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.
Taban Vektörleri / Basis Vectors
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En provenance du cours de Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
25 notes
Koç University
25 notes
À partir de la leçon
Doğrusal Uzaylar / Linear Spaces
Rencontrer les enseignants
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] a
Our previous session,
We examined the issue together linearly dependent and independent vectors.
Anti vectors i and j in this plane and its side
for example a vector derived from the sum of i and j may also be considered.
They would be dependent.
Now we start with the following concepts: Linear expansion.
Contact a e1, e2, e3, get confident team given vector.
These vectors c1, c2, c3,
You can get hit with gathering a vector x cm coefficients.
Obviously, the coefficients for variables c1 to cm.
We take different as we dealt with each change a new x
We'll get vector.
Therefore, by changing this set of vectors x c it is formed.
Here in this cluster formed by the vector x
We call linear expansion of the relative position with respect to this email.
This is the way to hit the vector x by collecting a set
and c are obtained from our newly formed you change vectors
expansion of the set of vector e, we call linear expansion.
Just following simple theorem is obtained logic behind that.
e1, e2, linear expansion of the team soak create a linear vector space.
Where x is a set of very clear that now.
Unless you change team every car we get a new vector.
The whole point of this vector x, so that x is the vector of gathering and
be open or closed under multiplication by a number of clusters.
Other words a more obvious if we take a consisting of a x c y we take it that d1,
d2, d3, whether DM, when we get these two balls
new z vector of the vector sum of the two also he finally e1,
e2, e3, is obtained from EM opening in this way.
This set is closed under vector sum of these kind of means.
More clear words, when we take two vectors
obtained after summing these vectors are within the same cluster.
That can be expressed in this email are.
Again, this vector x çarpsak with an alpha number,
Once the vector obtained alpha alpha times c1 c2
vector e is obtained by combining the alpha coefficient times cm.
So in each of these vector multiplication by a number
It remains under within the same cluster.
The cluster will be closed.
If you remember the theorem we have seen, under these two processes
which it closed sets form a linear space.
Of course, we see that linear space a superior structure of a cluster.
Because no transactions in the cluster.
There are only elements of the cluster, there are relationships between them.
Give an example of it right away.
The plane e1, e2 and e3 given vectors we take,
so hit them with their linear expansion coefficient
we obtain a set of vectors are vectors in the plane again.
One e4, e5, e6 again as we take the vector,
of vectors in the plane, their linear expansion is plane.
A linear space.
But this linear space are plane.
On the other hand we take any two of these,
e1 and e2 not all three of them if it works i and j,
linear expansion thereof is plane.
Therefore, the following question comes to mind: it's the economy,
How can we be the most economic one open space.
Then we call the base team vector tool makes this process.
If we consider the example I just plane and a plane j vectors
base creates.
Similarly, the E1 and E3 constituting a base.
e2 and e3 constitutes the base.
One e4, e5, e6 you take in twos they can create base.
I say you can create.
Because you take this e1'l e1 instead of three zero three zero vector does not create this base.
Because it is interdependent.
Now we call the smallest team in the bottom team in these vectors.
Or called in Turkish bases.
A, at a given space that can be opened
The base of the vector space is called a small team.
Just these two theorems follow one another.
It is linearly independent vectors of a bottom pack.
Because even if one of the others expressed linearly dependent
now you could have a smaller team because you can.
Therefore, if they were not the smallest initial dependent.
Therefore, regardless easily it is revealed immediately following the definition.
And yet it all in the following theorem vector base
The number of independent vectors of the base are the same.
This again comes from the definition of the base and team
the number of vectors is called the size of this space.
These definitions and theorems follow one another.
If they were so interdependent base team claims
Watch your team in the vector,
you expressed when one of öbürküsü dependent.
Smaller he goes.
Therefore, it would be less than the number.
Now, let's make an example again.
We saw in the previous example.
e1, e2 and e3 plane vectors are linearly dependent.
We take any two of them in their time
It is independent and therefore can form a base for the team plane.
All at any level in any two of them and two of them
It is sufficient to vectors.
The same vector twos to work for it
representation by vectors such new coefficients Y1 base,
Y1 Y2 base or two bases, including two base Y2 shown.
But here's the important number of vectors base team
the number of all vectors is two.
This also shows the size of the space.
Shows that two of the plane's size.
Let's take an example of a little more complicated.
We have seen examples of this.
We see here that it is linearly dependent.
Rather, we see examples like this.
When we look at them, c1, c2, c3 three vector write
c3 by the number zero in eşitleyin here,
We find the coefficient is zero.
If c2, c3 zero is zero.
c2, c3 If zero is zero.
c1 is zero.
If you look here, this is zero.
So the only solution of three equations for three unknown c1, c2,
c3 can only see the equation can be effected by not zero.
They are therefore independent.
Therefore, e1, e2,
e3 vectors form a base.
However, since their number is three, this r4
Although the three-dimensional space has reached the space, it produces.
So it generates a subspace of R4.
Here, too, there is an important observation as follows.
Often it is a mistake like this.
You count the number of components here.
You call this four-dimensional space.
However, defining the size e1, e2, e3 basis vectors
It is that many of their number is not defined by the number of how many.
For example, a number of them, although they are defined by the number of four three
because they produce three-dimensional space.
This is an important observation but a simple observation
It can also issue to be considered in the same function space.
Of course, this is not a function such as a vector to a cosine square e1, though,
Although sinus square e2, e3, though the cosine of 2x
2x right here from cosine cosine squared x
plus or minus sine squared minus the E1 and E3 that we have to know
we see can be expressed in terms of e2.
