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.
Uzaydaki Vektörler, Doğrular ve Düzlemler / Vectors in Space, Lines and Planes
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Do curso por Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
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Düzlemdeki Vektörlerin Öğrettikleri / Learning From Vectors in the Plane
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
We work with vectors in the plane of our previous section.
Of course we can draw vectors in the plane.
As we can draw two component, we can show with two numbers.
Therefore, before we start with the geometry.
Then, each time the drawing is not easy to do and is already looking forward
It is not a clear process, a good thing to grasp issues.
But the two numbers showing vectors in space
Or, find the location of points on the plane, so we see the guidance vectors.
We learned to calculate the length of the vector.
Here we know the Pythagorean theorem vectors
Create a steep triangle, observing that this
We have found that indicates the length of the hypothesis vector of the triangle.
Thus the square of the hypotenuse of length to each other
We know that the sum of the two sides of the square.
Now we want to move to three-dimensional space from the plane here.
In fact, something to pass the three-dimensional space is very limited.
Our primary objective was to pass away more than three dimensions.
But the transition from two dimensions to three dimensions and from there we see how easy it is
We can easily expand the size,
We can expand from there in the infinite dimensions, this is our way.
The plane is easy to see how the transition, I had j vectors.
The unit vectors.
One of the first dimension, one x1 x2 respectively,
unit vector in the direction of the second dimension.
These components can be very easy 1 and 0, 0 and 1.
We obviously a point three times in the past three dimensions here
We need to demonstrate three components in space.
Now of course this is starting to push a little space in our geometry concept.
Because we have to do a trick like this in three dimensions in space
We need to able to display the image plane.
This is not an easy task.
If you look at all the art that the genius of those who do good,
We see in these three dimensions to describe mislead our eyes in two dimensions.
Now here x1, x2 and x3 axes çiziyoruz
Considering that he, orthogonal axes.
We are trying to show it in the perspective plane.
A point x1, x2 we look at the projection plane
and from x1, x2, we are removing coordinates.
We are also looking at the projection on the third axis.
We find the third coordinate.
Therefore, we can show the location of a point with three count.
These three points of the plane again, this time inspired by the good,
j, k vectors which are able to show in terms of three units.
That any vector of the same length in the two Pythagorean Theorem
time to start applying for the position of this point
If you submit xy, x1, x2 Let the projection onto the plane.
Here we find a vector.
He also where we draw the vector x1, x2 plane at the base of it.
We know the length of it.
That length of a base.
This frame x1 plus x2 squared.
Now that a base point with a vector containing the
Considering the level that produces a right triangle.
This right triangle here.
One point from the initial point up to a point that we say,
One side of the triangle that is perpendicular to the base of the triangle where the other side is also one of the two base.
Here he squared it to the base of a square
We know that two plus the square of the base, see here
As written, but we knew that the base frame of the already x1 plus x2 square frame.
Here we find by adding x3 frame.
Supremely easy to switch from two dimensions to three dimensions, you see.
Here you can add a frame while also x3 x1 plus x2 square frame.
Here show with two numbers x1, x2 x3 you add a well.
If you want to consider their basis vectors,
longitudinal us because we want to deterred i, j, k will wish to print.
But conceptually useful.
Base vectors i, j k are adding the next one.
Therefore it easy to switch from two dimensions to three dimensions.
This could well mean that we now sat in four dimensions,
We can now also fifty size.
The purpose and aim of this chapter is how easy it is to take three to two
showing could be passed on to a higher dimension
intuition to transfer.
Three dimensional three-dimensional space we ourselves
able to draw on our experience.
Even if we wanted to get some tools they can show true as much in.
But over the past four dimensions when we no longer had the opportunity to drawings.
Therefore, to understand better geometry concepts,
but not a good method to account.
Especially because you never can not pass beyond three dimensions.
Even if you have difficulty in drawing in three dimensions.
Easy in two dimensions, but we have difficulty in three dimensions.
But to say that makes us optimistic that we can now seeing this transition in four dimensions.
Ellie size can now be even a little more size to the infinitely cesurlaşın
function gives the idea that we can move in the space.
Here we need not dwell on all these equivalent to writing.
It means the collection of two vectors to collect vectors.
Or their i, j, k how we want to write the use of the
We are gathering from each other in two dimensions equivalent to the number in the same location.
Here we collect the third number.
It means to multiply a vector by a number of gene vectors i, j,
If you write with the means to multiply the whole thing k c.
Here, we see immediately that the components come equivalent to shock.
This i, j, we can draw kbps hakkaten also even make some tools
This vector also may show the bar.
