[BOŞ_SES] Hello.

In our preceding session, a linear

We have seen that the space of four about transformation.

Description space.

This definition can be a subspace in the space all goes to zero.

We call space-space-space or zero.

T space from this definition,

The transformation vector T are carrying a target space.

But we can not achieve all of these goals rather than space.

We can reach him into a subspace.

It also can be reached, the call can be covered space.

Now, before we see to be able to internalize more definition,

We will see four examples to reinforce.

Our this example,

ie two-dimensional space of the three-dimensional space,

target space from the definition of a space the size of the space is small size,

definition of the size of the target space is the same as the size of the space, a well

We will see examples of definitions space is larger than the size of the target space.

By these two examples we have seen for the last two examples five.

Now let's get started.

Given to us as a transformation.

(x1, x2, x3) items,

ie, a vector in three-dimensional space on a two-dimensional space,

In accordance with this definition into a T.

In our case, to determine the space of about four T.

These four spaces: Definition space, we start with the aim of space.

Easy on them because the definition of what kind of vector space

If T must be acting immediately apparent.

Three variables, must be acting on a space of three components.

Therefore, three-dimensional space.

If the target is a two-dimensional space space.

Because the elements one and two.

We want to find them for base vectors.

There are of course infinite base for a three-dimensional vector space.

Each consisting of one common side three vectors.

Among the most simple Cartesian basis vectors.

So our earlier i, j, k vectors we describe three dimensions.

We can choose them.

But that others could choose the most simple ones to them that this

standard Cartesian basis vectors or vectors.

Similarly in mind, the target in two-dimensional space in the target space,

two-dimensional, as we see here, again, rectangular or standard

We base the vector is going to select the most simple solution.

After that there's two of our space.

One space shuttle.

This target did not shut down every place in space.

Likely inability to close.

We are trying to understand how this space is a space.

This requires a little calculation.

See, where x1, x2, x3 and x1 because,

x2, X3 because any random numbers,

who accepted them in terms of their multiplier

We are writing the two-component vectors.

See here only say that it seems x1 x1 (1.0) 's

When we multiply (1.0) x1 we are creating with this vector.

One x1 second element does not occur.

x2, both elements are there in the first and second elements.

With a circulation of x2 (1,1) we have to multiply this product by the end of the first elements of the vector

in the second element in the x2 x2 Let's get the second component.

x3 looks somewhere.

x3 of these (0,1) when we hit the first component of the vector x3 invisible.

But the X3 will appear once the second component.

As we have seen in any space transportation

If we recall that a year of transformation vectors.

(x1 + x2 x2 + x3) was transforming as T.

Any y vector wherein f1, f2,

we show the vector is a linear expansion of f3'l.

but the floor needs to be two things, two properties.

These vectors we call this base vector

A set of all vectors in space can be.

Here f1, f2, f3 doing this.

At the same time so that the smallest tool shall be independent from each other.

Here we see very easily that the vectors f1 plus f2 f3.

Therefore, although this triple vector team

It can open the space shuttle, all in the transport space

vectors can not produce a base, because it is not linearly independent.

But if we get f1 and f2, a base vector that

We know that and are independent of both.

We do not because the f3 f3 linear combination of the two.

So the transportation of two-dimensional space that we see is because two

All of this can create the vector space

elements and here we are have found a base team.

The fourth space zero space.

Two kinds of space can approach zero.

We have a promise so important and we have stressed,

We draw here in red, we showed we have identified,

The size of the definition space, the size of the transport space plus the size of the null space.

We know that a third of the size of the space E expression.

We know that both the size of the space shuttle,

so there will be a zero-dimensional space to space.

Without any of these accounts,

Using this theorem, we find that there is a direct.

But now, in this space here, and let's let's determine whether these base vector in space.

As a base the size of the space will also comprise a single vector.

Zero vector n to show in space

and the effect of these return T on any vector,

the sum of the first vector of the first and second components, making the first composition.

He is doing the junction.

The resulting vector sum of the second and the third,

The second component of the vector is doing in the transport space.

To our two

We find one number goes to zero.

Here n1 plus n2 equals zero,

n2, n3, plus it will be equal to zero.

Now as you can see here, there are two equations,

but three are unknown.

Therefore, it is more than a random one for an unknown number

we can choose as a parameter.

For example, just say the last T n3

We find that T for T N2 minus.

