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.
Özdeğer ve Öz Vektörlerin Hesaplanması / Evaluation of Eigenvalues and Eigenvectors
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À partir de la leçon
Kare Matrislerde Özdeğer Sorunu / Eigenvalue Problem in Square Matrices
Enseigné par
Attila Aşkar
Prof. Dr.
[empty voice] Hello!
Our core values and the core vector of the previous session
we see that we have identified is obtained from the equation.
Here unknown lambda.
E is unknown.
a matrix of a given matrix.
Buna lambda a e'nin lambda e'ye
Coming to the solution of this equation is not equal.
Now if this equation equal to zero
solution or a solution that is equal to zero at all times
When you multiply by zero always returns to where the matrix to be.
This is called an empty solution, but only if any lambda unknown
so here's a parameter to zero solutions
means to find the means to be here, because zero determinant
olamzs a negative light on the determinant is zero then the opposite.
With the opposite being the case, it's the opposite of hit equals to zero as the only solution available.
We like to say that we often only solution.
But the only way it works here just does not give a solution is equal to zero
It gives the vector no lack of anything that is space solution.
If it says no to an event if an event shows.
Or, trivial solution called it empty solution.
So the important thing is not empty find solutions.
A simple equation equation here.
This determinant should be zero.
If we're writing a matrix that visually portray,
We are raising the lambda across the diagonal.
Lambda is not yet known.
Similarly, the E is not known.
On the right side is zero.
Now we want to have a little further to say that if the zero determinant.
Here means that the determinant of zero to four
We show you a concrete matrix in a verse so.
In this case, the determinant was calculated as follows.
All product of this coefficient goes.
But a very important one for us here reset
those that do not are not multiplied in the same columns and rows.
If we go on the diagonal numbers on the diagonal
elements do not belong to the same rows and columns with each other.
So will this product.
In this fourth power multiplication of the lambda.
We do not know the means to get out of here lambda
Things will be a function of lambda terms.
Show that PN as lambda.
This characteristic strength core values function veyahut force
function is called.
Lambda lambda will be really special because they are a team
but this determinant will be zero.
Considering the overall size
square matrix nth
They force will occur in terms of lambda, lambda terms are unknown.
In the nth degree one would be the root of the power function.
This root to light a lamp in two lambda j, lambda n If these,
This sets a matrix of spectrum, or
because they can represent the spectrum is called a matrix.
They can show.
What if all the properties of the matrix can be stored in these lambdas.
Of course, this time we bring any lambda,
A minus lambda lambda i j times the determinant of the identity matrix
zero will not be here to be the only solution.
Enjoy at least one means that unknown means no single solution.
Here are identifying them lambda
This eigenvalues force
Here's the function's e vectors using these lambdas.
As an important feature.
Then here we are already late to the samples.
There is only useful to understand that better be strengthened.
Let's take two core vector.
E and E both a lamp and extract value from a lamp and two opposed them.
If a lamp and another lamp in this email is different from two
vectors are linearly independent from each other.
So even mention it self as opposed to different eigenvalues
eigenvectors are linearly independent from.
This is important for those.
We need if we want to create a base for a self-vector matrix.
This is independent from each other in essence vectors
An important aspect is paramount.
How do we prove it?
We write this equation in a lamp and the lamp in two opposing self-worth and self
vector equations.
Means that two independent and e e a c c e a plus
linear with two to two equals zero vector
c means that the only solution for a and c both zero.
We are trying to show that you mean it.
Let's multiply this equation by a linear relationship.
We take this equation.
c e a plus cruz hits with a two to two.
We know that at a time to one lamp to an e times.
A time to know that there are two sides of the lambda to two.
First, let's write this multiplication is open.
We separated when we wrote,
c c a a a two plus two equals because a zero
We had already hit zero to zero a'yl here.
As we place two at a time and to a lambda lambda
we get an equation.
Let's go through this again by typing the equation.
We know what this car is two to two of the linear relationship.
two minus two e c e c a.
C e will be marked minus one here if we had the right.
So here we have a c e,
here too is coming to terms with the latter duo came to the minus.
Taking this as one factor common to say that a car from a lamp in the first term,
two coming in at minus lambda second term.
Our theorem has a lamp,
We start by assuming that two different lambda.
If a minus lambda different from each other if they are it will not be a zero c.
Back a car is equal to a zero remains.
to a vector it can not be zero, because then it would be an empty solution.
So the car has to be the number one equals zero.
this will fall to zero if c is a first term.
Again with the same idea to the zero vector or two would not be empty solution.
This means that the car must have two zero points.
So we see that the two c c equals a single solution.
Therefore, apart from having different eigenvalues
eigen vectors have proven that we are independent from each other are linear.
This is important because we want to use as a base for these e vector.
One of the most important features is the basis vectors
vector of the cluster element is linearly independent of each other.
Now here we did for the two self-worth and self-vector two.
Where c is the same process if a larger number two plus two to three e c
three plus four car situated, we can prove in the same way as putting.
Therefore, these two know why a car,
c i is true we do with that usually prove each other
eigenvectors with different eigenvalues are linearly independent in general.
Now comes the problem thereafter.
I hope both of us said more concrete problems
both will make will exhibit in various scenarios that might arise.
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