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Hello, our previous session

the function of a matrix We have seen that there are two basic ways of calculation.

Any function using one of the Cayley-Hamilton theorem,

The 1st to n minus n is the order of the matrix

We have seen that can be calculated in terms of forces.

In this second method, the matrix diagonalization

at the same time any function in this matrix transformation köşegen-

and he is diagonal in the prevalence of asthma by reversing

matrix on

placing the functions of eigenvalues reverse a conversion,

one that is based on the Cayley-Hamilton theorem, to diagonalization.Lines one.

Now let's see concrete examples.

Our first example is a two binary matrix.

This matrix is 1, 2, 3, 4 has its eigenvalues easy.

We let the third force that account.

Now force a small order,

To that extent we can get it by multiplying three times with itself.

I said this method to zero.

This is actually not a serious procedure because of a higher order

though nothing or infinite series, though you can not apply it,

but let's do it just because we find other methods to provide solutions.

We see what is happening as well.

Now with a means to find a means to multiply the square.

We take the first column of the product known work according to the rules,

We invest, we stood to the first pillar.

1 plus 8, 9 gives.

Again, hit the first column laid on the second line 4 plus 12,

because we bumped 1 to 4, we stands 3 and 4,

16 and making this process as similar to the second column.

And we found a matrix frame.

The second by the Cayley-Hamilton theorem

force two binary matrix is a matrix

For matrix shown in terms of the first force and zero-force.

If you make this account a matter of fact, four times a plus five s

We still hold it, we found it.

So here's a detail account, you get a four times

1 on the diagonal, you collect 1 put,

Do you get it five times.

To find out since we got here a square cube.

We are writing a frame.

We found a frame.

We will take with a hit.

Again according to rules known product we get the numbers here.

Cayley-Hamilton theorem tells us again that this third force me

even if it is for two binary matrix of zero force

So we see a unit matrix and found that he hit it with his own.

If you calculate this number in fact,

twenty times the first unit matrix plus

twenty times that we find himself in.

Of course, because this directly with this easy cross to the third force

We have achieved.

But such a force in the twentieth though it still possible, perhaps in the computer

but you have to make repeatedly hit repeatedly size.

Even if you want to later calculate an infinite series like this repeatedly

You will not find a definitive solution to bump off.

For him, this method may not work in some cases zero said.

Of course, the small force in the process of functions.

Cayley-Hamilton theorem and using it directly

made a hit with the mixing method, this method said a base.

Because not directly use the correct only the Cayley-Hamilton theorem.

He wrote that the Cayley-Hamilton theorem,

If you write a function that roots instead of lambda, which gives zero.

Roots are writing a work that equation 1,

2, 3, are writing the minus lambda over 4 diagonal.

Calculate the determinant of this matrix, see there is one term in the square lamp.

On the other diagonal line on the diagonal here lambdayl

lambda lambda squared multiplied.

Lambdal terms with two lambda minus 3 and minus lambdayl to have

1 product, it also gives minus four lamps.

Lambdas terms of terms not found in the lamp

3 minus twice the product once work

Ie minus 8 minus 3 4 5 happening.

This is equal to 0 when you look at us saying it will give the roots.

They replaced a sheep of the same equation lambda

equal to zero matrix also provides a matrix.

Lambda square instead of a square minus minus four lamps instead of four,

minus 5 instead of five times the identity matrix.

Here, we see here that we see right now in terms of a square, and then

we can calculate.

Here we find really a previous page

plus twenty times as twenty times a unit matrix,

here the same thing pardon a square

We found four plus five for good.

Here I have found four plus five.

So a square that we have achieved with the right directly on the cross

We verify the results, we have provided.

Now a square with a multiply.

A're writing here, a square is here

five plus four plus four? we are writing five II.

Then we get a square four plus five then, with this hit.

Now we repeat the same thing again.

We know a frame.

Cayley-Hamilton theorem in accordance with these four plus five then.

So instead we can place the equivalent of a square here.

Four times a square, a square four instead of five we put a plus II.

Plus five, we are writing here in five au.