The cons would n1 n2 with the T T.

Therefore, the zero vector in this space

We see that as a particular infinite with a T parameter.

This n1, n2, n3 vector common factor T

carrying out (1, -1, 1), we find that.

There's no space here at the zero vector space

here we see that any vector (1, -1,

1) We understand that the multipliers of the vectors.

This is our last course earlier than we remember centers

the three-dimensional space coordinate assembly central

We see that the equation of the right.

This is true because the infinite point

We see that this direct formation of the zero space.

We initially this theorem we know more

We used to find the size of the zero space without making any calculations.

Here also it turns out spontaneously to provide here.

The second example is the conversion of a three-dimensional space from a three-dimensional space.

We had a previous transformation to the two-dimensional space.

We did sample.

Now where x x x that two of the three

We see that with the formation of vector conversion.

Again, we would like to describe these four spaces, four spaces under the influence of T conversion.

Description space still easy, while three-dimensional vector,

Or, again, Cartesian basis vectors of vectors, the most simple

We can get the standard two to three vectors e a e.

Target space, as you can see it in three dimensions.

The first component x wherein x plus two,

x two minus x three second component, the third component in x minus x two.

No need to think about here and more,

we're going to a three-dimensional space from a three-dimensional space.

Again we can get the standard basis vectors in three-dimensional space,

therefore, we are also found at the base of the target vector space.

As in the previous example we have to make a little space for the transportation account.

A typical vector in the vector space shuttle to the right.

This output again because an x where x is two x three random numbers,

A typical vector in the transport space,

x x x the multiplier two-thirds of them can express the vector is taken.

See, let x a, x looks as a first component

The second component does not appear, it means that we must take a zero second component,

As a third component looks.

x bir'l to fulfilling necessary when we multiply this vector a x.

x When we look at two, located in three different places; There is also the first component,

in the second component, the third component.

Wherein the numbers a,

minus one and is manifested as we receive a vector.

Indeed, in this vector x instead of x iki'yl has hit two of

We can bring.

x There is a three in one place as you can see.

No In the first, the first component of the vector we get zero.

We take a negative in the second component,

We get zero again in the third component to the x three available.

This really is the vector with an x, x two,

x We will obtain three balls we hit the left side.

Now here's a typical vector in the space shuttle again

T (x) of the three vectors of the vector,

for one of these three vectors for two and we call for three straight

We see that the expansion x x two x three by selecting random numbers,

Select the numbers you want and you can always generate a vector

You can produce all vectors in the transport space in terms of these three.

Again, the question is; This vector team is doing its duty,

that is able to produce all vectors in the transport space.

But does it constitute a base?

The answer is simple.

If any vector expressed in terms of others

If the base can not be so dependent.

There are different ways to do it this time, a car,

c f c both hit with three unknown, for two,

Let's take it for three to zero.

Here c a c two c, we obtain three equations for three.

See a car from the first equation,

plus c, we find two c do not contribute here three c c a plus two.

Secondly, no car, but the car had two, minus three car there; third

equation here as the individual components that we take the time a car,

minus two equals zero c, c not yet contributed three.

For example first and second of these equations themselves

When we collect will go if you get two c, c will have a two,

here it is understood that a car is zero.

If we are going to remove a car, two car will be two,

c is understood that two is zero.

two three zero zero c curse, that means that there is only one solution of this equation

In this linear connection

The only solution is that a car

c c equals that of two equal three, we see that zero,

This also proves that here straight independent of the selected vector.

Thus, the vector sets, the trio vector team

but also a team that opened this space

that together they form between themselves a basis for independent.

So we see that the size of the space shuttle was three.

If we recall our previous finding a target in space it was three dimensions,

It covers all means of transport space target space,

both because the three dimensional; and this is exactly

It shows that the correlation, one to show that the transformation T.

Judging by the zero space, so in the previous example

We can use this theorem, as we have seen two kinds.

Without any finding or provide account the size of the zero space,

or select the zero space directly from there

It shows the supply of the size.

Now here, now we know the definition of space

We found that the size of three, we found that three of the dimensions of the space shuttle,

so it is size zero zero space.

That's only just the zero vector in the null space.

These problems could end right here,

but we are also providing access to zero belirleyelim space,

There's no need to actually understand the subject, but merely hand

it would benefit both our skills also increase our understanding,

We can see what's happening when the size of the zero space zero.

