[BOŞ_SES] Hello. We will talk about the matrix function. If the type of functions just as a x, starting with the power function The simplest type is more algebra beyond We say we kind of exponential functions, logarithmic functions and more as well as the matrix could be here in a completely different function, but also able to describe the function of the square matrix. The square matrix has to be because we take the strength of the matrix. The matrix can be multiplied by their square but that We saw referring to the square of the matrix privileges. These concrete examples show this privilege again. So you get a rectangular matrix. There is no such frame of a rectangular matrix. Çarpamazs because you will get the first column, Rotate where you hit two three element while here. It is therefore incompatible with this product. Therefore, such a rectangular matrix defined frame. Force a square matrix of the same force as zero x is a unit matrix, as happens for the zero force. If you multiply the first k minus a force again in a'yl It turns left or right k'yınc force. Yet when we take the square of the matrix, When we take the cube of the matrix of the same size will be square matrices. For him, this product is always defined. Left or right to hit will give the same result. Yet how could an x k'yınc the jth force çarpsak force, which they Ranked multiply if we multiply x j plus k'yınc forces will come out of a number. For the matrix, it is also true for the square matrix. Yet the strength of the matrix if we take the jth a more k'yınc forces or k'yınc forces to take after we take the jth force j k'yınc force will still be surplus of the matrix. x is a square as obtained in this work if you take the cube. Multiply this matrix by a number of forces Summing up, we get a force function. Just as the product of x, hit by a number of forces have collected a force function we were getting, It is a numerical force function, this force is the matrix function. Again, an example how the sine, cosine, exponential functions, liter, where the exponential function of time X. this leads to infinite series expansions of Taylor series. If the matrix using the same function If this series converges in function identified that the matrix exponential of a matrix, I times a matrix of a matrix exponential sine function, defined functions such as cosine function of and such provides an important advantage in applications. The first thing that will determine we receive the strength of the matrix and what happens when matter is that the eigenvalues. Example of a matrix j are the eigenvalues lambda to get opposing eigenvectors ej, If we hit it right on a a'yl, see here will consist of a square. We crossed from the left. a frame will occur. A square on the right with the çarpılmış will be a time j, j. This vector matrix multiplication for a number of lamps It will fall out of the matrix. The product will also be a time i j j times lambda Returning to the beginning. So here is a lambda j already had a lambda j EJ also came to light in the coming j squared here. Here we observe just that. A matrix of the eigenvalues of a square is also the eigenvalues. eigenvalues of a matrix, eigenvalues lambda lambda j j squared eigenvalues are square. A matrix by expanding it A matrix of eigenvalues and eigenvectors of lambda j and ej eigenvalues and eigenvectors of the opposing forces k'yınc eigenvectors of the matrix stays the same, the eigenvalues comes k'yınc force. Here we do when we apply repeatedly for frame here we get the proof. This is an important thing because it's hard to force the k'yınc of a matrix. But k'yınc matrix number, Supremely easy to get the k'yınc forces. If we take it a step further this ain Consider the function of force composed of forces. We collect C hit their numbers. This function of the eigenvalues of force still remain those of group relations, eigenvalues of the same this Instead of a power function The force obtained by putting lambda It has eigenvalues. So as we can see easily. Whether you are a power function of the moment of order. We wrote this tidy collection with this sign, j k with a number of k'yınc forces. k is zero then the unit matrix from zero to force a short pth We know that the moment of force to the first and moment of. Let ij multiplied by the function of this force. As a result of this process, namely the process as a linear process We get into e j collection for the change. There are multiplied by e j and k. K'yınc of each product with a k lambda ij We know that the forces. J jth display here shows that the eigenvalues. Lambda come here, he also jth eigenvector. Lambda terms that are achieved in this function If this function is the same pin functions that are denominated lambda. So we see here that where the force is derived from a If you want this function from b, b matrix this would be a matrix, eigenvectors are still going on to eigenvectors e