See if we get the product here.

16 have this product, 5 A, there outside.

21 a he.

Four times I hit the 5 ~ 20 s.

Direct to cross to achieve this, as we saw earlier

we have 21 to plus gives 20 II.

21 to plus 20 s.

Now we are better able to manage a short-cut method in the renovation.

Here we have a main method this method.

Öbürki were the function can be calculated easily because it is a cube.

We know where any

f a c c plus times I can write as zero once.

This e 1 to see the inner product c is staying here when we take time to zero.

a time to a stay of lambda 1 to 2 times.

On the right times to remain lambda 1 cube.

Because we have lowered it to 1 with 1 equivalent of an e everywhere.

We've seen this before already in general.

The first equation c 0 plus 1 equals C once lambda lambda is 1 cube.

In the same way in the second equation lambda lambda's placing 2 instead of 1.

Now we need lambda lambda 1 and 2.

We saw him a little while ago.

We are removing diagonal minus lambda over.

We wrote a matrix, 1, 2, 3, 4 minus lambda was removed.

It calculated the function of this force lambda squared minus four

lambda minus 5 we find zero eşitleyin by the roots.

We know how to find the roots of a quadratic function force.

Here the results are given.

Let's make the mere sum of two root 4 happening.

See here 4 come from there, with a minus sign.

Both the product of multiplying the two root minus 5.

See here's minus 5.

In this way we have proved we can directly or indirectly proven.

Let's put minus 1 instead of lambda 1, 1.

This is happening because the lambda minus 1 plus 4.

1 plus 4, 5 minus 5, 0.

So the roots.

Lambda 2 instead of the five we put 125.

minus 120 to minus 5 than a 125,

it also gives 125 minus zero.

Now we are in this equation c0 and c1

The equation for the unknown in a lamp, instead of -1

Instead we get two equations what happens when you add 5 lambda.

See, there is less than minus lambda came into c1.

The right side is the cube from one lamp there are still -1 -1 cubes.

C0, c1, plus five times in the second equation because the second root was 5,

equal to the right

equal right

There are thus two sides of the cube lambda 125.

If we write this in a matrix of two equations with two unknowns

coefficients will be 1 -1 -1.

1 -1 -1, c0, c1 and unknowns.

The second factor in our equation one, five, and 125 future.

One, five and 125.

The easiest way to find the unknown c0, c1, and Cramer's rule here again.

Cramer's rule we calculate the determinant factor in the matrix.

1, -1, 1, 5'ten determinant 5.

There minus 1 minus 5 plus 1 plus 1 more for that 6.

We use the right side for the c0 coefficient in the first column.

Minus one and 125.

Right in the second column remains the same.

Wherein determinant -5

125 plus mark the second time because we are changing.

-5 Plus 125 gave 120.

We put the right-hand side for the second determinant of the second column.

The first column remains the same 1, 1.

The second column instead of -1, we have put 125.

From here, 125, -1, plus or minus 1 for the more advanced, we find 126.

Determinant for c0 determinants of these determinants,

determinate and determinant for c1 divided by 120. 6 20 6 We divide.

We divide 21 126 to 6.

Therefore, we find that the cube is I plus 20 times to 21 times.

This once more, we provide the results we found earlier.

Of course, here we have a significant advantage now highlighting.

This is not a third force at 103.

Nothing will change if forces.

It would change the terms of a single right.

Here's one lambda 103.

force, two lambda 103.

The force would be.

Therefore, the labor of the work,

It would be different from labor to find and write an email to the third power.

However, direct calculations you look at the length you hit the road.

Here you can use the Cayley-Hamilton theorem.

Here 103.

Though force will hit 103 times repeatedly repeatedly.

As I said before, here not to over a finite force as a

Even if the function requires an infinite series would be the same effort to be made.

We see this kind of applications.

The second method was based on the following.

It is diagonalization matrix, and vice versa

It köşegenleştiriy in any function.

For diagonal cube here

We are writing function in terms of lambda are over.

For a lamp, that is one third the strength of the lambda lambda two of the third force,

I get hit with three of the left and right Q Q minus 1.

