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.
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Doğrusal Cebir II: Kare Matrisler, Hesaplama Yöntemleri ve Uygulamalar / Linear Algebra II: Square Matrices, Calculation Methods and Applications
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From the lesson
Kare Matrislerde Özdeğer Sorunu / Eigenvalue Problem in Square Matrices
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
We define our core value problems of the previous session.
Of course, the opportunity to see all the aspects we have not yet had their full fine.
With four problems with the examples below we will do what we are doing,
We'll see what we encounter such cases.
We can summarize the problems of self-worth.
First there is the core business value.
We do it as follows: We are writing the minus lambda In that matrix, diagonal
We write it on the minus lambda We calculate the determinant.
A power function for determinant turns out to be a numeric value.
If the matrix is, if the size of the force function nth degree
a force function, and a lambda of the solution thereof, the two lamp,
We find the roots will die in the slide.
They have our core values.
Some of these complex eigenvalues
may be valuable, others can repeat each other.
These are exceptions.
Therefore, be real and discrete values as we can summarize here
As can be repeated values, the values can be complex.
This forces any function, even a minor force
Even if you think the function may be equal to two roots,
It may be real or complex can be separated.
Here we find the roots.
We find eigenvectors to come opposite to each root using these roots.
This made the process consists of two steps.
Again, we can also encounter different situations.
We are face to face with a simple equation here once.
Determinant for the right side is zero
It will be zero because the only solution.
The only solution is already bad, though.
The only solution would be equal to zero, if not zero these determinants.
We want you to be especially zero,
We choose it anyway these lambdas to make zero.
But in this case not only valuable solution will be very valuable solution.
This means that some released, it will be the value we can free choice.
Now let's go through our example.
The first example is a very simple example.
one, two, three, four, we start with the matrix.
We find a special function to find self-worth.
Eigenfunction can say.
This determinant for it appears here in two places lambda
We stood there in the first diagonal when we calculate
lambda lambda minus minus minus three with a second diagonal
We stood on relations, we are removing four twice.
When we edit it lambda lambda squared minus four minus five equals zero.
Here we see,
two core functions or special function for this dichotomy is a matrix
Or, the characteristic function is a quadratic function of force.
Here we see that the second moment function here.
The roots of this quadratic function forces the determinant zero
who roots.
We are doing this as follows: linear term
We can work half the formula for the coefficient is an even number.
This is plus or minus half, its frame,
C. four times minus
Wherein A, is a C minus five to plus is five.
As you can see in the bottom two, plus or minus the square root of nine
because it turns out the square root of three.
Here we find the roots of five and minus one.
So, there are two separate roots.
There are two discrete core values.
We will find the eigenvectors corresponding to each.
This second step.
Again, we are writing the matrix.
This time again we will no longer diagonal but on interests lambda
lambda are no longer certain.
We found five one lambda, we found a negative one.
We are writing this first equation.
Self vector writing just so easy to get X, Y, so let's write.
So these are the components of the vector e1.
But for now, they will find a valuable solution for X
Better to write a simple thing like Yi.
You see, when we organize this equation
As -4X + 2Y, the first equation.
2Y second equation in 4X.
There seems to be two equations.
But already this because of the zero determinant
not two independent equations.
Indeed, the first equation of the second axis of çarpılmış.
So here there is one independent equation.
If we calculate to control the determinant,
There are four less see here, here minus two.
Eight minus four to minus two is multiplied by minus two plus
eight minus multiplied by four, eight minus eight; zero.
When we find the determinant of providing zero.
Now here are two to say against the unknown equation.
Let other core values.
The second core value, we also find a negative.
We are writing again, we are removing the lambda value we found out the diagonal.
Now lambda particular, we are putting a minus.
See, when you edit here a minus minus one; two
2X + 2Y stayed so happens first equation.
The second equation is still here see it as two consecutive negative
plus it is giving.
4X + 4y.
Yet even seem like there are two equations,
We observe now that these two independent equations.
Indeed the first two times the second equation.
In both cases, also it means that there are two unknowns against an equation.
We can freely choose one.
We made it to be the easiest.
For example, say you have an X in the equation Y becomes two.
Here we find the first self-vector.
X is a second equation still say, we find a Y minus.
These two vector are independent.
Already the theorem.
Theorem says; separated from each other opposite to different core values
He says that core values are linearly independent.
Indeed, e1, e2 is not a solid,
so they linearly independent vectors.
Theorem we prove the theorem that the page they were four hundred forty-four pages.
Now, a slightly larger second sample matrix, a matrix three three-pointers.
We'd like to find self-worth and self-vector.
Again, we go to find self-worth, the first step.
Again, we are writing the matrix.
Diagonal from the lambda minus, minus lambda,
minus lambda lambda times I would remove the matrix.