j and eigenvalues from the same wherein the function obtained from the power function, lambda's going on in terms of the value obtained. We'll take it one step further. This function forces the same thing is true if the knock endless. These functions are the lambda functions. Of course you need something right here speaking of infinite series. Or it has to be cool is supposed to be convergent in a convergent zone. Thus also in the same area to determine the convergence matrix. We see that this theorem come easily and here's how succession the results will be very important that we do the proper numerical functions we transform the process functions, the function of the matrix. This is an important win. Which is known as the Cayley-Hamilton theorem theorem says that name. We are lambdas how we find these? Determinants were finding it a minus lambda calculated roots are zero. How lambda a, lambda two, lambda n eigenvalues force functions of the matrix If it provides shows that provide of a matrix. As we can easily prove: If a previous p a theorem for this özkuvvet If we choose the function that multiplies each time we ij force in the jar on the right side of the lamp function occurs. But lambda j were the root of this power function. So the right side is zero. As per definition because lambda are zero. Lambda and the p value as we choose to make zero on the lamp. So the left-hand side to zero We find, but it goes a vector of this vector that are for all i j If zero for all base matrix itself has to be zero matrix. This leads us on to something supremely interesting. Just to the right as a result of this theorem We come n times the size of Although we see how we can find a matrix An'Nur statement. Because that roots from a species If we account for these eigenvalues force function, n'yinc of his order, as we saw in the previous section, We see that there is a power function without excuse n'yinc degrees. Even where the coefficients are not so random coefficients. Coefficients we saw in the previous section, which does not change the matrix. But most importantly, this function will be zero n'yinc degree of force. It will be zero matrix. Here we see now when we solve an'y that moment, minus one and minus two units of a moment in terms of matrix It becomes a function expressed. Thus, less than the force of a matrix n'yinc able to account in terms of strong matrix. a is zero, first with the force of up to minus one. This now opens to us a very interesting way. If p, then from the greater inasmuch n'yinc in force in other forces, the other cons If I multiply once more a'yl to account in terms of the force to one. We know an'y. So an'y to take the time here last summer and a'yl When we hit a force will increase. Here's to begin with a n'yinc force, the less will be the first force. we multiply a matrix for a unit we come to the first force. But on the other hand, we know that a is n'yinc forces other small forces denominated. When we put it not only An'Nur, Cayley-Hamilton theorem we proved a bit in size before theorem, in accordance with the plus smaller than the force of the first in a We can find it displayed in forces. When we apply it consecutively, a n minus any force We see that forces can be expressed as a. This is a good thing supremely. Let's say there are two binary matrix, you find that the hundredth power. This is just a unit matrix, and is expressed by itself. We will see examples for this already now. If we take it one step further convergent Although a function for function, here we can think of it as a force must exist forever. Indeed, such has all forces to turn on a x e to the x series. But if of a matrix e'inc the exponential if we have account, then the series is not forever, until the first function will consist of the negative forces. This supremely good thing, because we do not know yet the coefficients. But we know the structure. Even if a series goes on forever, it's just the minus one so that the coefficient of the grain, which starts from zero to zero, one, we know that the coefficients indicated by two. This will provide exceptional ease of calculations. Not with an endless series always converges with an endless series of matrices We can show all functions. Now here is the c0 c1, cn2 an unknown numbers. This is the one known but algebraic numerical unknowns. How do we find them? Very easy to find. This fan function e1, e2, If we multiply one by one until the eigenvalues EN was, ı'yl the future because a number each time, ej'y to hit the matrix unit, but the unit will a'yl to ej'y to hit the lambda times to come. When we hit the first negative force, next to the first force and the minus lambda We are getting so many have obtained the equation. It's equations also comes in: See, In accordance theorem here on the left side, in accordance with the theorem we proved a little before any function that we knew j hit the eigenvalues that it