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.
Doğrusal Bağımsızlık / Linear Independence
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Из курса от партнера Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
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Doğrusal Uzaylar / Linear Spaces
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
We will continue our work on linear alien.
A cluster in the preceding section when a
We can see that the linear space.
The two operations will be defined and the space under the two procedures
Each cluster will be closed and we saw that the creation of a space.
Now creates space anymore, we'll pass the subject in the pile of linear space.
They gathered in the two species.
One of them will start now, linear independence and space begins
We will now enables the display of items in the subject.
In two dimensions, i.e., the vector is in the plane, there ij vectors.
They were building a foundation.
Show all vectors in this basis that we could all planes.
Now we'll talk about it more general.
Here the component,
We are interested in the vector defined by points.
We'll start with a definitions.
Does one of these vectors e1,
e2, e3, with the number c being sucked,
the number c, c1, c2, and I gather cm and multiply.
This process definition, because we know the multiplication of a vector by a number.
The premise of this process is supposed to provide.
We know the collection.
It makes the proposition of the total.
When we equate them to zero, we get an equation.
Wherein e1, e2, em given vectors.
c1, c2, and c3 is also unknown numbers.
At first we see the existence of an immediate solution.
c1, c2, c3, we also cm, zero, zero, we say that all of these factors,
We saw before we hit the reset any vector consists of the zero vector.
So where did one collection consists of zero.
Collect anything because it is again reset by zero,
We see that also provides vectors in this equation.
In an empty solution we can say in Turkish.
English called the trivial solution.
This is obviously not a solution much plant-available form.
Do not empty the main problem there are solutions?
We call them in non empty non trivial solutions or solutions.
As I close it from a different angle to this issue.
Any one of these coefficients are not zero.
For example, the latter to facilitate the work that is different from zero
I think.
O zaman bu c1, e1, c2, e2 cm,
em is equal to zero will be arranged as follows: cm drop in EM left,
Take öbürkü the right, we divide by zero cm because it is nonzero,
We get the following result.
This has an important meaning.
This means that the vector can be shown in terms absorb others.
So dependent on others.
This work brings to mind the idea of dependence or independence.
If and only if all these coefficients are zero, while the
we can provide the equation, so if you have just a blank solution,
If the solution is not empty, we can not do the division.
Therefore, we can not show any vector in terms of others.
Or, more comprehensive than it Here
In this way we define as little hesitant.
One solution to this linear equation independently
c1, c2, with zero being the CM
can be achieved, we say that the linearly independent vectors.
Meaning, as we saw earlier,
all of it will not be possible for a division by zero is zero,
We can not be expressed in terms of any of the others.
These are linearly independent, if such a solution,
are not all zero, then any one of them as above
We express terms of others.
We call also linearly dependent.
For example, let's do it.
You are given the following three vectors in the plane.
We see it very well with our intuition,
once e3 e1, e2 twice if we can create.
But far more than a hunch here
While we strive to make the question is: Can we expressed E1 and E2, E3 sex?
Let's write them.
e3, he was given vector 1 and 2.
a coefficient b1 we do not know, trying to find them.
Is there such coefficients b1 and b2?
e1 e1 times times given here, one still we do not know yet
Are the number of b2, we are writing.
When we collect them from here b1 zero, zero b2, b1 b2 happening.
Here's a b1, b2, we see that the need to have both.
So e3 able to denominated others.
This issue of dependence or independence
We have shown here in a way closer to the subject of intuition.
Really, here we see the e3 is dependent than others on.
But when we look more formal terms,
We ask that.
e1, e2 times c1 plus c2 times of c3 plus time equals zero vector e3
Is there a solution without multiple C are all zero?
When we organize here c1
Once a zero times zero c2, c3 two times.
We get the following equation: c1 plus c3 of the first component is equal to zero,
There is no contribution here the second component c2, c3 plus two equals zero.
It is now, we know we want this car bulmamık coefficients.
Now here against three unknown, c1, c2, c3, there are two equations.