Account also said.

We are writing to occur.

Q. We are writing to the left.

Q minus the one we're writing right.

This Q cons before we found the Q course.

Q is the sequencing of the matrix of eigenvalues.

See, here we find the eigenvalues of matrix problems.

1 and 5 turns.

When look at -1 2 2 staying here.

Here you say will be the second one to the first unknown unknown -1.

5 will be here when you get -4, will stay here 2.

First, second, 2 turns when you say unknown unknown 1.

Therefore, this first column

We put e1, e2 we have achieved in the second column put Q.

Q We need to reverse the determinant if you remember old lessons.

1 times 2, 2.

3 plus 1 will be here, determinants 1 divided by 3.

So we will divide into three overturned the transpose of the cofactor.

As we find the cofactor.

Diagonal elements correspond to the first column on the first line and threw it comes second.

See, there's two.

This is instead of columns and rows that are taking 2 to 2, 1.

We have brought here.

We bring the first row and first column instead of 1, -1.

Likewise it came 1 instead of -1.

These sub-matrices was matrix derived from the determinant.

We will change the signs for cofactors.

It was 2 plus 1 -1 1.

One also comes to this situation as it received the transpose.

We can also provide really.

Q 1 and Q old hit 2 times 2 plus 1 we invest will receive 1 3.

But we split 3 for 1 coming on the diagonal.

Take the second column where we stand second to the second diagonal line.

1 -1 -1 multiplied as you can see.

Plus 2 times 1 2, 3.

3 to interrupt will occur on the third diagonal 1pm.

We're taking the first column for all but the diagonal.

We stood in the second row.

See 2 times -1, -2.

1 times 2, 2.

0. So it was zero except the diagonal.

Again, we take the second column of the first line -1 plus 1 hit zero.

The second diagonal zero outside.

That means we have provided a product.

Q minus 1, multiplied by Q that the unit matrix.

Now we have the summer to occur here in the transaction, we wrote Q,

We wrote Q minus 1, minus 1 right Qi, Qi left in this challenge by multiplication,

We have a 3 compartment at the end of Q minus 1 1 3 so.

We're following this operation.

We wrote them.

If we start from right, we stood in the middle of the right matrix.

Matrix consists shown here.

Take it in - we multiply by the matrix Q -1 -2 1.

This product is also considered that we have achieved what happens when you have made.

For example, do the one we put away and we hit the first column.

-2 Plus 125, 123 output.

However, we have received the deposit, we hit the first column.

2 times 1, that was a plus for both minus 1.

2 125 250 times.

250 plus 2, 252.

Right on the case.

Of course we need it in 3 partitions.

Q minus 3 in 1, 1 to 3 due divide.

All the time we split, we find that the matrix before

we get the numbers.

Again, the second example.

This example is an example of a general nature.

Let's find a common solution in two binary matrix.

So a any f (A) is given to us.

This f (A) from this any function,

We know that c0 of the function and expressed as c1.

Because our two binary matrix matrix.

So minus 2, we expressed in terms to the first force.

We have two known, c0, c1.

İki özvektörümüz var, e1, e2.

Hit the left side of the kind we have achieved just now

f (A) is an eigenvalue lambda e a to f a,

here comes a car zero, once here c

The time to stay for more than a one lambda lambda is a e a.

e are düşürüyoruz because on both sides.

Similarly, the term comes from a binary lambda lambda instead.

If we say the term traditional structure,

so unknown to the left and right side are known to take if given function.

We assume that we know a Lambda Lambda Lambda two say that for one,

for two given lambda.

c c a zero unknown, also known dilemma in a lambda light.

If we arrange them in the structure of this matrix equation

There was a lamp in a first row, the right side for a lamp,

this was changed to that obtained by one lamp lambda i in the second line.

Again, determinant of coefficients according to Cramer's rule

lambda lambda two minus one, where you're writing,

c determinants opposing first zero

We are shifting the column to the right column, here for a lambda light for two already have,

for two plus a lambda lambda is multiplied, lambda lambda is also good for those of multiplied.