opposed the eigenvalues of this Instead of putting a lambda functions, it is achieved by the bay. Now is the time we have here in this process, as you can see the same force function here instead of lambda times for EJ fam here for the future. This te terms theorem I showed a little before. In the right side, this force as a function of n minus one We found much of forces. This ej j vectors are also seeing that there are two sides of the vector, It means that these coefficients are equal. It seems that many have obtained scalar equations, we're going to have obtained numerical equation. If we arrange this equation in a matrix structure, c0, c1, in terms of a negative coefficient cn The column vector of the unknowns that In terms of a coefficient, a lambda lambda two, one for the lambda lambda n minus two until the first force, lambda n minus, we get a matrix structure for the same. Square matrix, because there is the one lambda. It seems that there is one line. Forces from the one that zero force, hence the number, the less goes to the first force. It seems that there are too many columns. The whole point is that if it bitmese this endless, We would be necessary to solve the infinite coefficient at the beginning. But here we see that the grain is sufficient to solve the unknown. Now, how do we calculate this? There is a second way. Now comes the second way is opposed to this. Now that we have a left-q minus the merger, he hit the right q'yl We make diagonal, and we also remember the following theorem. A eigenvectors j are the eigenvalues of the lamp for any j j are eigenvectors of a function will be again. The eigenvalues to be calculated with this function. Now it just comes easily to this: The How q were köşegenleştiriy ai, here we take j as columns, vector q, q matrix that we create, on the contrary we hit it left-q minus combine, How's the same here as for a the köşegenleş We also see that for a köşegenleş. If we reverse this equation, so we want to know for ASU. This is not at all easy to know. This second method. N minus first method we have determined that the force to one. Here we are developing a second method is diagonalizing. The function of this lamp one on the diagonal two functions on the right side lamp easy, but we already we are looking for a. This equation is obtained q'yl left to see where we hit the unit. Right-q minus the merger we hit here consists of a unit, FA therefore remain open here. Q'yl and we do the same for the right q minus merger gets hit, here we find the path calculation. for which we can calculate a mean. lambda for a very easy thing, and q minus one easy. This means that we are able to achieve any function with such a cross. Let's look at this visual structure. Before you get any function a number to a matrix of a two what should I say, we do not know it yet. But theory tells us: If we multiply the left q minus combine, If we hit the right q'yl, this function in diagonal, lambda place and changed with the fan function will be created, köşegenleş. Why? Cause, but his eigenvectors with eigenvalues of any function. Therefore, the same q matrix, au diagonalization.Lines the same q matrix, any function will also diagonalization. The only difference is when we have a au diagonalization.Lines only on the diagonal in a lamp, I'll be in the two lambda lambda, where the function on the diagonal eigenvalues will occur. This eigenvalues lambda a, lambda, we find two effects that can affect the function ae. Also when we did the opposite, eigenvalues of work here on the diagonal q Put the right diagonal matrix inverse, q'yl has hit the left you get this function. We have already proved it. It may seem surprising, even for a very simple proof. We want to calculate this for ASU. So here we are putting a diagonal matrix. Right-q minus the merger, we stood on the left q'yl. But the opposite were true, a to start, a left-q minus the merger, carpacaktık qyla right. Q minus combine function left, right q'yl has hit the köşegenleşiy. But to go from full-diagonal matrix in reverse if future deficit. Left q'yl has hit the unit will fall, q minus units will be combined to hit the right matrix will fall. Thus the transformation sequence diagonalization.Lines see here left inverse matrix, inverse matrix right here. Direct matrix, the columns are eigenvectors right matrix; The full matrix that also left. We saw this in the case visually it consists of two pages. Now we will move to sample, The theory also proves very easy as you can see is actually not that complicated. Now here we get the matrix A, two, three, four numbers given by the matrix. We will apply two methods we have seen this before on it. One of them, this matrix c0, c1, type in, That is a once in a cube c0, unit matrix, c1 times a will because any force The cons were able to express the matrix to the first force. Pause a little bit before we will be in accordance with this example.