As you can see this will be a function of the third degree lambda
because in terms of lambda lambda cube next to be seen in three different places.
That any line can open by column.
Suppose we turn according to the first column.
Minus four remaining sub-matrices with the determinant of oxygen,
minus three minus five times lambda lambda, look where you're writing,
also it comes as a plus for a minus minus exit from this product.
This term brought the first element.
Second, minus the sub-matrix of the merger.
We are taking the line that minus one exists, we take column.
Back minus two, two, minus one, minus one for the remaining five.
So twice five minus minus lambda
As you can see minus times minus two
thus becoming a plus.
Here a number, but a minus,
but if you remember was going as signs; plus minus.
It's here you'll hit the plus value for this axis.
The final terms of the first column again
We open at this time last row plus one.
Because it does not change the sign, +
-, It comes to the + sign +1 come.
The first column and the last line of the cast, backward -2,
here are writing -2, 2 times 3 minus
λ, but it comes with a minus sign because this is the second diagonal.
We are writing it.
Now we can easily see here.
4-λ factor here; When we organized this for,
See here comes the checkered λ term λ'l comes two terms.
5 times λ, λ 3 times with a minus sign, then the minus sign, -8λ coming.
Here, considering the flat numbers, there are 15; 1 There is 16.
See the terms of the brackets in the second degree
It was the force functions.
When we no longer edit them in the multiplier λ them.
The common factor, see here, there are 10.
+2 -8 Wherein
and 2 at the denominator out there,
It stands out as you can get.
With back here 5 minus λ
+1 Staying for 2 as we receive the common term is 4-λ occurs.
When we look at here, there is still a -2.
We are writing to -2, as well.
See here, six there,
here's term, there λ'l terms,
When we collect all this we find in this way.
Putting them together, 4-λ is becoming a common factor as you can see.
This is a common factor in the back,
quadratic function of this force remains.
Here we find the roots immediately.
First stem clear 4.
Here the term second
When we find the root of the very term
We're finding now that there are 2 and 6.
If we look at the supply, the sum of the root, 8, and here comes the minus 8.
Multiplied by 12, it comes 12 as you can see.
Now we we put in place the roots, just put 4 look easy.
This will be 0.
We need to find when we open this determinant 0.
2. When we set, we have to find when we open again 0
This determinant, 6 0. We need to find.
Already this will reveal itself.
Let's start from the smallest to find the eigenvalues.
We write λ = 2. When we put the matrix again.
We are removing the λ value via diagonal.
But now certain λ.
There are three different values.
The first value to the two happening,
See here when we organize this equation consists of two numbers.
-2 Here, here 2, X, Y, and Z we multiply,
e1 of the components X, Y, Z call.
To facilitate the work, to facilitate writing.
As you can see, it turns out 2X- 2Y + 2Z first equation.
We look to the second equation.
-X -X + Y + Z + Y + Z,
We look to the third equation X -Y + 3X,
just once already this second equation,
We see that the first çarpılmış -2.
Çarpsak -2 to +2 will be the first number, where the number will be -2,
We find this number again here.
Another audit would also have the following:
X-Y + 3Z first of equations
-2 floor of the equation,
If you add a second equation that you find.
If we can achieve this, so we can take any two of them.
For any two of them, it seems easier to get that second and third.
When we collect this equation, we collect when they see X 0 turns out,
When we gather that they 0 Y turns out, 4Z remains, 4Z = 0.
So here we find that Z 0.
When we find that Z 0, Z If 0, these two equations
Even now it seems independent at first,
Z, U çarpılmış the axis to each other is 0.
So there are these two equations instead of one equation.
Or give X- Y.
E provide it, we will make a selection an easy way.
Because against two unknown, there is one equation.
For example seçsek X 1, Y will be 1.
Z is the number that we have already found that before the 0.
So we found the first eigenvector.
We're still the same process for the second eigenvector.
We write the matrix again.
-λ We put on the diagonal.
Λ is now no longer certain, 4 this time.
We organized this equation
When the E2 component X, Y, Z say.
Because the X, Y, Z of no value to them.
After that we set E with the values found.
As you can see in the first coefficients of the first equation it is 0.
So 0 times X, X does not appear.
-2x + 2Y, you look at the second equation
-1 stay here for the time X, -Y + Z
this is our second equation; Our third equation in X-Y + Z
As you can see there is in this equation.
So what is the first equation,
here again it seems now three equations to three unknowns.
But if done correctly account our appendix, not three independent equations.
Indeed, the first equation, the first equation,
readily seen that the sum of the second and third equations.
We gather the second and third equations.
I will go see X, Y -2Y come from; Z 2Z of the future.
Thus, it is not three equation, there are two independent equations.
If we take this first cut it in simple equation, comes -Y + Z.