But increasingly we see how difficult the job.
We can also process with inner product.
It is like the inner product of x and y in two dimensions.
To each other from opposite number hit, we hit collect components.
Here we collect a hit in the third for being the third.
Consequently easy.
So i, j, we prefer to k'sız spelling, we prefer.
Because this is an economical software.
Every time i, j, k in writing that they are using other
to reduce domestic product impractical task.
We prefer to work with numbers for him.
As well as opening a path for us in this n-dimensional space now i, j,
without the need to short terms of being able to show that it is possible.
Let's do an example.
This is a two-dimensional like we did.
This x, y, z components of the coordinates
I hang him on a weight by a point.
The weight of this
Next point A to point B on X
c point and tie them on to other points on the z
and what the future here trusses
How can I try to find a force to disperse this force as we have here.
Now a beautiful application of these vectors.
If you do not have this quite a process in itself.
While we strive to do this without vectors.
We think.
A, B bar that is how much force
We do not know but we know the direction of these vectors.
And we also know the B direction.
A given point.
Point B given.
The point C is given.
Point D given.
Thus we find the EU vectors.
The usual way to care
B coordinates to find the path that will be the EU vectors
We will raise the coordinates of the first point of the coordinates of the second point.
As we see here are the coordinates of B 1, 0, 0 because the x on the axis.
We extract the coordinates of that.
We found it because it minus 3 in 1.2 and below the hanging point.
Because we show the AC vector c point on the y-axis.
Its components are 0, 0, x and z components of the second distance y point.
Gene AC vector, wherein the carrying
We do not know what it is, but we know the direction of the force vector.
We know that the vector itself, we call it AC.
We also find it by removing the AC vector coordinates to the C coordinates.
Similarly to the z axis in this
We find that connects point D vector.
We extract the coordinates of the D coordinate again.
This means that we will disperse these forces, we know the direction in which the weight is distributed,
but we do not know how long it took the strength of each.
There also to force the x, y and z say.
So nok to the axis x, the force on the leading vector x,
i.e., the y axis of the bar to the point C
We do not know the load will be moved on
y, and z is also a voracious say, to the point that in our point D point.
We we're here as a force.
It shows the strength we win.
Repeat the same shape here now EU toolbar
x, we do not have the force of an x.
We know that this x direction, the EU force.
So b x multiplying the unknown number we define it.
X, the vector here in bold.
We do not know, but we know that the vector is a vector in the direction of this vector is
EU vectors.
But here the size of x.
We do not know.
This is a number.
We want to find out how many times the EU.
Similarly when we say things we do for our year and that x
the sum of y and z are equal to the forces w.
This will be offset by hanging this down to minus if we win these forces.
Here we find the EU.
Here rewritten: (0, -2, 3).
We found the AC.
We found AD.
Each of these x, y and z by multiply here we obtain a vector equation.
This will force even as we are also here.
It is also known term in the right side.
x, y, z bilinmeyenlerimiz var.
This vector equation produces 3 units.
We collect the first component of x 0 is coming times,
Next comes time -1, z -1 time comes.
So the first equations y z = 0.
Because there are 0 items in the right-hand side.
-2 Times the second equation x + 0
-2 times y times z.
We do this by writing.
As you can see here, no year.
The right side again 0.
This is the third
To write the equation 3x +
3y + z 0 times.
3x + 3y equals, non-zero number in the right side this time we are writing this.
A vector equation as you can see here, but that is part 3
we know it comes to opposing the equation 3 units.
Easy to solve this equation.
When we solve them x = 1, y = 1, we see that z is equal to -1.
So this is x, y and z force that we see in what direction.
Z means being negative by AD
where it not to D force is going to de D.
Because this is our minus value.
That means pushing the right side of this force will be needed.
Physical side of the business.
In fact, this also shows the rope with not tuttural.
Because the rope is only pulling force, these simple things that you see in the mechanics.
Therefore, here you need to put a bar.
Do not take our negative sign because you also check out.
When we put these vectors, we find this anymore.
It goes from A to B in the vineyard
As the value of x plus the force that needs to take it out.
If you think of it as physical, you will keep pulling it.
Y will keep pulling.
The Z got to keep pushing.
Because z -1 output.
This shows that interesting.
At the beginning we look forward to bağlatsak it also shows that the system will stop.
Because this is our negative output.
If we account forces here,
it forces the length of the vector.
Because it was no longer a force vector x for us.
The EU is a vector x the amount of this force, shows that the EU many times.
0 The frame of the frame components here
+ -2, + 3 the square root of 13 squared off,
This comes roughly opposed to 3.6.