Yet zero vector n to show in space.

We think it's really the same from the main definition, no different from the main definition so far.

Only, in this way will we want to find that the zero vector of outgoing.

When we found the way to go to zero, resulting in three equations

We are: The first equation is equal to zero,

The second component is equal to zero, the third component is equal to zero.

Here we see now see that it is equal to the one of the two,

The two from the three is equal, so the one at the two,

The three equally among themselves.

Put a plus here we bring a n n n a a plus,

The two and the three, because the one is equal,

We therefore find the one that is zero for the three zero.

n would be zero when the two zero zero when the two would be in the three to zero.

Therefore, there is zero space as we have seen only the zero vector.

That said, here is an account confirming that theorem.

Let's continue with our example.

Consider again a conversion into a three-dimensional space of the three-dimensional space.

This transformation T still an x, x two x three is transforming the way we saw it right.

The first two steps still very easy; definition work space

T means that impacting upon definition three-dimensional three-dimensional vector space.

There may be, in spite of the infinite set of base

We prefer to choose the most simple, the most simple of these standards may base floor.

Target space in three-dimensional space again.

This standard vectors as the most simple base vector for this place,

We choose the Cartesian vector base.

Ie, for any year in the transport space

T (x), we are writing here is equal to the conversion year,

again this was done in the previous examples as the vector x a, x two,

x will be a combination of the three, we want to write it that way, because xa,

x two x three numbers that any one of these vectors for this emerging space,

team opened the transportation becomes a vector space.

In the same way we're looking at an x here as it seems,

As a second component looks, it looks at three in the third component,

means that x is a one, one, we are forming a where x stands for UC.

x two, one, minus one and minus one seems to be.

Here we take in this vector.

x three-one, minus two,

minus three is obtained.

As you can see in this, we are creating in this vector.

Therefore, compounds of this right vector

It produces all vectors in the transport space.

Therefore, there is a base for these.

The first task to produce possible all year, doing it.

The second feature is that they are linearly independent among them.

I have to do here with this account, but you can find details of these accounts with ease.

Write f3 as a plus b times a times f1 f2.

And here a negative one, three, you can easily find that b divides the two.

This means that the vector f3 f1 and f2 can be expressed in terms of.

As a result, we understand that f1 and f2, f3 is dependent.

There needs to be an independent base vector.

Then choose this duo.

To understand whether this is the duo's base, it is independent of f2 f1,

We see it easily.

If a solid f2 f1, f2, so get slammed with a certain number

If you could, would f1 to f2 dependent.

We see that happen.

f2 f1'l we do not get slammed with a number.

Therefore, we find that where f1 and f2 of a base.

as f1 and f2, f2 f3 well if we had a base.

These are independent.

f3 to f1 if we could create a base but if

that we're one or two, take one or two of the more simple choice.

Now here we encounter a new phenomenon.

Prior to the three-dimensional transformation in three dimensions, the three dimensions of the space shuttle.

Because three of the size of the target space,

all of the space shuttle tended to obscure the target space.

Here is the target three-dimensional space, but transportation is a two-dimensional space.

Thus, the target space is a subspace of the space shuttle, it can not cover them all.

Because the two-dimensional space.

Second, we see something that needs more attention.

Generally, if a three component if

vector team seems to have three-dimensional space.

But here in this space forming the base team

not the number of components is the number of basis vectors.

While three component shown therein,

The number of basis vectors for which there are two, it's two-dimensional.

Or show three component does not require the three-dimensional.

Indicates that only two vectors in three-dimensional space.

When it comes to zero-space, we can do the same things again.

We determine the null space directly.

Or using this theorem,

drawn with lower theorem, using the theorem that the size of the zero space,

both of which bring us results.

We called for a change here, let's determine whether the zero space directly.

Zero means that T-zero space transportation,

Thanks to T, this N1 N2 N3 zero vector of means of transport,

means are driven to zero, it means to be converted to zero,

all terms in the same context.

Here we get three equations.

The first component will be zero, the second component will be zero,

The third component will be zero.

As you can see there are three unknown here.

n1 n2 n3 and three against three unknown equation.

Although it seems a bit, but three equations

challenge of examination you get the second equation twice,

We get that third out of the equation as the first.

If we see it twice, that n1 twice.