In the second equation, the X-Y + Z looks.
When we collect this equation,
See more precisely when the second equation we have out here -Y,
+ Y will have to take; Z there will be take-Z; only X remains.
X=0.
We found wherein X is zero in time,
see these two, it's completely identical to each other remaining two equations.
Thus, again against two unknown, we have only one equation.
So, again, it will be infinitely many solutions,
An easy solution to say again Y 1; Z is 1.
We found it here at the X 0, Z would be 1.
So the second eigenvector eigenvector is going well we saw.
The third eigenvector, we find choosing λ'y 6.
Because the roots we found that we remember the beginning, 2, 4 and 6.
We have 2, 4've done.
Now came the third eigenvalues.
λ on the diagonal rather than -6
The equations obtained when we sort what you see here.
Still unknown seems to be three against three equations.
But first equation is doubled,
minus two plus hit,
rather, the first equation,
We see that çarpılmış minus two in the third equation.
Therefore, this equation also düşürünce,
We can work with any two-one.
Second and third equation to work here because it seems easier
No number 2 here.
These, as we have seen from, went down to two equations to three unknowns.
Inmesek already, we need to understand that this is a mistake.
Determinant is zero for three against three unknown
not one independent equation.
There may be two, we can see in some cases even a genius could have landed.
Now here we solve that equation,
for example, I gather them.
X are taking the time we collect see one,
Z is taking his one.
Here, four Y = 0 is happening.
Therefore, Y = 0 turns out.
Y be zero when Z
where Y is zero because the rest of the equation
X minus Z is equal to zero, we fall into one of two equations equation.
If the data is also an X-Z, we find right here.
You could otherwise be solved, whereby three unknown against three equations,
Z would choose any number, you can choose to present such a Z, Z
Y is zero to one, you are found to be otherwise.
Summarizing these results, as follows; first stem
lambda vector essence we found against two equals here.
We found the second root lambda is equal to four
self-vector and vector essence we found six against the lambda value.
In accordance with these core values are still discrete theorem
opposing them for being different from each core values
are independent of each vector in, you can control it.
For example, if you think e3 e1 and e2 to express terms, you can not express it.
Now again three three-pointers
We will find self-worth and self-vectors of a matrix.
We can go a little faster now.
We are writing again to the minus lambda calculation of the determinant
Before we put on the diagonal, and we calculate the determinant.
This determinant of any line, but you can turn the column again
According to the first example, let's open the line, the more we open earlier compared to the first column.
If we do this process, as a third
We are faced with the very function, the power function.
Lambda squared minus one, minus lambda gather five.
As the largest force will be lambda küplüce as you can see here.
We now face a special situation here.
A root yes five, the other a root.
But we see that the second root is repeated roots.
So this lamp in a matrix of core values,
five; lambda two, one; Lambda three, he is also.
Because a power function of the third degree will be the root of three, how
The second degree function of a force always has a root two.
Now when we place them before feeder start,
here again we put minus five on the diagonal.
We are looking to opposing this matrix equation.
Yet now we see that the third equation
It can be obtained from the first and second.
I gather these two first and second
Gather equations and cons Let checked.
This happens twice in the third equation.
We can already find three unknown against three equations.
If we find we have made a mistake, you're wrong, we calculate lambda.
Therefore, there is no need to say which of them every time something in our search,
We can take any move or two.
This is because the two equations to three unknown if we take for example a Z,
As an easy choice, X and Y, the rest of the two can solve the equation.
Wherein x is a, in which Y has one.
You do interim detailing anymore, you can easily find.
Indeed, a Z. When we put balls in this equation,
If you remove the X will disappear, you will find the right Y.
You will find immediately after finding the Y. Xu.
In the second eigenvector a little more complicated,
We put a minus, not to be mixed, or rather a new case.
Hatirliyalim'm a repeated root.
Thus, both lambda two, one; lamp and the lamp in two thirds, equal to one.
As minus one equations obtained when we put it,
see here one, two one, one, two, one, one, two, one turns out.
As you can see all the rows of the matrix is equal to each other.
In contrast, when we wrote the equation, the unknown still three against three
It seems to have the same equation, but these three equations.
So there is not these three equations, we have had one equation.
Because here is a lambda repeated roots.
It may not always be the case,
but it can go two free degrees,
UNKNOWN until an equation can have two to three degrees of freedom.
The reason therefore is that the lambda repeated root.
Therefore, we have two free choice.
For example, X is equal to one, Y is equal to zero can take.
X represents a Y equals zero as sheep; Z is equal to or less is happening,
and we find the essence of the vector, or X is zero, we can choose one of Y.
Xu Y. When we choose when we choose Z in a zero
We see that minus two.
The first of the core values that we find our means
We have one self against the vector.
