Doğrusal cebir ikili dizinin ikincisi olan bu ders birinci derste verilen temel bilgilerin üzerine eklemeler yapılarak tamamen matris işlemleri ve uygulamalarını kapsamaktadır. Cebirsel denklem sistemleri, sonuçların tekilliği ve var olup olmadığı, determinantlar ve onların doğal olarak nasıl oluştuğu, öz değer problemleri ve onların matris fonksiyonlarına uygulanışı vb. konulara derste değinilmektedir. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler: Bölüm 1: Doğrusal Cebir I'in Özeti Bölüm 2: Kare Matrislerde Determinant Bölüm 3: Kare Matrislerin Tersi Bölüm 4: Kare Matrislerde Özdeğer Sorunu Bölüm 5: Matrislerin Köşegenleştirilmesi Bölüm 6: Matris Fonksiyonları Bölüm 7: Matrislerle Diferansiyel Denklem Takımları ----------- This second of the sequence of two courses builds on the fundamentals of the first course, is entirely on matrix algebra and applications. Specifically, the studies include systems of algebraic equations including the existence and uniqueness of solutions, determinants and how they arise naturally, eigenvalue problems with their applications to diagonalization and matrix functions. The course is designed in the same spirit as the first one 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: Summary of Linear Algebra I Chapter 2: Determinant Chapter 3: Inverse of Square Matrices Chapter 4: Eigenvalue Problem in Square Matrices Chapter 5: Diagonalization of Matrices Chapter 6: Matrix Functions Chapter 7: Matrices and Systems of Differential Equations ----------- 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: Köşegenleştirmeyle / Solved Problems by Diagonalization
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Из курса от партнера Koç University
Doğrusal Cebir II: Kare Matrisler, Hesaplama Yöntemleri ve Uygulamalar / Linear Algebra II: Square Matrices, Calculation Methods and Applications
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Matris Fonksiyonları / Matrix Functions
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[empty voice] Hi.
In our preceding session, where
We saw the third power of the matrix calculations.
You will recall that we made in various ways.
To the mean strength of the three, to three times its own
It means to multiply; hence the third force that indirect
methods; eigenvalues, eigenvectors of
You can do without direct methods based on the use.
Eventually you'll hit the city three times.
But the; Of course, if you wanted to find a higher force,
It would be a very long and tedious road.
That case is now in the process that we go through here soon,
We were able to account indirectly.
They especially like the exponential function here and sinus functions; you open the series,
infinite polynomial will be laid, will account functions.
These, as we do to the third power,
multiplying several times not able to find his own.
Therefore, the first and second method
We will calculate the exponential and sinusoidal functions using.
These methods are as follows; Cayleigh- Hamilton theorem,
He tells us that; function,
until the first force minus any function of a matrix,
As the series A, we can show as a finite series.
This is a very important thing, especially here functions as the infinite series
for it; n minus one, including the size of the matrix,
square matrix size to be able to account.
N our example, since the two; it
infinite series and infinitieth force
The function of the term has been found,
just that it's zero-force unit matrix,
and we can calculate its own.
Because so much of the first negative force, the terror it has this opportunity to show us.
Here are a reference to page 254 and 255.
Using this theorem and eigenvalues we find a general solution for n = 2.
Therefore, all these functions even very little effort
and many other functions, we can compute more easily.
We can still have the same result in the diagonalizing,
there also were having viewed occur, which function calculator
of that going to the matrix,
that function to calculate the eigenvalues, we put on the diagonal.
Obtained by the self-vector Right
the transformation matrix, the matrix we stood.
We stood in taking the inverse of the transformation matrix from the right side.
Therefore, application of this similarity transformation source with
In here the page, 247 and 249.
given in page.
Now let's move to our example.
Before we provide in this reference
254 and 255, two binary matrix we find a general solution.
We say that; any two functions with a binary matrix,
zero force; zero force also becomes the unit matrix,
A unit matrix; and we find the multiplying itself.
We have two unknown coefficients, c0 and c1, given that they also function
calculating eigenvalues with values in the second value in the first eigenvalues,
We stood with them appropriately lambda2 and lamda1.