We find one in the AC cord.
Minus the name is checked for your tutturamazs forward to it.
You have to put in a bar tried to go here when you gather accurate.
AC cord, we find the name of the force.
We can not do this job if vectors such as you see here.
Now I want to give message here: beautiful work in two dimensions,
we can draw.
We can revive again in 3 dimensions.
Ourselves because we live in a 3D space.
But even here vektörsüz it does work very hard.
Here's also not possible to draw further away when you move to revive in size
not possible.
We can only try to understand a number of things.
You can already consider how abstract,
so we can move towards our dreams.
Just what they saw.
We can go beyond business.
It is very important that the size of these vectors generalized sense.
After a new
still subject lines in space and planes.
We know the truth on the plane.
n equals x plus y.
When we go into space where one can pass towards the second point.
You're given one point per correct
and from this point it is given a direction, vector.
We are here, it's hard for us given point p0
We call it a point x0 for her.
Vector given.
Complying with this component, v, w say.
There is also a floating point pr.
General moving from the position of the vector x, say,
also to its components x, y, say.
In the same direction as we are describing,
in a plane as explained in geometry we can also disclosed.
A neck, the point P0 there again.
This is a given vector in the direction perpendicular to p0.
So we want to find the vector perpendicular to this plane.
Again, we'll choose a floating point.
P point is also its components x, y, z.
x vector.
Now it is prepared in a very similar way in both thereof.
We say: (P0, P) in the direction of the vector geometry
as this means that (p0, p) vector u solid.
p0 given, granted, we do not know t, we can say it's a solid.
For P x, y, z given variable are given.
As you can see we were able to transfer it immediately to geometry,
the equation in space.
In the equation of the plane of each of these floating point,
Composite the fixed point given for each of the moving point.
(p0, p), we obtain the vector.
N it will be perpendicular to this vector is perpendicular to the plane of the point in all this.
N vector of a given point.
So once we take the inner product p0 p n
this means that that two vectors.
Internal multiplied when it comes to equal 0.
So the geometry math in a language
He also told at the beginning of this language in the preamble.
The event icons in the language of geometry we translate here.
But this is not enough.
However, you can not do more than one job.
You still need to quantify it.
See right through algebra equations,
We are writing from algebra through geometry as follows: (P0,
p) means the position vector of the vector p,
x We have shown with the unknown vector x p zero
Remove the pan position as we get the zero vector p.
This is given of a solid.
This right here is the equation, the equation in space as vector equation.
If we want to find the components of it,
because we have the numbers to make transactions.
x x x minus zero zero Let the right.
This means that the x component will be equal to zero plus X times u t.
Because the components of the vector u, v, w we did.
u, v, w we did.
The second component will be times t r y minus y is zero,
The third component is equal to z minus z is zero at times we will be t.
Here again, the right side known Passing x is zero, y is zero, z has zero right.
Z of the plane thereof is correct if zero, w is zero.
Z is zero and zero when we zero in z.
Here's a T between x and y.
We know when we have no means equals x plus y t the equation.
Some of these side geometry is not essential for many linear algebra.
Again, we want to look at the plane of the equation,
the fixed point zero zero pa pa pa.
Here we want to find the plane.
This given that the vector perpendicular to the plane.
Thus every point we get on this plane by combining zero p p
We find the zero vector p.
This should give the multiplied by zero.
We show that with inner cross to steepness.
P Zero P still have the same right as the zero vector x minus x.
This is multiplied by n to give zero.
N given case.
N The components of a, b, c he get call it convenience.
N. I am writing a, b, c, vector.
x minus x x x minus zero zero vector components,
y minus zero y, z minus z is zero.
Domestic product means mutually
components in the equivalent position means to multiply.
A negative x will hit zero by x, y, y is zero with a minus b, c and z z minus zero.
All these are unknowns in product cost.
x, y, z are known.
a, b, c, x is zero, y is zero, z is zero.
See coming by here comes the ax, CZ comes.
These negative mark on the left side sum
the ax is zero, zero by coming CZ zero.
Also say it because it is known d.
If we pass in front of such an equation we get less.
Now you have to show it in the following.
We will work with equations.
The few things known about it see in two dimensions here
as x and y indicate when a correct equation.
Three dimensional x, y, z shows a plane equation.
If you're in the correct equations to solve two,
We know that the plane two lines intersect at a point.
Here is the correct cut-off point,
It becomes the solution is going to solve the two algebraic equations.
This geometry means that the intersection of the right.
This confirms knows that intersection.
If there is no common point of these two parallel right.
No solution means that algebra in terms of means.
