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[BOŞ_SES] Hello.

In our last session we briefly.

Now we are doing the conversion.

This similarity transformation, obtained by multiplying the left and right Q-1 Q

Q The conversion time is a major transformation because we choose convenient,

The columns of the matrix of eigenvalues

When we create, we obtain a diagonal matrix.

Now the question arises: to have one element in the overall n².

n rows, n from the column.

But this time we make diagonal,

the only one information becomes zero because all the others.

Maybe it is reset but also carries information

mainly information on the numbers of non-zero diagonal carries.

Could we in on this a number n²

We compress the information it carries one of the item's information.

I wonder if there was a more casualties?

It is not just here to prove that theorem.

Already, though he would not have a sense of the eigenvalues and diagonalization mania.

Proof is very simple.

B Let's say we have achieved by converting from the matrix.

Let's compare the eigenvalues of B eigenvalues.

Then we find the determinant of a matrix eigenvalues then-λ.

This is the determinant of the matrix but also new

We do not know coefficient of λ 'or equivalent future.

Μ s time to give him the opportunity determinants

we found that the eigenvalues.

Now our aim is to show that it is equal to λ μ the two.

This is so easy.

Now that B, Q-1 AQ yazalım B

Value -μ times.

Now that Q-1 to Q matrix multiplication unit,

If the edit-μı see that the matrix

If we keep the middle, minus -Q'yl

Q. Q. force on the first left if we keep to the right,

When we opened this product's Q-1, and the conversion of Q'yl,

The formation of the unit matrix Q-1, Q unit

Q-1 and conversion matrix data again Q'yl unit matrix.

Therefore, this inequality, we provide this equality.

Now here's a triple product.

We know that one determinant of a product

determinant of the product of one term.

Thus, Q-1 determinant,

A middle-μı determinant and Q determinant.

But Q inverse determinant is the inverse of Q determinant.

The product of these two means that an offer.

Therefore, we are proven viewed B-μ

then the determinant of times, once by the Q has fallen here,

A-μ time units determinant of the matrix.

But you also say it μ, λ as you say, the same determinants.

Therefore, they were the roots of the eigenvalues obtained here,

B will be the same as the eigenvalues.

μ λ will say that the roots of the same equation.

Therefore, under this transformation eigenvalues remain unchanged.

This is an important thing, that essence,

We can learn something starting from value added.

Det A-λı'nın determinantı λ

n'yinc degree will give you a strength in terms of function and

This force will function in the n'yinc λ'n force, the force will be 1st,

n 2nd will be the force and at the end of the λ'n

Let's call it that way first, and the first coefficient of zero force.

I1, I2, IN-1, IN,.

At the theorem says: This In the

value is also invariant under transformation.

Indeed, the determinant

In opening the A-λ, it's roots λ1,

λ2, including initial -1 by n'yinc forces that λn

here the power function of the nth degree

radicals can be factored in this manner,

We can always find it easier to simply eat 2 of 2 equation.

Root AnAysA two t, n = 2, which had a 2 for 2

bir matriste L-L1 kere l -l2 nin çarpımıdır.

That is why we are already seeing that this time the root of 0 is λ1 and λ2.

These coefficients means that the eigenvalues λ1,

λ2 can be expressed as λn.

2 of 2 easier to eat, we see here that the first number of cases, but in general,

We also I1 λn-1 multiplier,

The sum of the root.

Why?

See the terms of this series, open the product, λ'n n'yinc force of the future.

He once had such term of each λ's.

λ'n n-1st forces will be multiplied by λ1,

will be multiplied by λ2 it will be multiplied by λn'yl.

2 of 2 ate it easier to see.

We already saw him as an example.

So this second factor,

I'm sorry to be the sum of the first coefficient λ.

The second factor for this is λ'n n-2nd force

root of the sum of twos will be multiplied.

The latter number also appears here,

not just root λ λ where λ number of non-λ1,

λ2, λn'n will be multiplied that we will see that this could equally determinant.

In all the examples we have already made

in this, did we arrive, we have been very conscious to ensure

but because we have not encountered much emphasis on it.

This n'yinc degree a function of the overall strength of the matrix,

Independent feature of the eigenvalues.

n'yinc degree a function of λ force.

λ1, λ2 with the difference of course variable λ λ1, λ2 are also the root of it.

Coefficients here are not so random.

The stem has a relation given in this way.

M, whereby λ did not change under the similarity transformation,

so that the roots of these coefficients are in so n'yinc degree

function of the coefficients in force, they can not be changed.

So we bring them to the definition of I1, I2,

In the coefficients of the matrix it is called invariant or immutable.

I1 has a little feature.

Im, have a little feature.

I1 very easy to see the sum of all the roots In the product of all the roots.

Which is the most simple.

Now we will see them on the examples.

But we do say that if we start with, there's so n² number.

Each information carrying because of these numbers.

We bring this analogy transformation under the diagonal structure.

As you see we only have the number one.

But with all we have learned that the eigenvalues of this to the

The next transformation matrix from the same eigenvalues.

Because you call the eigenvalues of the diagonal matrix,

It respectively λ1, λ2 λn will appear.

On the diagonal -λ, -λ, -λ summer is already here now

This will give you roots.

I also come from the literal to be called so because of this I have this

λ1 and λ2 unchanging, immutable and therefore themselves in terms λn.

Let's make an example of 2 2 eat immediately.

For it is more concrete matrix benimseyebilme

the elements of a matrix as given here,

two, two, one, two indicators defined by two.