Here comes the difference of these functions in the denominator lambda2- lambda1.
Now it's not like this miracle, it step by step, step by step,
We have very detailed way out before.
Matrix given in this example, 1, 2, 3 by 4 matrix in which,
We found the eigenvalues of our previous application.
No feature of the matrix,
very simple and consists only of the number of eigenvalues also comes integers.
Self vectors are also very smooth vector for the oluşuyor.o
account to plunge into turmoil,
It provides the opportunity to see the merits of the method.
For example, where f (A) for the exponential function,
We'll do a sinus function.
Before working with this function lambda1 = -1,
lambda2 = 5, we know that.
251.
We examined in detail the matrix provided on the page,
To do this again accounts; the plunge into business details,
We use this method to see the results we want to see here.
There lambda1 Lambda2- in the denominator; so,
5 (-1) and 6 occurs in the denominator; c0
For the f (-1) 'I will multiply lambda2; lambda2 =
5; f(lambda2)'yi de, yani f(5)'i
We will also multiply by minus one, minus one so there will also be a plus here.
So for these matrices, we function,
If we've come to the situation which function to compute the data.
c0, c1, and will put in place after finding.
To embodied more, here we write formulas.
But we have examined was derived from this formula, 1, 2, 3,
4, consisting of two binary matrix for now,
but for any function can be customized functions.
f (-1) i mean work,
Instead we put the first eigenvalues,
-one; We will put the second eigenvalues, then the fifth.
So c0, wherein the insertion of the formula f (-1)
and f (5) we find good, we find similar c1.
By putting them c0'l 1 matrix; A hit with c1,
two binary matrices results we've achieved have collected this is happening.
Here with a simple streamlined; because all of a factor of two,
We come here because it is a common structure in each term.
That work is not difficult as you can see.
There may be difficulties, if A is a very large matrix
then the eigenvalues, as here, by hand
We can not find, but then we get the advantage of a suitable computer program.
Therefore, to understand the method brings immediately after application.
The second sample, sinus (P divided by 4) was the function.
Again and c0
c1 the type of expression that the selected function.
This means that it is time for this sine function; minus one,
Instead we put -1; sinus (- pi divided by 4).
P divided by the mean of four 45 degrees here,
In this sine 45 degrees,
1 divided by 2 divided by 2 or root root 2.
But (- pi / 4) means the following point, where it will be sour.
We already know the sine of x, x minus x koysam you instead of changing the sign of the sinus.
f (5), when we arrived; so instead we put 5,
(5 times pi over 4); this is pi divided by four; Four divided by two p;
Four divided by three p; four over four p; five pi divided by four here.
Sinus doing here still 45 degrees in the horizontal axis, but down.
He for- root or 1 divided by 2 divided by two is kok2.
We can replace them immediately.
As you can see, the root of -1 split second common factor;
f (-1) and f (5) t A; When we take him out,
6 are in the denominator here, work easy.
So the 5 coming here, coming here in the first,
six; six from each other, refine, root -1 2 remains divided.
When we look at c1 (f (5) - f (-1)); that they have the same value,
means (1 1) taken out by partners as multipliers, we find 0.
So this c0 + 0 matrix in this initiative.
We also found a split root c0 minus two.
Matrix unit you see here.
This gave a particularly simple result.
We chose these numbers on purpose anyway.
Well, c1, c0, or in some cases, it may be 0.
Here, c1, 0..
There is a feature.
To show just might be such an interesting case
The multiplier unit matrix occurs.
Ala sine function.
When we think of six divided by P pi
180, 6 divided by 30 degrees.
30 degrees here.
The sine of 30 degrees in one half.
But here we will take the value -1.
Sinus, owing to the negative pi divided by six.
Sinus minus pi over six divided by the mean minus six per sine, so minus one half.
f (5), when we look at, because the second core value 5.
Instead we put 5.
5 six times pi divided.
6 Even if I am, at this point, we come to 180 degrees
but a missing 30 degrees, we are here.
It's still 30 degrees sinus sinus as one half of it.
So here we see the values.
One minus one half, it turns out one of the two split.