We can choose to enjoy one's means.
For example, t seçsek c3, c1 minus t, c2, minus two t happen.
For example, T
minus a seçsek, c1,
A c2 can be found in two.
That approach, in other ways
supporting, we arrive at a result that allows.
Here, c1, c2, c3 zero
Many also see that the solution is not infinite.
t five as you say, we find c1, c2 we find.
It means that there are solutions.
On the other hand these vectors we need here should be the previous
If we had received approaches in twos.
So I wonder if e1'l e2 linear dependent?
Did linearly dependent on e1 e3?
e2 and e3 m linearly dependent?
If we look at them we see that he is linearly independent.
Because of these two vector obtained linear combination
We observe that it is only just and blank solutions.
Thus, any pair thereof
constitute a linearly independent team.
Let's make a somewhat more complex example.
So in the space defined by four the number of r4 space vectors,
Let these three vectors: e1, e2, e3 Still more
close to intuition solution e3
Did others say you're looking at can be expressed.
So e3 will allow it to b1 and b2
we can express the coefficient?
When we write, we write here e3, e1 is equal to b1 times,
e1 given here, b2 times e2, e2 given again.
When see them organized as the first component
b1 zero here, only b1.
From here, b2, b1, plus this.
The third component is zero plus zero, zero in Makiki.
and hence does not contribute b1, the first vector.
If the second vector has a contribution b2.
Now is the time we left it equal to 3 is equal to b1.
B2, b1 plus or minus 1 equals.
0 is equal to zero, a third identity fourth
We're looking at minus 4 equals b2.
This time against two unknown what was now but a four equation
one already had three equations is equal to 0 is 0.
B1 to have found here.
Putting the second equation here b1 passes to the left of the sheep would be minus 3 and 4.
This is the fourth equation in what is already compatible b2 minus 4.
It follows that where b1 and b2
0 from which we can find different solutions.
So this vector team is dependent on each other.
When we take the same action in twos
We see that, independently thereof.
For example, I wonder e2 e1 connected with the understanding Midiru
way is to look for a solid E2 is E1.
Here we're looking at e1, e2 are equal tee times
When we get four equations for t.
0 equals t 1 equals t,
0 is equal to 0 and 1 is equal to 0.
Now we are faced with contradictions, as we saw here.
1 is equal to 0 where such a conflict with karşılaşıyoruz.d
t can not be found in a number that will allow these equations.
Therefore, we can determine the terms and these two vectors e1, e2
independently from one another.
Linear composition with the same result
If we try to answer c1, e1,
c2, c3 plus the sum of the e3 e2 are writing is 0.
Vectors herein wherein the vector e1, e2 wherein
When we gather, including writing the e3 you see here again
As it turns out three unknown three-fourth of the equation 0.
But although three separate equations
seem, these two independent equations,
the second equation
When we see that we collect from the fourth to the first grant it.
See the second equation
Remove the fourth equation of c2 will disappear here
plus minus c3 c3 will be there so it will be the same as the equation.
The three equations that seem like they say
only two are independent.
Therefore, solutions can be found not empty.
Because there are only two against three unknown equation.
Anything can choose one of them.
For example, a C3 Do you find c1 choose not here.
Here you will find the C2.
Here's one solution is not empty.
Also possible to find the general solution but to find one solution that counts.
Similarly, the E1 and E2 is linearly dependent
that can be viewed in this way.
0 equals one comes here three equations for c1 and c2, including 0.
See here c1 0.
From this last equation c2 0.
c1 and c2 in the equation as to ensure that it is consistent 0
We see and we see that it is the only solution.
Therefore, e1 and e2 are independent vectors.
Now this independence
There are important because again
we remember before i and j unit
Define the vectors we could all unit vectors in the plane.
We say that our unit vectors, vector was going on our floor.
Now we try to generalize this concept further.
For now, let's also take a break.
Base and base vectors of vectors after the meantime
We also get to watch a description of the components.
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