[MUSIC] Okay, as a concrete example, let's consider the initial value problem. x prime = 3x-y-z, y prime = + y- z and z prime = x- y prime of z. And the initial condition is x(0) = 1, y(0) = 0 and z(0) = 0, right, okay? Okay? Then we can rely to the initial value problem into the matrix form, okay? Say the capital X prime = 3, -1, and -1 and the 1, 1, -1 and the 1, -1 and the 1 and times X, right? We are definitely the capital X is equal to the column vector x, y, z, right? Together with the initial condition x(0) = 1 and 0 and 0, right? That's the problem we have, okay, okay? So, okay? We need to find the eigenvalues of this coefficient matrix A and the corresponding eigenvectors, right? Think about the characteristic question of A, in other words we have to consider the 0 is equal to determinant of A- lambda times the i, okay, okay? It's easy computation, it's easy computation to find determinant of this 3 by 3 matrix, okay? Then you will you will get this is equal to minus of lambda -1 and the lambda -2 and the scale, okay? So that we have two distinct eigenvalues, okay? 2 and 1, okay, okay? Among them, first, say lambda 1 = lambda 2 = 2, right? This is an eigenvalue with multiplicity 2, right, okay? And another eigenvalue say lambda 3 = 1, this is an eigenvalue with multiplicity 1, okay? We have 2 eigenvalues 1 and 2, 2 has the multiplicity 2 and the 1 has the multiplicity 1, okay? Now let's try to find the corresponding the eigenvector, okay? Then for the first the eigenvalue 2, right? Then we have to compute the following, right, okay? So, okay? A -2Is of 3, okay? And times this eigenvector then must be equal to 0 vector, right? So this is equal to A -2 then, okay? We have to subtract 2 from the old diagonal entries, right? So this is 1- 1 and- 1, okay? And 1, 1- 2 there is 1- 1 and -1, again 1- 1 and -1, okay? Unknown k has 3 components k1, k2, k3, okay? That must be equal to the 0 vector, right, okay, okay? So we have that equation, simultaneous equation, okay? And this simultaneous equation, right? Is equivalent to okay? k1- k2- k3 and that is equal to 0 okay, okay? That's the only one equation, okay? We can have from these 3, okay? And that means what, okay? That means the vector k must be of the following form, okay? Which is the same as k1 = k2 + k3, okay? So that we have the vector k must be of the form k1 = k2 + k3, and the k2 is arbitrary and the k3 is arbitrary, right? There are no restrictions on k2 and k3, right, okay? We'd better to rely to the k in the following equation, right, okay? Some component to reach as a k2, okay? We have 1, 1, and 0, right? Times 1, And 1 and 0, something containing k3, okay? 1, 0, 1, okay? 1, 0, 1, you can rewrite it in that way, okay? And we can say that, okay? With the k2 and the k3 are arbitrary constant, arbitrary, okay? Any vector, okay? What's the conclusion, okay? Any vector of this form, okay? Is an eigenvector, unless both k2 and k3 are equal to 0, right, okay? Any such nontrivial combination, okay? Nontrivial combination is an eigenvector corresponding to eigenvalue 2, with multiplicity 2, right, okay? Now can you pick up the linear independent eigenvectors corresponding to eigenvalue to them from this, okay? We can conclude, then let me erase this one to have a space, okay? If I'm choosing kt = 1 and the k3 = 0, you have 1 nontrivial vector satisfying this relation, that is an eigenvector, okay? On the other hand, if you choose kt = 0 and the k3 = 1, then you have another such non trivial vector 1, 0, 1 satisfying that equation, okay, okay? So by those two special choice of k2 and k3, you can conclude that, okay? You can conclude that the vector k sub 1 which is 1, 1, 0, okay? And the vector k2, okay? That is 1, 0, 1, okay? They are 2 linearly independent, okay?, eigenvectors for eigenvalue, lambda 1 = lambda 2 = 2, right, okay? There are 2 linearly independent eigenvectors, right, okay? Corresponding to the eigenvalue 2, multiplicity 2, right? That's the maximum possible number, okay? All other possible