We write the determinant of this matrix diagonal matrix that core values

minus lambda over're enjoying it and we are organizing in terms of lambda.

See, there is a single term in light squared.

Only the multiplication of lambda where lambda square on the diagonal.

See terms Lambdal where you stood with lambda a1 minus 1,

where you stood with lambda a2 2.

No other terms of the lambdal.

1 a1 and a2 with the lambdas term multiplied by 2, a1 and a2 2 2 multiplied.

It is also here.

The sum of the first term on the diagonal coefficients as you can see,

The second determinant factor in this matrix.

Let's go a little further.

We force this second order function, where the course

two in our case, but the one that usually root.

This simple function, this second moment

We find that we know when we reset the root formula known roots.

The first term of plus or minus four times,

once the first term, the second term.

That's where we're writing clearly.

B squared minus times four times C

As you can see the roots minus B, so plus is coming.

D. Remove and we divide him into two primary root, because here,

here we add the square root of D, we divide into two.

Now we see right here in Lambda Lambda combine two of toplasanız

As you can see by the square root of the square root of minus D plus D will cancel each other out.

Other numbers are the same, a1 a2 1 plus 2 to be collected, but each

When combined because half a1 a2 1 plus 2 will be.

As you can see here, here we see the underling total at the roots.

So this matrix köşegenleş if it would have a lambda lambda two on the diagonal.

After köşegenleş conversion with numbers on the diagonal of the matrix,

The numbers on the matrix diagonals köşegenleş would remain the same.

This is the first observation.

Our second observation still two unknowns,

We know from the second moment function solution sorry,

Lambda Lambda combine two of çarpsak see a split of the four will be released here.

1 a1 a2 squared minus 2 here

First squared minus second squared plus B because BA.

II Square de D.

D in summer we're here.

So this time we remove the frame from D to go to the square,

It will return to the plus minus sign for that.

First, second karesinden minus Square.

This is a split of a four future we crossed a divide by two,

where he will take on four of them.

As you can see behind itself it remains determinant.

So here,

The roots of this matrix even something it also said something to think about the core values.

Only a quadratic function of the power function

You thought the relationship between the roots of coefficients.

Indeed, one root of the sum of a1, a2 2.

But now, now that we need to connect to the matrix.

The sum of those over the diagonal matrix.

The product of the roots is going determinant of the matrix.

Let's make one three three-pointers.

Of course, we can not take as general duality in two-thirds of the three.

Let us do it on concrete examples.

In this example, select the appropriate roots.

If we calculate the determinant of A minus lambda I,

so we know we're writing the procedure matrix,

We also put on the diagonal minus minus minus lambda lambda lambda.

As you can see we do so in terms of the expansion of lambda are multiplied three times

The cube will come to light, even less forthcoming,

minus the lambda lambda lambda minus multiplied by minus.

Lambda squared term future.

If you see that it is six open struggle.

If you look again Lambdal term future lambdal terms

here three three-pointers

When you open a lambdal terms of a determinant in the future when it comes to them.

Minus six times and coming lambdas term.

The root of these terms one, two, we find that there are three.

We observe that the immediate, here again, this third coefficients

degrees function of the strength came from that and just forget for a moment

to tell him we're looking for the roots of such a function.

See the sum of these roots plus two plus three; is six.

We see that the issue here.

If it roots for this product.

Once or twice he goes three times six.

That is not a feature of the emergence of these six roots because you get a simple,

two, we get three.

And then we see that the six also see here in terms of lambdas.

A minus is plus or minus signs are going well, but the gist of it.

And now we take the twos product turns out there is something interesting.

Two, the product of two and three; six,

combine three multiplied three.

So once twos lambda lambda two, three lambda lambda twice,

three times a lambda lambda.

The total of these eleven.

This need to surprise us.

This wherein coefficients.

Now let's go into this matrix.

The sum of the numbers on the diagonal matrix of four plus one plus one.

As you can see this one going the same six.

And this is not a coincidence that none of these.

Immutability of the core values underlying it because the roots.

And that the force function of the core values

it comes from the root of the coefficient.

Now we take the determinant of it.

Of course, this may take a little longer.

When you find that you üşenmeyip you account that it was six.

It comes to the multiplication of the stem.

Twos determinants if we see the slides on the diagonal

two of these have a binary matrix,

one, minus two, four, we get it.

One four, minus one, four, have one, we take it in,

This duality in the second half.

We take upon that the diagonal one.

We're going to the other skips his neighbors.

One, minus one, minus four, consisting of one matrix.

Now when we see them in four plus two accounts; six.

Four plus four; eight.

One minus four; minus three, which when we look at them on an off.

That is shown on a concrete example of this general theorem.

Ensuring.

That this theorem given in these pages that our main theorems.

Theorem tells the constancy of these coefficients.

And here we find it in a more general n'yinc ranges.

When a total of three triple course,

here it will be a total of three double dual.

The product of three only two-thirds will remain a lambda lambda lambda.

These are many

housing is fundamental properties.

Because if according to my arbitrary choice

then he would change some things can not be a credible theory.

Nature rules me, it is not affected by any one's personal preferences.

Do you find the same results in coordinates so that if you receive the ALSA.

Coordinates of the month after you make the calculations, if you get too down to earth,

Mars will be independent of the coordinates you ass team.

That's what I said of course this is a bit far invariants

but the basic features including those issues.

Now we are making a standstill again.

Then again, these matrices,

the implementation of a new core vector of self-worth

We'll end up seeing more of this market.