If the top two right there is overlap infinitely many points.
Because this is true,
All points on the first DC are also on the second true.
This means that there are infinitely many solutions.
There are infinitely many common points in terms of geometry,
There are infinitely many solutions in terms of algebra.
Similarly, we can get these two planes intersect.
The two planes intersect along a straight line.
There are infinitely many points on this right.
These two planes may be parallel, may have no common points.
So it means that there is no solution.
These two planes may overlap with each other.
Again, this time a two-dimensional infinity, it is an infinite number of solutions.
They intersect at a point right in either the plane or the two parallel
He is no overlap would not be a solution or two, he olşuy infinite solutions.
When we look at the space they have.
In addition to these there are also two towards each other's position as the cut-off.
This shows the situation with no solution.
Here we will deal with this equation.
Also at the root of this kind of simple geometric equations
verify the objects and the intersection of the plane.
You can also see an example of a correct intersection with a plane.
Intersect at a point.
Intersections can, if it's true or plane parallel to the right
If the plane is completely in solution also can be infinite.
Here is a geometrically opposite of these equations
I do not remind me.
Here are two examples.
Simple examples of this.
P zero given.
Components, coordinates a,
two, three, said one vector given here.
These components are one, one, get one so easily.
We want to find the right equation and this was the case for the right of the equation.
x minus x is zero, was one floor u.
As you bury it here, we find a representation in terms of the vector x t.
When we look at the components in equal first component t x plus x.
x is equal to the first Enthousiast field.
The second component of x plus y equals two t.
The third component of our three plus X equals t.
We have found a number of equations as you can see.
Four are unknown here.
x, y, z ve t.
The following questions can be asked.
Wherein c and d at the points of two given points.
Are they on the right is not it?
Here we see that they can provide the equation.
The sources of this kind of equation issues.
Again, a simple problem.
X zero given.
Again the same point.
One two Three.
A given n vectors.
the components of the vector minus one and two.
This vector perpendicular to n and
We want to find the last plane of the equation x scratch.
Here is an example of the emergence of new equations.
As you can see X minus x is zero, zero components x x minus one,
two, three x, y, z from one, two, three would remove.
We are writing this inner product is zero because it is given in the other vector.
As you can see here once x minus one,
y minus y minus times a minus two times, three times z minus two equals zero.
These are when we say x
y and z on the left to collect as unknowns.
Known, too, you have a look here there is a minus,
He got here plus two plus one,
minus minus six here have had five,
by five also came as five plus put right.
We have obtained equation.
So now we're here in this equation but two more advance
receive one equation, if we plane each, algebra to show an equation.
Questions like these intersections team will take us to the equation.
That's also one of the fundamental equations of linear algebra topics.
And behind them regardless of the geometry, but there are also a small number of equations four
Though it was not against the geometrical variables.
A plane would be in a very abstract space.
That's more than the unknown,
Twenty-two-dimensional space, there would be a plane if unknown.
To teahhül it's not easy to embody the dream
but here we see already in the success of mathematics,
we can not keep things to reach the vehicle.
Thus, this section
As this example toparlıyoruz finishing.
Again, there are a number of examples.
These are not so much.
Again, the usual summarized as a finish.
Here we see the fundamental concepts and methods in this section.
It's a good thing to understand the philosophy.
Before we begin with specific topics.
We can see, we can keep, the vectors are able to draw on the plane
We start and we go to three dimensions, it is difficult to draw here.
Already the previous, already in our boots in our next section
be able to work with the vector is not possible.
We have seen that out of the equation at the end.
We have seen several transactions in the plane.
These include addition, multiplication and inner product with a number.
their equivalent of n-dimensional, summer, we dokunamadğı, we see,
It can not be in the draw size will be even function in space.
One of the important things that makes the inner product to each other.
This concept is important because the two show each event vector
If you thought these two vectors it is in line with each other
in which event indicates that the relationship of each to each other.
Indicates whether the orthogonal relationship with each other.
So all these concepts in a very abstract space
We have made great progress when we can examine.
There homework questions.
I encourage you to review them.
These issues will be further strengthened by remitted.
They already as short as quiz questions
We will call these and the like.
For now, here we finish this chapter.
Our department can appear again next two simple two unknowns
They taught us the equation.
Here almost all topics in linear algebra here too
account concretely visible again without confusion,
You can draw, we will see in a structure can be felt.
Determinant of how we see the emergence and
We'll see how that arose in the context of this simple matrix.
In the next issue will move a little more work soyutlaşarak
but we came to these parts of the factual information that abstract understanding
I believe it will be easy.
Bye now.
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