N1 have collected here by 3N1.

N1 here.

We struck twice in the second equation.

Two missing n2, n2, then the negative is happening.

The second equation from the first equation twice çarpılmış

Collect, the minus is a minus four plus three.

So, although it appears here if the three equations have only two equations.

We can take the example of this third equation.

We have seen in various applications.

If connected to the other one of the equations,

If you do the Gauss-Jordan elimination, this third

You encounter reset to zero instead of the equation are equal, does not contain any information,

it is not wrong, because we assign this equation is consistent but does not contain any new information.

Reverse therefore remain two independent equations.

But the three are unknown.

To solve these three unknowns, for example, say x3e 2t,

When we gather here these two equations n2 expenses,

interest expenses to and from the time we n1 n1 t

We find that n2 minus 3t.

Indeed, if we settled in after them, t n1, n2 minus 3t,

plus minus 2t've x3, it is provided that the equation 2t.

We easily see that others also provided.

This means that we collect these components as a vector,

t be a common factor out, the back of the n1 1,

n2 minus three, N3 de 2 stays.

As you can see,

all vectors in the null space vector is multiplied by t here.

t may be for any number of random, there are infinitely many n.

But all these are not the random eternity,

but vectors which are obtained by multiplication of the vector t here.

If we remember from geometry, a direction in a three-dimensional space

gets hit by a parameter vector, may be even a constant that,

We know from the constant zero if the center is the right one.

Here are our last line of zero from the central space of 1, -3,

All points on the right 2.

E the infinite number of them.

But, thereby creating a space.

But because the whole point of a one-dimensional space,

this infinite point, each one creating a vector,

here are able to account as a multiple of the vector.

We see the provision immediately.

Zero space than the size of the space dimension to the definition of the size of the space shuttle

removing happening.

Description I have found that a third of the size of the space, where already was immediately obvious.

A three-dimensional space.

We found that two of the dimensions of the space shuttle.

Using these two, we find that there is a space the size of zero

Here we have found itself in direct null space.

Because it is one variable parameter

We have provided just that one-dimensional.

Now we are looking at a somewhat different example.

In previous examples a transformation T

taking the space from the definition of the target space

or less dimensional him or he had taken the space in size.

Here's taking a bigger-dimensional space.

This transformation T four-dimensional space vectors carrying the three-dimensional space.

Let us define the transformation T has said.

Here x1, x2, X3 consisting vector,

components as it separated by commas wherein the first, second,

We carry a four-dimensional space for the third and fourth components.

The dimensions of the space again with the aim of definition given space, three and four.

The most simple is also the basis vectors to be here.

This standard basis vectors, Cartesian basis vectors.

Transportation accounts, such as a space still earlier, we need to opening.

Here, although the four compositions in y,

not only in terms of three variables x1, x2, x3 is expressed in terms of,

so vectors in the transport space x1,

x2, x3 consists multiplied by three vectors.

Let's check again rapidly, x1 will be a.

No second component x1 in the second, zero.

The third component in a time x1, one.

The fourth component x1 seen once.

Referring to x2, a first component in a second, no third x2.

As you can see, this vector also comes as a third.

The x3, x3 not the first, second component x3

once it comes, it comes at a time x3 third component,

The third component X3 also comes in fourth once again.

As you can see not the first component (1, 1,

1) As we create this third vector.

So this year, the typical vector of vectors which transport space,

that led them here vectors g1,

g2, g3 they call the vectors.

Now, does this constitute a base vector that?

It is creating a base for the space shuttle?

Let's see.

For example, G3 can be expressed in terms of the other two

We understand this is a g3 vector dependent.

If it is not is independent.

G3 g1 equals a times we say here; wherein g1,

where g2, g3 g3 to be here

It will be here for four equations, two unknown.

See the first equation is equal to zero, a plus b.

The second equation will be equal,

This seems to be just the second, not in the first.

So one will be equal to b.

The third equation will be equal to, where is will not be an equals.

The fourth equation a plus b equals one.

Now we've got four equations but we have two more.

Just clear from these two unknowns; We've found somewhere a = 1,

b = 1; Is that a problem for other equations as a valid solution, is not it?

b, and we put the first equation,

0 = 2 encounter with such an inconsistent situation.

Similarly, a = 1, b = 1, the fourth equation

We are experiencing a situation such as inconsistent again put in place 1 = 2.