If we calculate c0 with them, they still have a common factor of two split.
There is a split six already in the formula.
5, which is divided by a sine minus two
as well as taking out the common factor for both -5 here,
1 -4 happening here.
Here there are 12 in the denominator above -4 below.
They simplify consisted of minus one divided by three.
When we look c1, still common outside factors
1 and -1 occurs here because we get f (5) there.
f (5) divide the two were a plus.
And two minus one over here,
2 occurs as you can see here, we get out of one half.
These two are going to split a six after another.
So sinus, c0, where we found the unit will hit the matrix.
We hit the c1 we found here, will collect.
That's a negative split, as shown in detail here Matrix trilogy unit
We stood, we stood in a sub division of the matrix.
When we collect the results we've achieved.
It turns out.
We can do the same jobs diagonal matrix.
We said to the diagonalization by similarity transformation.
Q and Q minus the diagonal matrix obtained by one.
The Q and Q minus 1, as shown here we use it,
This is the function you account lambda1 on the diagonal,
because the values are still lambda2 lambda1 matrix 1 and 5,
1 and 5 are writing on the diagonal,
lambda values.
Where Q is the left-Q
This is achieved through the use of a vector as a column opposing the essence.
We saw them all behind.
Unfortunately, without knowing where the rear of difficult progress but also to be known
There's nothing too much.
We put the column unit vectors.
Conversely needed.
We know that to account and vice versa.
If you do not provide it to take the Q minus 1 column
bring it horizontally 1, see 1's going to hit the 2 plus 1.
3 but a split in the denominator will be 1 because the diagonal from the top three in this product.
Second, we multiply the second row.
Again, this is my first column as yatırııp Let us multiply the second row.
See times 2 -1, -2.
2 times 1, +2, 0.
Q minus 1 Q'yl continue as you can see more here
whether you can give to ensure that the identity matrix.
Q. We put the left.
Q matrix here.
We put Q minus 1 to the right.
For it is a split of a number three on the front we can put it anywhere.
We got a split of three here and we got here in the matrix of Q minus 1.
f (-1) f (5) I put here will multiply.
You know that to calculate any function.
Here we write this formula, the exponent of the matrix,
here means to calculate the exponential function f (-1) instead
Because -1'l 5 core values, they are not subject to random numbers.
f (-1), instead of putting -1 is obtained.
f (5), instead of putting 5.
When we put these numbers on here left diagonal
We find the right Q'yl we do hit the Q minus 1's e to the AU.
The results we found before, and the Cayley-Hamilton
theorems with c0, c1 coefficient is the same pattern of results we found.
Of course, also it has to be this way already.
We find this in two different ways, but both methods from different areas
According to another advantage, there are advantages.
Likewise sine divided by p
minus four still are calculating function f () which we reckon the value of -1.
f () 's are account the current value of 5..
We have found before.
Q'yl in from the left,
We stood on the right side of Q minus 1's.
For the forth herein (-1) and f (5) i would put.
But f (-1) and f (5) we can get it out to be equal to each other.
Inside the first, it remains 1.
All the time we made this product, let us root out already two to split this one.
There are obvious unit matrix.
The unit matrix to the right or left
again hit the matrix remains the same matrix.
Because a neutral matrix unit matrix.
It does not make an impact.
Making an impact but leaves.
Q minus 1's Q. This was also turns again to hit the unit matrix.
Therefore, a negative result divided by two as root again before
We reach the results we found.
Minus one over root two times the unit matrix.
See here minus one half times the unit matrix output.
Already it is also supposed to be.
P divided by six in the same way that function
Instead we put the first core value is -1, we put 5.
It is also the second core values.
Again we find them.
As you can see we are putting these values when the unit is not here,
It looks like a unit, but as it turns out a diagonal matrix.
We reach the conclusion that we found this product to the challenge of still earlier.
These two methods also, but I repeat, they are important,
each at different application
The advantages and both may need to be known.
Basic is still self-worth and self-vectors.
Now we're doing a pause here.
I think that an application will show three interesting.
After doing this, we are also finished with this section.
These applications are really [blank audio]
It is selected from the basic problems encountered.
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