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.
Matrisler ve Vektörlerin Dönüşümleri / Matrices and Transformations of Vectors
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Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
25 classificações
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İki Bilinmeyenli Denklemlerin Öğrettikleri / Learning From Equations for Two Unknowns
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
Our preceding section,
The simplest equation of our session
We have two unknowns may be viewed two sets of equations.
We learned quite a few things here as well,
This might be the only solution of equations,
We learned that no solution or an infinite number can not be the solution.
This is the only case for a much larger equation may be the only solution,
may not be the solution, the solution may be infinite.
Now we will look at a different size.
Here are arranging a factor,
We have achieved a two-dimensional array, We call this matrix.
Now we try to understand the nature of the matrix, the matrix What are they doing?
In fact, the matrix is doing that: There is a matrix,
We bumped it with a vector x, it also finally
It may look like matrix; We achieved a year.
If we read it otherwise: y equals Axe; if y is as fx,
We can see such a function.
But these functions are very simple features,
the first force is a function of only x.
In the name we give a transformation, to separate the functions.
Current sense; x give you time to transform this matrix that x is a y.
Let's see this with concrete examples: We have a zero instead of x
Let the vector; We have seen that this is a unit vector in the plane.
Let's or other unit vector; Let's take a zero; or as a random vector
Let; is not it important to have one, both of which are non-zero numbers.
And it does not matter now, even without knowing where it came from as well as,
Let's see how well this vector into two.
We're doing it this way: We are writing to: one, one, four, one.
We write a zero on the right side.
As was the case in this factor, combined with one as in the equation
We stood, we collect combine hit zero, one turns out.
We're in the second row in the same transaction, we stood four to one,
Combine zero hits, then four plus zero: four-giving.
So this is a zero matrix
It has turned into a four-vector.
The process is similar when we did the same matrix zero merger
We see that the second vector into a vector.
If we take one vector,
One product of this gene challenge of merger,
When we combine the balls hit one of two tactics.
Similarly, combine four of the second line,
When we gather again in five hit one off merger,
This matrix is a vector that is still turned to two five vectors.
Navigate to me, this e a vector
Consider the two vectors and then e; When we multiply this with the E vector,
once you see one, minus one times two: negative is happening.
Here, too, when we hit a four times
minus two times: two gives.
All the time we multiply by e Two
When we hit a third six off two of the similar process.
Now why would you say it has what feature did.
But it did lead to something interesting, now that our observation: matrix
a first vector has transformed four,
A second vector was a convert, as a third vector into two five,
vector minus a couple of fourth
If we show a minus times a minus two
transformed into the same vector: minus to a solid.
Here are three six, one common factor in all three if we take two vectors,
again it turned a two to three times a vector of the two vectors as you can see.
There is a feature of the first three vectors,
If data matrix which transforms it into vector.
But there is a feature of the second vector,
first it turned its direction vector is a vector,
minus once she turned to him; The second vector is also turned three times to him.
They have a feature; because the same direction, transforming the vector,
It will be called to extract vector and
It will play an extremely important role.
We figured it was a zero vector as the first visual.
A'yl a zero slammed it into this vector.
The second vector is a zero vector j see you see here.
j is A'yl was when we hit into a vector,
It has transformed these vectors.
More generally, when we multiply two five-vector with a vector,
It transformed with this vector.
As you can see you can also change direction, height can be changed,
In both previous ones, it has changed both in size and changed direction.
However, we receive and e e a e a feature of the two vectors, an x direction,
y direction minus two components; When we multiply by it
at the same rate but negative tactics that combine çarpılmış the opposite.
But it is important, enjoying a vector in the same direction.
Similarly, when we get to two of, and it has hit A'yl,
two to three times, enjoying, enjoying the same direction.
It says that there's a feature I already put it as a question.
But let us also suppose that these features probably one
We have to be vectors that, so let's see.
Indeed, we will encounter a significant eigenvectors understand,
When we examine the more general.
As we have seen,
these vectors to multiply our A'yl
When giving a vector in their direction, but size may vary.
Currently we do not know until you change the length lambda,
an example was less than the previous embodiments, the light at a time,
minus once he gave e was also give to the other three times.
So this kind emerged in response to an equation.
This is not the essence of the existence of random vectors.
In this we see that the features in the lambda, which is trying to extract value.
I gave them their money's worth in English literature,
If those who want to follow, it would be of benefit.
We solve it as follows: the time to the product, e xy
Let's think like that unknown components; coefficients in a one,
A one two, one two, a two equals lambda xy times.
Lambda times XY can also wrote: Suppose again that such a matrix,
diagonals on lambda, we multiply it by x and y when lambda
x times the first element, because zero times next year;
The second element will be zero y times x lamp; This means that
we can write the equation lambda times vector multiplication in this way.
