linear space
We call the first sub-space of this linear space.
As a theorem comes immediately to mind.
If under these two operations addition and multiplication by a number
a subset that is closed is a linear space.
Hence the lower space of the first space.
Let's see several examples.
Let's start with the most simple.
Let the set of real numbers.
When we gather again to obtain a real number two real numbers.
When we collect the 7 and 5, both real numbers,
We still get a real number, we get.
One number is 5, when you multiply any number by 1
You multiply by commas 2, however, came a real number.
So that is closed under addition and multiplication by a number set of real numbers.
Thus, the set of real numbers form a linear space.
A linear space but the most simple linear space.
You can say something similar to complex numbers.
When you put two complex numbers out of a complex number again.
When you multiply the number will still be complicated by a number.
This means that the complex numbers set off under these two processes and linear spaces.
Now we see the non-linear sub-spaces, we will see spaces.
Now a linear space of all real numbers, but if we take the natural numbers
so we call the natural numbers plus the value of whole numbers.
Because they do not form two linear linear space
When you are really still it turns a linear sum of the number.
This means that under a closed collection.
When you receive two linear t2 you collect points, you add a third
You will still be collected as changing a linear space.
The number of such transactions would be closed because under this process
natural numbers, but still we get a
According to the collection of natural numbers but has the opposite negative sign.
This is not because of natural numbers plus the cluster will be valuable.
This is where you will find the opposite to the natural numbers,
is not to be negative in this cluster are numbered.
Therefore, it does not create a set of natural numbers.
There is also a natural number that gets hit with an unnatural number 5,
7 1 2 decimal places with your emerging çarpsa
5 number 2 and hit the first one comma
6 gives the exact number, but not the total number of 7 1 hit with 2 decimal places.
Therefore in the bundle with a number against the number of all
It shows that this sets off a multiplication process.
Thus, the set of natural numbers is not closed by the two main processing.
Accordingly, a space not form.
They only remain as a cluster.
Geçsek the set of integers,
The number of natural numbers that only consisted of a cross hair.
Also if we have even a sour coefficient of 0 in them.
Now we find the opposite could not find one before.
Therefore, there is a problem at all.
According to the collection is completely closed,
because when a full count by an integer
Do you collect or three collect and gather them in the exact number of which interests the same thing.
JOIN with 0 again because when you collect t3
we said no neutral element is provided.
We find the opposite, because for every integer including negative numbers.
But also it provides about four strikes,
BOŞ_KAYIT with hit
You can not provide relevant propositions; because an integer
When we multiply this number by one count each time a non-full
exact number is certain to rise.
For example, I'm going to jump in the number-this example.
All numbers between minus 1 to 1 we think.
They also constitute a linear space because the two in this set
There is no guarantee that the sum of the time you stay in this cluster.
0.5 to 0.6 to 1 comma 1 toplasanız interests.
It is also not included in this cluster.
Thus, by collecting these sets it is not closed.
Similarly, for any number within this cluster of any number
gets hit again, not in this cluster.
For example, 0 comma 1 is in this cluster.
Bring it with you çarpsa 100 remains inside the cluster will be greater than this one.
Therefore, this cluster is not a closed set.
It can not be a subset of the set of real numbers.
Let's look at vectors.
Let the following set of vectors.
The first component x1, vectors having from 1 second component.
A vector of this plane.
I wonder what this set of vectors, and this is not closed under addition,
It is closed with a crash?
Let's see that.
x and y, I gather.
The collection also now get a valid collection,
so first we will collect items,
the second will collect items, we will find the total vector.
As you can see, when we collect two x1 plus y1
but the second component will be from 2 to 1 plus 1.
When we say that this two-vector, the second component of this structure
1 we obtain when we collect vector which vector is not within this cluster.
Similarly, though we multiply this cluster of any number;
You can see see first component will be multiplied by the alpha,
The second component of that element, but this is supposed to be 1,
The elements of this set will be any alpha.
Therefore, the shock, the shock is not covered under this cluster with a number.
If we think of it as visual, the second component
Vectors with 1 as follows: Are there any x1,
but x2 component 1, 1, 1, 1, will be one.
Now we take the two of them,
for example if we take the second and third sample vector with them and gather them here,
The second component of the total of the two as you can see,
here we do the same process, it will be 2.
