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.
Çözümlü Problemler / Solved Problems
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From the course by Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
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From the lesson
Doğrusal İşlemcilerden Matrislere Geçiş / From Linear Operators to Matrices
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
Our previous session was very simple but very basic, our observation.
Show the number of a transformation.
In this issue we are in a two-dimensional table on knee
When we say the matrix structure obtained.
We do say, e1 vector in space definition,
e2, ej, that the transformation we transform the EM individually.
Then each of the target vector
We write f in terms of floor space.
The resulting number is the matrix of the elements we put in place relevant.
So we take the j'yinc e transformation.
For it in terms of first, second, third,
we put here as j'yi'inc writing component.
Now we will give concrete examples related to them.
[BOŞ_SES] First
Our example is the following: three dimensions,
conversion to two dimensions.
We can say now already.
He comes from the number of column space definition will be three columns.
The number of lines in space meant the size of the target.
Here the target space will be a two-line because it is a two-dimensional matrix.
Now we need to work with the base team here.
Standard definition on the basis vectors in the vector space let.
Let's also standard on all but Target vector space.
But of course, three-dimensional vectors for the three dimensional definition space i,
j, k In heef space, f1 is a two-dimensional
f2 are each given two both have two components supplier.
Now before I show it.
ie the first team, we take them as a couple.
Then we get different pairs team.
How do they change the vector in base
If we get different components, here we will get different matrices.
But the same transformation.
We will see that the numbers vary according to the base.
Transactions made now as follows: This fundamental transformation given here.
We E1, E2, E3, we find the transformation.
e1, (1, 0, 0), respectively.
instead of x1, x2 instead of zero, zero instead we put x3,
As you can see one here, we're getting one.
When you get the e2, e2 (0, 1, 0), respectively,
zero means that instead of x1, x2 instead we set a time,
Instead we get a x3 anyway when we put a minus zero.
Similarly in to e3.
They will take for them expressed in terms of cost.
They were in our standard vector for vector.
Once a zero plus zero times, one.
Because f1, f2 (1,0) and (0, 1) was the vector.
Similarly, the second base
We get a minus for the transformation of a vector.
Minus one and three as this
We find the transformation of the third vector.
As we write these columns after the,
but we get a matrix with two rows and three columns as you can see.
Here, we use the first method.
Because the primary method for finding these coefficients,
those obtained in the left f are made to match those derived from the genus.
The same thing we said he could find in my inner product.
We have two methods.
Again T (e1), T (e2), T (e3) we calculate u.
T1 said for the first item, we stood e1 of this merger,
We stood and we divide f1'l f1 neck.
Again we stood, we divide the length is doubled e1 f2.
We stood f1'l to e2, f2'yl We stood,
T again (e3) f1'l ü We stood, f2'yl We bumped.
So what we have achieved this, each of the vectors f1, f2,
f1, f2; f1, as you get hit with this time line f2.
See here, for one there
If we see this happening in a row, but if we look at here, one, one, the first column.
A negative one, the second column; minus one, three, third column.
We find the same thing we find with the previous match.
Now here, this second method can be seen more complicated.
You are right in on that when done manually.
But when you consider them to write a computer program to calculate,
this by itself as a routine hit-hit places to find them.
But here, a job matching process more difficult to encode computer.
It is useful to know both methods in this regard.
So this mapping is based on a very simple idea.
Components, left components, we expressed with the right components.
The main cause of the problem here is not easy,
We have standard base on both sides.
But when we take this job very non-standard bases it is not equally easy.
Let's take a second sample.
See, here again, the definition of space, we get the standard base in three-dimensional space.
But although the other for the target space
not standard vectors, not the standard base.
These two vectors perpendicular to one another, so we take this inner product
once a minus is a plus once, giving zero.
We find it again in two ways.
Here, E1
When we finally get a good transformation vector e1,
i, j, we work with a K in the standard base, one, zero,
We have put in place again zero when x1 1,
x2 and x3 when we put in place a zero, we're getting one.
Here is a minus one minus one here, we three get.
Now they are not so easy to write in terms of these vectors.
Standard base for them when they are one, one, minus one, minus one, three,
We wrote immediately.
We need to match them.
Here the pair of them, one, because this one is easy.
This is a one when, here we need to take a zero here.
When a negative one, zero, zero, here we have a reception here.
Minus three so it does not look easy, but it's a once f1 base,
b f2 base he once wrote that we can find.
To combine them by minus three, minus two combined tactics.
These are situated, is that you get this matrix representation.
This matrix representation because we changed different base than the previous one.
If we make the same process of matrix multiplication, alignments like it
No online BuLi need to be completely put hit hit.
It's easier than maintenance.
As I said, it also made the computer much easier.
In the same way, E1, E2, E3, we find the transformation.
We found the same right.
Now we stand f1'l after them.
Here's an f1.
f2 a negative one.
f2'yl hit the hit the f1'l We find these numbers.
hit the f1'l, f2'yl We bumped.
We find these numbers.
hit the f1'l, hit f2, we find the number here.
So when we put them in the first column,
here comes the first line.
The second column, wherein it comes from the second line.
The third column is inserted on the third line.
These bring situated, we see that we have found the same matrix.
Finally an application
in the definition of space, do not take the standard base.
Not take the standard base in the target space.
Again we'll get to the bottom we changed the team
in the matrix will be different.
Again, as we find with the first match method,
here is the transformation of the first vector
We will write this right in terms of vectors.
To the left vector e2,
wherein the base denominated again, we can match it with.
We achieved this way.
Also with internal matter Multiply, where you no longer need to think more,
that pairing will replace hit-hit.
First, in the same way the first,
We find the transformation vectors of the second and third floor.
Three of them.
