Hello preceding two

an objective definition of a space section

who have studied the properties of transformation processor space.

Among these goals is a subspace of the space,

an accessible space with the definition of space

the subspace, we saw the features of the zero space.

Here, when given a vector x T

We were carrying it to a vector y.

Now here instead of the vector y,

Let that be a known vector b.

Let's change also with a T, then to highlight the differences in the two intervening simple.

We use the transformations,

It takes us to a known b.

In the T it became a general rule to move the x, y.

Because y, x, y which x is a general moving?

However, when given where y, x will no longer be public.

We say that x to provide the solution to this equation.

It is known to be in the right because we're creating an equation.

So in fact we are going to equations of transformation equations

also it is known vector conversion is a conversion, but reached.

Now here, as we know from the algebraic equation

Gaussian elimination at the very beginning of this course in two equations with two unknowns

X can provide a single equation as we have seen this kind.

X may or may not be available at all x infinite.

We meet with three different base case.

We will now try to understand them in terms of conversion and

We will give you the basic criteria of the existence of the solution.

When these solutions are endless when the only solution when

We will try to understand the criteria that is the solution.

There were also three cases where it emerges in the equation.

If b is a vector in the right-hand side here, it's algebraic equations.

If we show to be a function of differential equations or

but this may be integral equation or a functional equation

In this course we will examine the algebraic equations in the book.

If we visualize what we are saying this,

We are experiencing the following three conditions: A x T

When we convert it turns out to be a (x)

We have shown that with equal time be given special status in that right.

If it be, if there is to say that the solution to the transportation space.

Because this b reaching x we can find.

If the T x of the space shuttle

Although we carry, not be in this space that transport

it is not possible to reach the T x b.

Therefore there is no solution to this equation, it is the visual expression

not be in the right space shuttle.

There is a third case again, we get x T

We came to T b and b's we carry in the space shuttle.

That means there is a solution, but there is now the question is, did odd solution or infinite solutions?

For example; If it is a vector of the null space x

When we move we collect X. x plus n is obtained

which will be then we will move in infinite solutions.

Therefore, if only the zero vector of the null space that x is no longer b,

There is only one solution because it is not something we can get any x-added,

but if these are also the nonzero

you add them all again went to b, b is in the transport space,

therefore, as we find in the infinite

zero if other solutions are also zero

Obtain solutions in infinite space.

As you can easily see how these spaces

It shows whether solution.

The task is to determine whether the transportation solution space.

The task of the solution, if there is zero space,

to determine that one or an infinite number of solutions.

We will see examples on it.

But I emphasize that the right to go to the sample; If one

When the effect of X on a solution if it goes to b

but bearing any n vector to zero

In this also constitutes the elements of the zero space.

If we add a layer of this X on any of the zero vector space,

of this solution

Due to the linear transformation,

A positive effect on the x tn, x-lowering effect on

Çarpılmış the effects of the n t and collect,

This definition of linear transformations, the definition of linear processor.

But (n) as you can see here only because it is 0 (x) remains,

A (x) we see from the TF for that x plus b is equal to b.

This means that; a ÇÖZÜMSAN x, x plus tn is also a solution,

immediately clear if the state 0,

If you are in the null space of only 0,

There can be only one solution, but the non-zero vectors,

As you can see t be what it takes to be the solution if there are infinitely many solutions.

Here is the information provided to us immediately on these linear transformations of space.

And besides that determines how much of the solution is as easy to use.

Here we compile the information presented here may be in a slightly different way.

b has the solution in the transport space,

so; b transport space "for the Range" is called

This terminology is used in English in the world

we also have a lot to show him the space shuttle with RA,

I hope those used in international terms,

follow international publications can gain a little more ease.

For this we give these internationally recognized terms.

This is the dimension R "order" we say in Turkish, English in the "Rank," it said.

If that is not right in that in R, we say there is no solution.

One of our visual detection,

here we are now trying to make a statement,

If the space shuttle to be the only solution in this case either or

There are infinitely many solutions, but this is not the space shuttle in b, then there is no solution.

Now that we tabulate our findings,

if the size of the space shuttle,

It equals the number of unknowns;

The size of the definition space, the number of unknown

we know the size of this space is equal to that famous theorem definition

The size of the transport space plus the dimension of the null space theorem

here means that there is zero null space of the size of the implementer.

So there are only zero-zero vector in space,

so there is no longer anything to add to any X solution.

If you add zero again x interests, then there is only one solution.

size of the order of English r transport space "rank"

if it is less than the number of unknowns; then

We understand the dimension of the null space is greater than 0.

Because theorem that says; r plus zero equals the size of the space.

In this case the zero space so forever

we add the number of each of these elements to be in

When we get infinitely many solutions for a solution that would be the solution again.

The practical meaning of this is that the value R us

It also orders the size of the transport space R

The number of equations containing information from each other independently.