Second, repeated root, we find it self against the two vectors.
Now we will see a slightly different situation in the next example.
We see this as concrete can be repeated roots.
This repeated root, the root of two against repeated twice
we see that we can find a vector independent of each other.
Theorem does not guarantee that the two vectors is linearly independent
because in the forty-fourth page of us face theorem says
eigenvectors for different radicals from each other automatically,
independent self-interest, but in terms of e2 e3 as you can see here,
E2 is not solid, and therefore independent of the e3 e2.
Thus, despite the repeated roots, we find independent vectors.
Recent examples are as follows: this is quite simple matrix.
Moreover, the roots that were chosen specifically so simple Let easily.
A triangular matrix, the diagonal numbers below zero.
Here again we are writing matrix, via diagonal
We are adding lambdas are negative.
Of course, because it is a diagonal matrix triangle
determinant of the product of the numbers on the diagonal.
So five minus times a minus lambda lambda lambda times one minus.
A minus lambda is happening in the square.
So again repeated
There are roots and stem off the first five as in the previous example,
Examples of particularly I configured in this way.
Two well worth repeated that there is a core value.
Find eigenvectors according to five
We put in place for over five diagonal lambda,
The first term turns zero as you can see.
Two, one; here minus four, one, zero,
Z'l term zero negative tactics.
We see that the term Z'l zero.
So Z is zero; Z becomes zero, see where Y is zero,
here you found in the first Y zero.
Already two equations is zero when Z is identical,
going twice as the first of the second equation.
Therefore, when zero Z, Y, X has zero one back,
X is also no restriction on the number we can choose what we want.
The simplest choice would be to choose one anyway XI.
So where Y and Z zero turns out spontaneously, without an election.
But here we can make a choice for X does not appear anywhere.
We take the simplest choice.
When we light the second root instead of a place,
As you can see here, it consists of zeros on the diagonal.
4X + 2Y + Z gives the first term.
The second line on the diagonal, this is going to be zero only at Z = 0.
When we look at the third line zero times X
zero; Y times zero is zero; Z is happening zero times.
So the third line is 0 = 0.
There is no contradiction, consistent.
But he does not give any new information.
Thus we have two three-variable equation against us.
Since there are two equations one zero already happening Z
Z we put almost zero.
Here we can do X and Y for free elections.
X minus Y turns out as we get instead of a two.
But now here we observe it.
Only one response to this, we find the essence of self-worth repeated vectors.
Whereas the previous example, the repetitive self-worth
We find it self against two vectors.
Here now we are making an important observation.
Eigen vectors as the number of repetitions when repeated root
we can or can not find it.
That equation, depending on the nature of each
It can occur in both species.
Now I want to touch upon another issue.
This can not be real value of the essence.
For example, here we are, one, two, three, four matrix of self-worth and self
In the first instance we find the vector.
Here we choose number minus three minus two, that we get this matrix.
We're still writing the matrix.
We are lambdas on the diagonal and removing negative determinant account
We are.
This duality can determinants of two said.
Lambda comes from the two squared multiplied by the square lamp.
Lambdalı minus the revenue from this product trilogy
and income from the product once or minus lambda lambda,
therefore it equals minus three lambda lambda comes in two lamp.
When we look at the C-only numbers lambdas
minus three plus eight when we look at the term; Coming five.
So in essence we function characteristic function
also that we have zero determinant
This function is equivalent to that force.
If we look at the root of this, we can still use half of the formula
We work with Dr. double the number when half the number of ABC,
minus times a minus plus or minus the square of one C.
We can see a C minus that four times.
So this would be plus or minus two good.
So we have two complex roots smell but worthwhile.
Because this is our strength, our function from the real numbers
The numbers are getting too complex value equivalent of each other.
A plus or minus two and minus one or minus two.
Now if one evaluates the following examples,
Our first example was that; It was a two binary matrix.
Our last example also two binary.
Others also three three-pointers.
Now we find two real roots in the first instance.
In the fifth example of the duality in two,
In the example we can find a little bit before we do two complex eigenvalues.
The second example was found in all three core values
It was found in a separate vector for one self.
Three three-pointers in the third and fourth sample matrix, but a separate
We met while the two repeated root roots.
Opposite each of the third example a repeated root
We find independent eigenvectors.
But the fourth example repeated
We find the opposite is only one eigenvector roots.
These were selected as examples summarize occurring conditions.
We should not be surprised because I digest them thoroughly why she turns complex,
Or, so why not find sufficient self-vector.
The number of self-vector is repeated when the Roots
the number of repetitions may or may not be so.
The third and fourth examples show us that state, respectively.
Now we finish so general discussion.
Single crucial problems in the core values of the symmetric matrix
There are privileges.
It seems appropriate to take a break before them.
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