Summing these two matrices,
because the xy common factor, then a
We will remove one of the lambdas, will remain as a two,
Two one that will stay as it is, will still be in the light of the diagonal.
This equation, i.e. the left side in the Ala terms
XY multiplier; In the right side on the diagonal
When we combine the lamp we get such an equation.
If this equation
determinant zero
There is only one solution does not exit.
But it's the only solution, according to Cramer's rule,
to fulfill the right side of a one-to two to,
zero and therefore we do not want to give a single solution here.
We usually think that something good solution is the only solution,
but the only solution is wherein x is zero and y is zero.
If there is no determinant zero, it is called an empty solution that does not work, again
I give the English trivial solution is also used in Turkish trivial solution.
He says that a solution useless.
But if we do this determinant is zero, but now we do not know the lambda coefficient,
lambda and so choose that we also get zero determinant,
then it becomes infinitely many solutions but also have different solutions from scratch.
This includes also the vector obtained in this manner are also eigenvectors.
See determinant for lambda appear here twice
The term means to multiply on the diagonal, with a minus lambda.
A two minus two lambda product,
negative on the second diagonal cellar, where the lamp disappeared.
As you saw when we opened these terms lambda next frame,
In the next frame lamp.
Lambdal have two terms.
They will again come together and interesting terms and lambdas
We see something, lambdas still determinant in terms of the output itself.
So this is something we encounter neck.
You can ask the question, justly,
it turns out that in two years, but is generally available binary equation?
In general, it would not be available if already need us to see them.
They are really the determinant, no matter how many dimensions this longitudinal matrix,
No matter what we face a number of our concept of unknown length.
Let's do an example.
Let the following matrix: A matrix with a four.
As we wrote earlier this:
Let's write in a times x times x equals lambda vector.
Second that these coefficients
degree of force function,
They are obtained by placing on the diagonal minus lambda.
If we get a minus lambda veyahut write here on the diagonal,
When we opened it will be a minus lambda squared minus four.
So we also open if the lambda lambda plus a squared minus two,
minus three to minus four as well.
If you use it to solve this half of the formula, one,
wherein the coefficient for that half of the pair,
plus or minus one minus the first coefficient of the first term
lambdas lambda terms by multiplying the square of the coefficient, it gives three.
So as you can see a plus or minus the square root of four,
We see that the two solutions, both as a lamp and a lamp.
So we see that there are core values.
Eigenvalues, the diagonal of the matrix
lambdas are writing on, minus lambda, we write the determinant of
We account; determinant gives a quadratic function of this force.
We found the two roots of this second moment function indirectly.
Of course, the root of two discrete output of the secondary power function here.
Could be the same root, it may be too complex roots.
We know that the roots of quadratic function force.
The roots may be real, or may be complex repeated real values.
This is the nature of the subject.
Self vectors How do we find?
Self vector lambda equals minus one, and when we received three lambda equals,
Let's put in a rather negative light, it was the first of a four-equation.
From here, minus one off here.
Three were in the second equation lambda,
Let again three and four on a diagonal in a matrix.
See if we say them two and consists of a coefficient in the first equation.
Four and consists of two coefficients in the second equation.
Two x plus y is the first equation; second equation
x plus four is two years.
One of them actually seem like two different equations as you can see
There equations.
Because the second equation, obtained when you multiply the first two equations.
Therefore, we have had only two against the unknown equation.
It has infinite resolution.
Because there are two unknown but there is one equation,
we do not have sufficient constraints.
If we choose a free one, choose such an x, y is obtained as two,
minus two is obtained as, x a time we choose.
The second coefficient, the second unknown is obtained minus two as well.
In the second equation, however, it seems to be two equations,
x and y; but these two equations each in solid,
Multiply the first equation minus two, minus see by the hit with two four minus
hit the minus two by two; He was twice second equation that the first minus.
So, again, we have a single equation for two unknowns.
We can do it here because there are infinitely many solutions in elections,
We find the one we want from them.
The others will be the same each floor.
we would be put in place of x and y is two,
means that the second equivalent,
solution is also achieved in this way.
Indeed we have achieved with this factor
Multiplying the solutions a,
he hit a combined minus two to minus two, minus a happening.
So as you can see is a minus once eur.
In the second case twice
a third and still
four plus we get six to two.
This shows us that there are two to three times.
Thus eigenvectors
We see what we get; Now of course these very two
The results we obtain two equations of motion of unknowns.
We do not yet know the meaning but operationally it in a corner of our minds
Something must have, because now we have been very general
in front of our eyes will be next examples.
Now we give a break will be good again.
Because I'm a little SINDIRELA them.
A new clutch, again how the two equations with two unknowns
obtain clear results of a generalization
To see that we will.
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