These three each one, f1, f2, f1, f2 hit,
Take the inner product, we obtain the columns of the matrix.
I repeat here, but a significant difference, no need to bother to match,
here you can directly take the place the number of inward product.
If you think you do it on the computer,
the computer's ability to think no more,
even how well programmed here, you say hit the stands, you can call collect
gathering; the computer to program them is a much more simple process.
In three dimensions, it does not matter how you do if in two dimensions,
but when you go into the higher dimensions, that then arises based on this difference.
Revealed the superiority of obtaining the inner product.
You think that a space be infinite-dimensional function in space,
then you can not do otherwise internal çarpımsız.
Now we get a result from this example, that you change databases,
unless you change any of these, we get various different matrix representation.
Although these matrices are different, the same transformation underlying them.
As in the same vector.
Vector is the same,
by choosing to base numbers showing the vector changes, but the vector is the same, because
When you create vector combine again with that base it consists of the same vector.
Here, too, from this matrix, you return to again transform,
each using different databases in the same conversion is obtained.
This is important information because of a linear matrix representation
we see that the transformation and representation in my chosen
We see the future with a different base structures.
This basic information.
Yet there is another example.
In this example three, going to three dimensions.
Go to the two dimensions of the three dimensions in the previous example, we achieved a rectangular matrix,
here we will get a square matrix.
Now in terms of convenience, there is also a feature that provides three-dimensional.
For the same dimensions, you get what base the definition of space,
You can get the same base in the target space.
Therefore, take the same f1 and e1, e2 and f2, f3 and e3 can get the same.
to base the definition of vector space, are also targets for
in space, but the dimensions are the same, we can get the same vector.
Here again, T (e1) say,
instead of x1, x2, x3 means to put zeros instead.
a sheep instead of a minus four x1, we a get.
Four, minus one.
e2 get a say x2, which means others get zero.
Again, it makes it in, (-2, 3, -1), we see that.
Similarly also, as last x1, x2 is zero,
x3 one taken by, here comes zero, two coming from two x three.
this second element x1 and x2
He comes to the X3 will fall again x1
x2 for falls, the X3 also comes five.
Because we're getting a x3 it is.
And because the ceiling are likewise shown in the standard,
the resulting matrix columns, four, minus one,
a so T (e1) we put in the first column I bring,
T (e2) 't take in the second column T (e3) in the third column Alp
We build.
Now, this time a bit different bases we choose to follow the base effect.
However, because the target space and the definition space of the same size,
because it is in three dimensions, the base vectors have the same opportunity to choose,
This time base, (1, 1, 0), (0, 1, 1), (1, 0, 1) we choose.
Let's place after them.
Instead of a definition x1, x2 time again we get one instead,
See the values obtained here as what you see here.
More details of these interim accounts are not important, important conclusions.
As you can see only it includes the conversion of e1 e1,
E2 transform contains only e2, e3 e3 contains only the transformation.
Therefore twice that e1 e3 plus zero plus zero times e2 times.
Zero times e1, e2 four times in the second row e3 zero times.
Zero times in the third row e1, e2 e3 plus zero times or six times.
When we take them places, as you can see (2, 0,
0) (0, 4, 0), (0, 0, 6) off.
We see here is a very interesting structure, a large number zero,
but those on the diagonal zero, they are 2, 4, 6 turns.
Now this kind of matrices, we call diagonal matrix.
Now here we are e1, e2 we give this self
make sure this worked out, but reversed earlier accounts; it
We have to get out of the diagonal to choose what is more important question,
E because of this, there are infinite sets, when you select them,
this will not always diagonal, diagonal will be a very exceptional nature.
This is the basis of the types of problems at work.
These eigenvalues are called, just staying in their neck after conversion.
As these visual tanımlasak a conversion,
like a machine, you give something that brings you into something else.
When you give ej'y, the definition of the matrix,
a linear combination of the overall plug.
Wherein the coefficients a, i, j constitute the matrix.
But as you can see in this case, only e1 e1,
just because it contains e2 e2, in which case you give ej'y,
only ej'li tactics, ajj'li tactics, öbürkü no.
Here's how to find these eigenvectors something very important.
From major maintenance: this n-dimensional space,
The unit offers the possibility of a reduction-dimensional space.
Again, a complicated problem, an n-dimensional problem, then online is that interact,
these intermediates that the interacting is, they all interact with each other, because aija,
i j is the coefficient showing the interaction between fi.
But when he came here, we see that e1 is just e1'l to interact,
e2 e2'yl the only, only e3 e3'l is interfering.
E them a special structure because the n-dimensional space where n
We come down to a one-dimensional space.
So there is a special place to understand the problems of these eigenvalues.
So consider the following practical thing: a full
square matrix, the line of the column, though, the number of frames need?
For example, in 1000, though dimensional matrix of 1000 lines of 1000 size,
Be able to show the number 1,000,000.
However, we choose the basis vectors to be diagonal 1000
We can show the count.
Therefore, a supremely self-demoted more
we are acquiring the knowledge, when we diagonalization.
This diagonalization fundamental event, for his mentioning it here
We are wise to see what it means in terms of conversions, so we're talking about.
Calculation of them in a later section,
One key issue will be seen in two parts.
Now in the same algebraic operations
we do, show the number of transactions, the
We can show in the differential and integral processor processors.
How the ax equals b because we're getting an algebraic equation,
one of differential equations, d, d dx as to indicate the derivative, n'yinc derivatives,
from the negative to the first derivative, on a processor that adds a fun equals g.
This is a linear equation but of differential equations.
Similarly, where f is unknown,
v ax is equally similar,
there is the problem of solving this differential equation.
Transformation shows.
Similarly, the differential may be integral processor rather than the processor.
A k, x, x consider the base processor,
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