Let's say three to five of them but you have known,

means that not enough information to determine the five unknown

If you have three equations for, that is to say underemployed No information

In order to identify these unknowns squeeze.

So there's a lot of slack,

There are a lot of relaxation are endless so in a sense solution.

This is a table presenting a visual brew that these determinations.

Examples will do now.

These examples are also two dimensions to three dimensions, as we did before,

different examples, including also two dimensions to three dimensions, the three dimensions and three dimensions.

We examined this case,

for example, we find the basis vectors of the space shuttle,

We have the state of memory space zero.

Now T (x) = y is not T (x) = b say,

i.e. a known vector, the vector is a zero if we.

Now we can solve this equation, it is finally the two equations with two unknowns.

Excuse me, are two equations with three unknowns, we can think of solving it.

But we have to solve it by the size of this space

We can not know what kind of solution examining.

This is important information.

Sometimes there are very complex problem, there is no able to solve this problem

Or, you may not be easy to solve, but do not have a solution before solving,

We try to figure out what kind of solution can have them.

Here the transformation x where x is a two x three,

x plus x two x two plus x We turn three and the

right side is a zero vector.

We have a base here we find the vector of the transport space, zero,

zero, as it will remember more detail referring to the previous example.

Where x is a kind of x and we separate the sexes in terms of the vector x three

When a typical vector in the transport space is located here

We know that the linear expansion of the vector.

Right here we see that if the right one,

there is zero e1 = (1,0) We know that it's now been denominated.

So here's a solution.

If we look at the size of the transport space in terms of size two,

description of the three space dimensions,

means that we understand that in a space the size of zero.

We found that solution means that by looking at the space shuttle.

Looking at the zero-dimensional space is a space in the null space,

we see that in the non-zero vectors in that

We understand that an infinite number of solutions to this equation, solving than ever.

Let's find this endless solutions.

T x A x two x three to one, he came to zero

We know what we found here

solution x1 = 1, x2 = 0,

x3 = 0 is one we can find a solution immediately,

x x two zero one a sheep a sheep here, zero, zero.

When we came here xa; a, x two; zero, hence,

It is zero and so we find that it is a solution.

Does it matter if all solutions is also the only solution that can be added to it

absent; We find it in the null space.

Let's determine whether the zero space; This transformation of any space zero zero

It is achieved by moving.

Here we find two equations to three unknown

therefore n3 = t as an arbitrary value

seçsek just the two of minus t,

We find also that the t minus the two.

Just here we get that the structure of the zero vector space.

As we have seen, there are infinitely many vectors in the null space.

T we change because you get a new vector.

But it's not an infinite number of random, all in one, minus one,

we see that the vectors in the same direction as a vector.

So we any x solution,

here we have found a solution that observes a zero-zero,

that by attaching a zero vector of the space that we find a solution.

We achieve this is that the general solution.

We just have to make it, so we also provide the

We know that traditionally come by sifting through these equations.

Which of these equations?

See, there are two on the right side of the equation; x1 + x2 = 1 x2 + x3 = 0.

So, three against two unknown equation.

s value as a random variable X we put the three, because before t

Whether we call it independent to say, we put any arbitrary value s,

You can see right here that when the x s x two minus three water it,

When we take here it will be minus s also a plus p.

So our solution x is a plus h,

two minus x s x s as three well here.

Here see the separate terms and not the terms of s'l the F s'l

The term a non-zero, will be zero,

The water taken s'l terms of a common factor, minus one will be the one.

This completely from obtaining the solution, rather traditional

to achieve the elimination find any solution to zero in this space

We see that we have achieved the same solution by adding solution.

There may be cases where it is difficult to find this solution.

Imagine that a large number of equations.

They are able to find, but you'll find that puts the computer in the finding that

Is there no solution, it is important knowledge to understand the qualitative nature of the solution.

Second example: If you remember,

There were two previous courses in three-dimensional space conversion from three-dimensional space, we also solve.

They are one of the right side of the two, zero, let's solve zero.

Why two, zero, we get zero?

He said uğraşmayal their account details.

Otherwise, no one had any difficulty with numbers,

It must deal with a little more calculated.

Let 's understand the essence of confusion account almost zero download

so we choose such simple figures.

Looking at the structure of the equation, x is a vector multiplier,

x two multiplier vector and a vector multiplier x'll get three.

x a multiplier; The first one has a x,

x There is a second component, a third component.

As you see, one, zero, one.

x two of multipliers; a first component, a second component,

not the third component, one, one, zero.

If we do not see in the first three components x x three,

There is a second component in the negative, the third component there is still a minus,

If we multiply x x minus three then because we stood with two of the three

because it is minus one, a vector in which the two is zero, coming to zero.

Now that we think in terms of space and the right side

any vector in the transport space in the left side.

Now that we know the transportation space, basis vectors of the space shuttle

If we choose these three vectors for one, for two,

We find three of transport for space but their linear expansion

they form a base to which it is independently three vector.