the eigenvectors for the eigenvalue 2, are nontrivial combination of these two vectors, okay? Which is given by exactly this form, okay, okay? Then divide, By the claim I made before, okay? Which means that, okay? So that you have X1 = k1 times e to the corresponding eigenvalue to 2t and X2 = 1, 0, 1 times e to the 2t, they are two linearly independent solutions, right? These two are the two linearly independent solutions of the given problem, okay? Finally we have one another eigenvalue, okay? We have one another eigenvalue say lambda 3 =1, right? So for eigenvalue, for eigenvalue lambda 3 = 1, okay, okay? To the same thing as before, right, okay? A -1 times I3, Is of 3 times k, that is equal to, in fact if you remind what the coefficient matrix A and the subtract 1 times of Is of 3 then you will get 2- 1 and -1 and 1, 0,- 1, 1, -1, and 0. And sum of unknown vector k1, k2, and k3 and that is = 0, 0, and 0, right, okay? Rewriting this simultaneous equation, okay, okay? If you solve this simultaneous equation then you can easily see that, okay? Solution will be k1 = k2 = k3, okay, okay? You can obtain this conclusion rather easily, okay? What does that mean for the k, okay? The nontrivial vector k satisfying this is equal to 0 vector is of the form, okay, okay? Just the k1 and k1 and k1 in other words k1 times 0, 0, no, how come 0, no 1, 1, 1, okay? Where the k1 is arbitrary non-0 number, right, okay? So that this third eigenvalue 1 has only 1 linear independent eigenvector for example given by the 1, 1, 1, okay? So you can say that, okay, okay? Corresponding the eigenvector is, okay? You can take it to be 1, 1, 1, okay? And that immediately means, okay? We have a third solution, okay? We have a third solution say, okay? X3 that is = 1, 1, 1 times e to the t, right? That's a solution, okay? This is another solution, okay, okay? We have a system of these 3 equations for the 3 unknowns, we now find the 3 solutions given by x and x2 and x3, okay? All together, okay? They are 3 linearly independent solutions, okay? Why they're linear independent? Can you see that, right? I'll leave it to you as a simple exercise, okay? These two are the linear independent because their corresponding vectors are linearly independent, okay? And X3 are linearly independent because okay, okay? These two vectors and the last vectors the k3 their eigenvectors corresponding 2 distinct eigenvalues, right, okay? Any eigenvectors corresponding to distinct eigenvalues, they are automatically linear independent. So that combining those things, you can see that these 3 solutions are 3 linear independent solutions of the problem, in other words, there 4 more fundamental set, okay? Fundamental set of solutions, right, okay, okay, okay? Then you can immediately conclude that the general solution of our original problem this one, okay? The general solution will be linear combination of, right? Arbitrary linear combination of these 3, right? So now we have a general solution is except key that is equal c1(1)+ c2(2) and + c3(3), right, okay? Finally, okay? Using these initial conditions, okay? Using these initial conditions we can find, okay? Using the initial condition, okay? Say x(0) = 1 and 0 and 0, right, okay, okay? Using this initial condition then you will get c1 = c2 = 1 and the c3 = -1, okay? I'll skip the actual simple computation, okay? Then plugging these 3 numbers, plugging these 3 numbers, c1 = 1, c2 = 1 and the c3 = -1 into the equation, okay? The solution will be, okay? Solution to the initial value problem, to initial value problem is, okay? X(t) = 2 times of e to the 2t- e to the t y(t) = e to the 2t- e to the t and z of t = e to the 2t- e to the t, okay, okay? So that's the solution for the given the initial value problem, okay? That's the example for the case when we have an eigenvalue of rate, okay? The multiplicity and there are exactly m linearly independent corresponding eigenvectors way, okay? Another case we are considering is another extreme cases. The following case, okay?