So, this vector provides the equality is impossible to find a and b.

Now, one of the resumption, does not make the other one.

So all of them provide no more than a and b vectors.

That means there is no solution.

So it is unable to express the G1 and G2 G3 currencies.

Hence g3 is independent of the G1 and G2.

These are independent of each other, so for this space,

we find that the vector space for the transportation team is a base formed.

Of the substrate because the three elements of order,

The extent of conversion, the size of this space is three.

You'll recall, the order of size of the space shuttle was deniyo.

The extent of what?

The extent of this transformation T, three have.

It now famous theorem Us

If we put in place, the size was three of the definition space.

Reached the space, the size of the space shuttle, we found three it here.

So it is size zero zero space.

We also find it again now.

But find they do not need.

Saying that because of the size of this theorem is zero zero space for us,

and it's only just proved that the zero vector in the null space.

Last and five ornegimiz,

We are going again in a four-dimensional space to three dimensional space.

This transformation is similar to a number of previous amount.

But, there is a small difference.

See here, all coefficients were one.

They have the same components.

We had made only two here.

Here, too, we have done a minus.

Now the space definition of the target space, easy as usual.

A three-dimensional space, we're going to a four-dimensional space.

Here we can choose the standard basis vectors.

Transportation space x1, x2, x3 multiplier When we create,

x1 factor at work again, one, two, minus an excuse,

zero, not because the second component x1 will be a two and a minus.

x2 multiplier vector, one, one, one, one will be.

x3 x3 multiplier vector is not zero for the first component,

one, minus one, two, zero, one, a former two.

Again open space shuttle, we have achieved a vector knows the team can produce.

Simple as whether they are connected to the base bi criteria.

We will take two of their three vectors is dependent on them,

The independent three.

In the previous example, the G3

We saw an example of this is independent of the two.

Let's see here again, g3, the g1 and g2 as yazalım composition.

If a and b allowing it, g3, g1 and g2 Abiliyordur produced.

If no such previous example to provide these equations a and b,

g3 is independent.

Just let us write them.

We take a once g3 g1 is equal here, we get b G2 times here,

here the equations that we will get four equations for the four components.

Two unknown against four equations.

The first equation 0 = a + b; second equation

It is not equal to b 1 a; third equation

minus one equals 2, the third component, plus b.

The fourth equation a plus b minus two equals.

Now we find here as a b.

b to think as one,

right from the first equation, we get a minus.

Now we have to solve two equations in two unknowns.

But there are two more of our equation.

We get a and b, that the two remaining

We want to determine whether it satisfies the equation.

E them in place, we immediately see that party.

See, back in the third equation, a, was a minus.

He came here minus two.

It was a plus b.

So two plus minus bit less it provided a third equation.

If we come to the last equation, a minus one, that was a plus.

b two birdies.

Here it is seen more clearly.

He is also a.

So 1 + 1, is provided in the last two equations means 2.

So all four equations that also two

We find the coefficients alpha 1 and b = 1.

So g1 + g2, g3 = minus happening.

Here, we see the supply again.

g3 is plus minus g1 g2.

This means that the two vectors,

but consists of a base is independent of each other.

Indeed, here also rewritten g2, g1 is not solid.

Therefore it is independent.

So this space, transportation space is two-dimensional.

Although wherein g1, g2 four vectors

If the components of a four-dimensional space

There is a two-dimensional subspace in subspace.

So here the results obtained, there are two observations.

One transport space, two-dimensional.

Target of four-dimensional space.

So it can not cover the entire target space transportation space.

It creates a subspace.

And there are two vectors in the lower space, so two-dimensional and size

in front of our eyes to determine which component is necessarily come from,

determining a sample as we are treated more concrete.

Not the number of its components, is made with a number of independent vectors.

Zero space, we can get the right gene theory.

We know the size of the space definition, three.

We found the size of the two reached space.

This means that the dimension of the null space is obtained as one of the three minus two.

If you pay attention,

As we saw earlier in this theorem does not appear the size of the target space.

Importantly, accessible space,

you can do the conversion on a 1000-dimensional space.

But you may not cover every part of it.

What is important is the portion of space that you reach your destination.

His goal for the size of the space available in this theorem.

To find out how the formation of the zero space

again this vector in space T

Turn it into vectors to zero.