Now the problem b min in this space?

Is it in the space formed by these vectors,

You mean it can be expressed by the vectors of this base?

Here again we are writing these base vectors.

b where the basis vectors can be expressed in terms of looking at mi

indeed, when we collect them,

e two, zero, zero, we see that.

a plus for two for two, one, one gives,

also for attaching it is not a thing of two to three.

A negative one around here, leading to a negative bir'l.

Indeed, we find immediately that such a solution.

So b is in the transport space.

That means there are solutions.

Now Is the only solution, we can respond to it is not.

Definition of space size is equal again right here,

access to space dimensions, plus the size of the theorem we use the 0 space.

Definition three-dimensional space.

Because x1, x2, x3 contains.

The number of unknowns.

In practical terms.

We looked at the e transportation space.

He also saw that the three-dimensional.

0 means that the size of the space is 0.

So 0 0 vector only has space.

Therefore, to add to any solution

There is no other solution 0 from space.

Thus, he found the solution solution

We found the only solution.

Now in terms of space it 0

We can also examine

but the theory has shown us already, just 0 0 vector in space.

Here we found that this transformation only when we equalize 0

0 vector is here to provide the solution.

Providing essential solutions now.

We also have a single solution and this solution can also find through traditional qualifiers.

This is equivalent to the following equation: the first component, the first component equal to the right.

x1 plus x2, equal to 2.

The second equation x2 x3 minus, right is equal to 0.

Again x3 x1 minus the right is equal to 0.

We are writing these equivalents.

Just look at last summer when

We understand that here in x1 is equal to 3x.

x3 x1 x1 place here as we understand that X2 is equal.

x2 and x1 equals, here 2 times x1, going 2.

So x1, 1.

x2 is 1 to equal.

moreover, it was equal to 1 x 3 in.

So the solution of 1, 1, we find that one.

We have found that the coefficient of the transportation space.

Burada x1, 1; x2, 1; x3,

We have already determined there is one.

So every time those providing space

It is understandable in terms of building size.

Let's take a third prefix.

It looks like the previous part.

1 1 0 taken right-hand side.

Considering the equations, x1 plus x2 is equal to 1,

x1 x2 is equal to minus 1, plus 2x2 x1 is 0.

So here are two unknowns, a two-dimensional space,

a three-dimensional space conversion, and in this way it is given the right.

Providing three against two unknown equation x1, x2 there?

Now let's look at the structure of the equation immediately.

We need to find the space shuttle crash x1 through the material presented here.

x1 1, 1, comes first.

The first vector.

x2 1, minus 1, we need 2.

This forms the base of the two vector space shuttle.

So here e1, e2 solution space now do not have the answer.

Vector in the right-hand side if 1, 1,

0 can be expressed in two vector species, there is a solution.

No solution can not be expressed.

Now see that we write.

b, m can be expressed in terms of E1 and E2?

Denklemleştirel it.

b instead of 1, 1, 0, c1, e1 times; 1, 1, was 1.

See also: e2 where e1 and e2 are here.

E2 we put in here we find two

unknown c1 and c2, we'll get three equations.

How this equation?

c1, c2, plus 1 equals.

We are synchronizing the left and right term components.

C1, c2, minus the second component on the right.

It is also equal to the second component of the known vector.

The third component will be c1 plus 0 equals 2C2.

As we can see there are three equations.

C1 and c2 we solve the first two equations for two unknowns that.

When we gather here 2 times c 1, 2 would be, would c1 1.

c1, c2 is 1 becomes 0.

We find here also 0.

We find here also 0.

You think to take place the third equation, if there is a solution if it provides,

If no solution provides.

When we put 1 and 2, 1 is equal as you can see

We face a situation as incompatible 0.

This means that there are no coefficients c1 and c2 provides this equation.

E1 and E2 therefore can not be expressed.

Thus b is not within the transportation space and there is no solution.

Now let's try to figure it solve the traditional screening method.

Our equation follows.

T x1 and x2 to see here Transformation,

x1 plus x2, x1 minus x2, x1 plus 2x2 equals 1, 1, 0 equals the future.

This summer we zamn the first equation,

second and third equations of equation.

This is equal to the first time we have gathered from the two equations 1 x1, x2, we find it equals 0.

When we put here not provide third equation.

So in solution by traditional sieving these three

equation that allows multiple x1, x2 we see is not found.

If we can provide twos.

But providing no more than three.

But they are given us the three equations.

Because here the transformation equation equals three.

Coefficients of the right side of the first equation,

second count of the number of right-hand side of the equation, and the third equation right

side.

Coefficients x1,

We are not the solution of the equation x2 can not be found for the unknown.

And this is a checksum is not included in the transport space b.

Now here again we take a break.

I urge you to reconsider your analysis of this equation.

To understand thoroughly.