This time we use the parallelogram rule
The second component of the obtained vector if the first,
This will be on the horizontal line, there will not be.
Similarly, any vector x alpha çarpsak get a new vector
It will, of the second component of the vector thus obtained will be one as you can see.
Thus, the vector set is not closed under both.
Therefore, they do not form a linear space.
Oluşturmayın of linear space, you've described on this cluster,
You no longer have the opportunity to continue with the process that you define.
Let's take as an example again: the vectors of length 1
Let; vector with length 1, of course drawn in a limited number wherein
1 set of starting the start radius vector of bits on the perimeter.
This vector set is covered by picking me?
Now let the two of them here, we moved here, we take these two vectors.
When we receive the total end of the vector obtained as you see it
the radius of the circle is not on the first.
So in this cluster, the long hard one,
1 had reached the same conclusion, though not any hard,
these vectors are not covered by the set of the collection.
Likewise, this vector with any number of alpha,
You çarpsa with a different number of alpha 1, the length will be 1 again.
So multiply under this cluster, a set of these vectors,
size is not set off the first vectors.
Therefore, they do not form a linear space.
Therefore, transactions may not be very useful; because the place
then you can not process.
Let's talk a little function of the space.
Suppose that the continuity between the functions JOIN with 0 to 1.
For example, the function is coming, making a splash, continues,
There is a discontinuity.
Or a combination of undefined.
After going again defined.
In such instances, you will find many examples.
This time we collect the two functions, it may not necessarily be discontinuous.
It can take two functions that,
When one upward bounce, bounce make the other one down,
discontinuities can stay when you collect them.
Therefore, a set of discontinuous functions, within a given range,
not covered under collection.
But JOIN with 0 to
If you receive continuous functions from 1
the sum of two continuous functions will still be continuous.
You can change the order, before picking up a f g
If you collect the function g f you find the same result.
t1, t2, t3, t4, you will notice that provides propositions.
Again this range is a continuous function of alpha
When you multiply the number up with a continuous function.
Therefore, a set of closed sets, continuous functions,
thus creating a space.
Creating a subspace is, if you consider all public functions.
There is also a piecewise continuous functions,
ie until a permanent place, making a splash, then again continuous,
maybe doing another leap; continuous progress.
Piecewise continuous function again when you collect this kind of functions
there will be.
Therefore, it is closed under set-piece permanent collection functions.
When Similarly, you multiply by a number,
a continuous function will again be fragmented.
Therefore, because it is closed under both operations
This continuous, piecewise continuous function set,
continuous function such as a broader family
It will still create a space in a broader family of his family.
We ensure that with these two theories, t1, t2, t4 Our supply
You do not need; If we only allow the closure, because they
the upper parts of a space,
Parts of linear space, so the linear space
When we received subsets only to provide closure,
there is enough space.
There is not much to do at least a processing space; because
You carry out immediately a process set by the challenge.
After not possible to proceed; it may be important to the maintenance space.
Yet something like this: if we n'yinc degree of force functions,
which does not form a linear space; because the two, such as wherein
Why not get a secondary power function,
but if we take it that the first reverse of the second coefficient.
When we gather force would primarily function.
But we want to, not necessarily just in this cluster n'yinc degree,
that are composed of second order function in this example.
Then we come out of the pile at the end of this collection.
Not a function we can obtain the second degree; hence this
not set off.
But if we say; degree of n or n small set of functions,
This generates a linear space.
In the previous example, only 2 sets of power not include,
2 and smaller forces can be so when we think,
When we collect the two functions can not occur more than 2 once,
but it may consist of two small force.
This is for our cluster, so by picking off a cluster.
In multiplication by a number, any force function
You çarpsa by a number, same interests, if the number is different from 0.
If you multiply JOIN with 0, zero-order function will be released;
but our definition, it accepts the definition of clusters smaller than 2.
Therefore, it is closed under multiplication JOIN with 0.
Here we reach the vectors in the plane,
We define the two numbers; If we expand the number of them in the finished structure,
As you can see we no longer had the ability to draw, n is greater than 3.
But just as the vector operations defined,
i.e., opposite each other, the sum of these two vectors
It consists of the sum of the components from the elements
as defined, it provides all propositions.
Similarly, a vector, defined by the number
A set of these elements in vector
any element of a car
we hit the number, as tanımlasak, all components will multiply by c.