We write them like that again any equations for the transformation T x.

Here it consists of four equations.

But because we have three component T's effects space

on three-dimensional vectors.

Wherein the first

component is zero, the second component is equal to zero,

The third component is zero, the fourth component is zero.

If we solve equations in one and two,

n1 n2 minus right here,

n2 minus the right solution in terms of n2, n3 we find.

After placing them instead

We find the whole issue here but two,

sorry there are three to four equations the unknown.

I wonder if these four equations are independent from each other equation?

You can easily see that.

This n1 is equal to n2, n3 is equal to N2 with the second and third equations

When we place, we see that these equations be provided as identical.

For example, there will be two times two minus n1 n2.

Here plus n2, n2 remained negative.

N3 south eksi n2'ydi.

This equation is provided as identical.

This equation also provided identical.

So just we two, and this equation by equation

do not bring us anything new, because they are provided as identical.

So, this information is sufficient information.

minus n2 n1, n2, n3 sufficient information to be minus.

Even we at T n1, n2, T becomes negative,

n3 is also a minus, because it would have a more negative T.

Therefore, these vectors the vector in the null space

Vectors are converted to zero by T.

This one, minus one, is the multiplier of a vector.

Again, as we said earlier, a correct equation in three-dimensional space.

Here are the coordinates of the vector equation of the center of the team from the center.

This means that zero space is infinite on the right

Combining the starting point consists of one vector.

Although we have the H vector in the base, there is a dimensional vector.

Although still consists of three components, we see that the size of one.

Now there's a change.

For example, we have given an arbitrary value of T if we n3,

N2 lieu of a T value,

They are like a still different

Even if we see that the same base vector.

It was minus s, with the need to multiply.

So nothing has changed in the base vector.

So far we have five

We should seize together as a comparative example.

Transformation from a three-dimensional space on a two-dimensional space.

We found both the size of the space shuttle.

So, again, we need to remember always to ensure that our theorem.

Description of the three space dimensions,

The size of zero because the space shuttle in space would be a size two.

He also becomes the space R1, a space and one-dimensional space composed of real numbers.

The second conversion is a conversion from three dimensions to three dimensions.

Here we find the three dimensions of space transportation.

Again, we look at these theorems, definitions size, definition of space size üçtd.

Reached space size zero because the size of the three,

dimension of the null space is zero.

We are going again in a three dimensional space to three-dimensional space in the third example.

But here is the size of the space shuttle,

We call it the order of conversion, we found that there were two.

Thus arises the need to be a zero-dimensional space.

Here we see the following: three dimensions three dimensions

We can go out us different scenarios.

First, the transport size, the transport space size three,

Three extent of conversion.

Three dimensional target space.

So, as a result of this transformation

All the objectives of all the space, we can get them all.

So, a space transport space covering the target space.

The size of the zero space zero.

Again, although the third dimension of three dimensions, transportation space

We see that it is possible the target space is a subspace.

In this case, the null space will be an empty space, a space will not be trivial.

That is not only the zero vector, containing the infinite element

but one-dimensional, that is arrayed on a right

We see that there is a space composed of points.

In the fourth and fifth example, there are four dimensions to three dimensions of transformation.

But we face two different scenario.

In the first case, i.e., in the fourth embodiment,

The size of the three space shuttle.

It was about the size of the target space.

Therefore, not covering them all.

But despite this too, it consists of only zero zero space vector.

Size zero.

Here we see the following: In both cases

The size of the space shuttle three, but one of them

in a three-dimensional space, a three-dimensional space, thus fully covering.

On the other, because the destination is four-dimensional space, his only three dimensions

can turn, can reach a conversion.

Here we also see the following.

Örtmes even zero space shuttle space completely,

target space, the zero-dimensional space can be zero.

Here, too, because a size zero but does not completely covering space

As we saw again in this theorem, the definition of the target space space,

no place in the relationship between space and achieved zero space sizes.

Because you target well into a million-dimensional space

You can get in but you can turn it in place in a two-dimensional, three

You can close a lot in size, you can close a lot in size lakh.

But the size of the target space does not work in providing this theorem.

But if the equation, there are targets

We will see that the importance of space.

Now I want to take a break here.

This we have identified four space on a linear transformation.

Our two previous session.

A preliminary, that in our current session we saw several examples about these spaces.

Let them call for a review of.