This means that three linear expansion of a function space
Although the three constitute a base.
Because one of them dependent on others.
If you take any two of these vectors in which a two-dimensional
function space, space that creates a special function
base of infinite infinite-dimensional function spaces
It produces a two-dimensional subspace.
Again, let's take an x and x squared function,
they are separate from each other because all possible
x values for the c1 times a plus
times c2 and c3 x times x squared 0
it should be a given c1, c2, c3 You may be able to.
In the end, all right, but a second degree equation c values
To solve Cantabile possible to solve for x c values do not
and it is also the only solution to 0 c1, c2 0, c3 is zero,
Therefore, this function is a three-dimensional
producing a space.
Second degree in this space
or less, e.g., one of the following
If the primary function can produce zero zero.
Function of a three-dimensional space can be produced from a combination of these coefficients.
Again, a three-dimensional subspace of an infinite-dimensional function space
We see that.
Go to the infinite dimensions of the situation somewhat more
confused because I explain it with an example.
Let's give the JX j cosine function data we had zero zero one,
j equals 1, j equals two, plus, minus, can put forever
Let them go so that linear spaces,
When we take the linear combination can produce an infinite dimensional space
but can not produce it because all functions and its cosine x
the floor is a double function for x in multiples functions,
x minus x to be just the same for exchanging a couple of functions that
whereas the general functions of symmetric functions can produce them
There are infinitely many functions due to the non-symmetrical function in space
If we find that the base vector has found independent of each other vector
Or, in this case their function if we find that if we base
You can create as anti unsymmetrical
ALSA symmetric functions that you will not denominated in cosine
But if it manifests itself in problems for you
just because we want to produce a truly symmetric functions
also set us all a dual function of the cosine function jx
data excludes only functions but also ruled out because of our function
What if neither symmetrical nor anti-symmetrical double excludes this does not create a database.
All vectors in order to establish base in all functions
You must be able to show.
Now that immediately follows the concept vectors
whereby components of a linearly independent vectors
They are working with the team because they form the base of any
We express Exorcist them from a vector x vector.
Given that this issue immediately following the team base
by the x1, x2, xn coefficients is one.
It's easy to prove this, the vector x x1
x2, xn GB coefficient and also the base x1, x2 base,
Show the factor we can assume that the same base as x3
x vector to show two different component, I can assume,
As you can see them from each other çıkarınca çıkarınca left side to give zero,
where x1 x1 minus minus x2 x2 base as base
As we get this equation xn xn minus base.
E1 E2, e3, that is independent at each other linear
The only solution in the composition is zero coefficient e1, e2 of the coefficients is zero,
E3 coefficient is zero coefficient of EN.
EUR X, which indicates that the coefficient is equal to x üssül coefficient.
So our initial assumption is not correct.
x1 x1 and base have to be the same.
x2 x2 base and therefore have to be the same x in only one
There representation.
Since this is a single number that we give a name BUNLARADA,
We call it anything, the component must be considered as components
According to the team, but only the base under the floor of one team.
Components of the base team will change you replace this component
Let us explain this with an example as something quite natural.
Let the vector X one and three.
No matter their rectors plane E1 and E2,
independently therebetween and two-dimensional vectors
one and three components of this vector can show them all.
Because we write in terms of x when they're getting it.
This plane
we show the vector in the plane, as you can see here a third vector.
This two component toward a j direction component of the three.
But the team that we change the base e1, e2 instead of f1
and f2 if we are now easy to see that they are independent from each other.
They form a base plane because two
a base vector, they may be in line with each other as long.
Independently from each other when the two vectors form a base.
According to this base see show xi team
the same vector in this new vector tool here
We see that this time was different components.
This b1 and b2 are previously've found that the components of one and three,
where the components of the two, and we find it.
This is a tricky thing, something amazing,
According to one but not anything surprising because the base components of the team,
According to the team to have a single, one and three; According to one team still had fun,
components of furniture, as we have an account here, and they are two.
This account is very easy as you can see we f1 and f2 of x
We want to write it as a component.
given an f1, f2 given a negative one.
To them, when we held the first component b2, b1 will put a minus.
B1, B2, plus three to give.
Two unknown two equations.
It is going b2, as you can see, when we collect, four would be twice b1,
two would b1, b2 b1 after finding the
here as a minus
obtained.
[BOŞ_SES] [BOŞ_SES]
Let's do an example for similar functions.
FX function of x squared plus you get three,
this second moment in the space of quadratic functions
and a function in the lower-order functions.
We have this base here, we take a xx a base frame,
components in the base of these three functions,
For that there is a zero and a square where x is x.
But we can also select a negative xx x squared plus team well.
You can easily show that they are independent from each other.
This base set of three terms plus
here it is a time of x squared plus x b1,
b2, b3 times one minus x times x squared plus he wrote
If we find these three parts of the two coefficients b,
We see that two and three split out, in fact we can do to provide here.
That means the same function in this new base team
Although the components are different from the first one.
[BOŞ_SES] An example is given here again.
We can proceed.
Now again we would like to take a break here.
So we reached the base concept using the concept of linear dependence.
We reached the base concept of using components concept.
So far we have seen a number of the collection and processing of shock
We also know from the plane, the angle between the angle between two vectors
We had to find a new processing requirements, the inner product that cosine
Open the middle gave, even to this public space forever
genelletil the dimensional function spaces in the inner cross to still be with.
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