Given that the determinant here.

c is the one to the right side, we change places with the second column,

where there is one, on the right side for a lamp and the lamp for two there,

means that the product of two combined for minus lambda lambda combined for one product,

As you can see here, we find that the determinant.

c c a determinant of zero and zero if it c

coefficients determinant divided by c determinant for a

We find one of the determinants of c divided by the coefficient matrix.

So here we have found you what else is given the function of the overall solution.

In some special cases oxygen may be equal in both lamps.

This is an interesting situation.

This physical money there,

events such as resonance occurs in such cases.

Of course, if one lambda lambda equals two denominator terms that will cancel each other out,

f A minus lambda lambda lambda lambda one for one, will occur in the denominator is zero,

so there will be zero divided by zero, where it will be zero divided by zero.

This situation is not something yadırgadıc,

You've seen it before on several occasions l'Hopital's rule there.

L'Hopital's Rule says zero zero if the share split in two lambda

Take the derivative of the denominator by Lambda taken by the two derivatives,

then insert one instead of two lambda lambda.

If this is done by two lamps in the first term of the derivative f

for lambda is a constant, for making hard task,

a derivative of two lambda, where lambda for a stay.

Here's the place the lamp in two different in function.

Lambda are writing one fixed lambda, it lambda

We take the derivative of doubles for the two lamp base, lamp instead of two in this

We are placing a limit for lambda lambda lambda one to two.

According to two in the denominator is taken by a lambda derivative turns out,

it means that we need to take a compartment.

Similar situation is for a car,

According to two variants of the lamp base for light received by two going on,

for a fixed lambda,

According to two for lambda, that is to say just above denominator

We see that for the base, instead of two lambda lambda have a summer.

Yet the denominator by two turns a derivative of Lambda received.

That means we can solve this interesting situation in this way.

Also found a similar overall solution for three to three.

However, we will not have achieved much, but in terms of implementation

There are benefits.

Both methods we are treated once more.

The terms will be up to the square for three.

According to the terms of this equation is found to square one,

ACE received two e and e trilogy domestic product,

for example in any ej,

the function of the eigenvectors ej.

Here lambda future cooperation, for a lamp,

ej just come here, come here lambda once ej,

EJ times in one frame of car two times lambda future here.

Here, as you see the equation arises in this way, we've seen before

c there was no cooperation terms, where c it's reached term cooperation.

c c i with that you change one of these, and c trilogy c lambda

Lambda Lambda is one of the trilogy Change we get this equation.

The first column as you can see we write is always a form of a matrix,

one, because a car is always a zero coefficient,

because it was coming from the unit matrix.

The second column lambda a, lambda two lambda three in the third column

Coming square of lambdas, right at this for a lamp,

lambda two, three lamps are unknown and zero in c c a c two.

Again, this solution is made by Cramer's rule.

Here is an elegance, you do anything for a long time by hungry

You will not find it, but if you make a partial direct Gaussian elimination,

If you leave this one to the second line of the first row

future instead of zero, where lambda lambda two minus one,

where lambda lambda minus a square next two frames.

If you take the same action again the second row from the third determinant

It will not change, yet this one instead of zero future.

Then one, zero, zero and some coefficients will be altered here.

It hit the very symmetrical, it consists of an elegant coefficient determinants account.

c is zero for the first column,

We are placing the right hand side, to the determinant of a car

we put right the second column to find,

c we put right again in the third column for two.

That's standard procedure to open them.

After making zero corresponding to these determinants of c

We find the determinant factor, it had been quite an elegant way.

See on this lamp in a circle,

Lambda Two, three lambda dizse, that with this eco able to find the right formula.

This zero c c a c known in two

as we know them theorem goes off

Because any function but to square up to the forces necessary

coefficients consciousness, situated, we find the solution.

Now I want to take a break again,

After this break we found a half

Using the general solution that duality directly

exponential function, a sine function can not find how to hit,

We will find functions such as the matrix cosine functions.

After the three three-pointers in one matrix,

EA will calculate some functions.