And we need to define a 0 element, it will be like this:
See 1 provides the premise, so it's
we are in the plane of the vectors has reached a very simple way to generalize.
These are vectors, and they will also form a linear space.
Of course, we do not have the ability to boot; but we review our subject,
encountered issues are always more than two components Or,
3 components can encounter more than the size of the components.
For example, consider an equation of 100 unknowns, it is unknown X1, X2,
You can define a vector as a sort X100.
Thereof in a linear space
We see that form, we show it in the Rn.
We have yet to define the size of this space but we know from the plane,
If there is a second component of the two-dimensional space it is what we know.
Now already will bring us to our next concept study of this size.
Before there alone, we have obtained some of the direct consequences of this proposition.
If we remember this propositions; vary as the sum of x and y, x,
y and z vary as the sum of.
The impact can vary by a number sequence,
We değiştirebl the order of the two numbers.
After you get hit with a number two vectors, we can change the order.
So before, with numbers, to collect two hit with a number of different vectors,
A number of the unit.
We all use them on some basic results,
We use them anyway since time immemorial without hesitation.
For example, the zero vector is one, not two different zero vector.
We also know that the number is one number zero.
Reciprocal collection of a vector is odd.
Two opposite can not be an X vector when it is given,
Or, the opposite can not be two more opposite number from it, as we know it.
There are number 5 by the inverse of the collection, minus 5.
Another has no opposite,
so we used the number of each of them without hesitation.
We use the vector.
If we multiply zero by any number of vectors, we get the zero vector.
This definition, it is possible to prove it using these propositions.
Here I am only result.
In subsequent sections to be left to the attention of course,
prove all is given but do not want to spend time with him here.
When we hit the zero vector with a number, yet we get the zero vector.
For example, do we know when we hit the reset any number again
Remove the number zero, Or, when we multiply any number by the number zero,
You çarpsa 1000 with zero interest zero again.
Çarpsa 1000 zero out reset,
the number of properties we use them without hesitation.
But these propositions, they are given the results obtained from these basic propositions.
Reciprocal collection of a vector minus
One that comes with these brackets are also equivalent to multiplication and
We need to put would not be possible to prove.
The collection with a number of missing a number here
Among the proposition that no lead extraction.
Using them, these propositions can be obtained proof.
There is also a generalization as follows: çarpsak with a number of alpha,
beta çarpsak, picked and picked and hit with another gamma,
No need to use parentheses.
This is the beginning of, not something that is between propositions.
See here, completely separated condition; with balls,
A number of cases divided by the impact.
Here we see that; When we get connected, hit both by number,
then when we bring the collection; Using these brackets Us
We see no need.
They have individual evidence, proved to be,
for example, begins as follows: For example, two if we assume that the zero vector,
Keep track of these transactions, it turns out that these two should be equal to 0.
Similarly, two of each other of an x vector
Assuming different reverse, it is also equally
You should be able to remove using only propositions.
This is an example of the power of mathematics.
Very simple proposition off after 4 plus 4
See; How it can become more powerful and just so
we need to do can be simplistic assumptions.
In this way, another example; a
reset vector multiplication,
again that the zero vector is only able to achieve using these propositions.
No such proposition.
But we use them so much; if these basic
As we accept as given.
In fact, these are not propositions.
Theorems and propositions, each of which so little theorem,
The difference is, propositions are accepted.
Ancient Greek for "I postulate", "accept" means, "I accept" means.
You can not query this, one audit,
To reach conflicting results in linear
algebra minimum number sufficient
Using these propositions and propositions, including any
contradictions go proposition, their strong structure for him.
I pass them in this way, but they are a nice work,
I suggest you study them one by one to understand.
Now here we are closing the first part; So, we have a cluster,
a community with a common set of features
Common characteristics of a community which consists of items.
But each set or each sentence, is also used,
the set also called on Turkish as well; do not form a linear space.
There are two processes need to create linear space; One collection,
one that will allow the shock to a number of propositions,
then there needs to be closed this cluster.
Then we arrive in linear space.
We can no longer process in space and operations can continue on forever.
Because the item at the end of each transaction in the cluster
We understand that, and so this process is repeated,
these, we can also take action on the new items,
ability to make the process so that we sağlayabiiy skills.
Now comes a second fundamental issue, which is "linear independence," he said,
IR in this plane and that Jia's generalization.
Currently we adjourn here.
These little work,
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