[MUSIC] In this case I'm assuming the following things, okay? The coefficient matrix A has an Eigen value Lambda 1 Multiplicity m, which is greater than or equal to 2, and the less than or equal to n, okay? And There is only 1, okay, only 1 lineary independent Eigen vector, okay? Say k1, okay, for that Eigen value lambda 1, okay? Another extreme case, right? Remind that, okay, for any Eigen value lambda 1 with multiplicity m, okay, there are k linearly independent Eigen vectors, where k is greater than equal to one, and less than or equal to m, right, okay? That we call there's the geometric multiplicity of the Eigen value, okay? In case 22-1, I consider the case where the m maximum possible linear independent Eigen vectors. Now, I'm considering another extreme case. For this Eigen value, there is only one, okay? Minimum number of linear independent Eigen vectors for 1, right, okay? How can you find then, okay, from this Eigen value lambda 1, how can you find m linearly independent solutions, okay? M linearly independent solutions, okay? Here's a claim, okay? In this case, okay, there is m linearly independent solutions of the phone, okay okay? It's rather a little bit the complicated way, okay? Say, The x of j, okay, and the x of j is of the following form, right? This is some of the k is equal to 1 to the j, t to the j minus k over, j minus k factorial, okay? Times some vectors k sub of capital case of jk, and of each of the lambda 1 of t, okay? Where the j is moving from 1 to the n, okay? There are m linearly independent solutions of this form, okay? How can you find those vectors, okay, where the vectors case of jk will be determined by the following equation, okay? Matrix A minus lambda 1 times I sub n. And applied to the K subject jk is equal to, okay, K sub j comma k minus 1, okay, for k is moving from 1 to j, okay? To find the very first one, okay, when K is equal to 1 over what happened then, okay? One I, Kj1 equal to K sub j comma zero, right, okay? So what should be case of zero? We have to say something about that one, okay? Kj sub 0, this is 0 vector, okay? This is what I have here, okay? That's the conclusion, right? That's what we did again, right, okay? If we have an Eigen value lambda 1 with multiplicity m, m is greater than equal 2, less than or equal to n, right? Corresponding to this Eigen value lambda sub 1. Unfortunately we have only 1 linearly independent Eigen vector K1, okay? Then I claim there are m linearly independent solutions of the problem, okay, okay, of this form, okay? And where those are the vectors K sub j coma k, is determined by this relation, okay, okay? Because the K sub j comma 0 is equal to 0 vector, right, okay? When K is equal to 1, this is a zero vector, right? Then look at this a solution. Look at this equation. That is an equation to determine the Eigen vector for the Eigen value lambda sub 1, right, okay? So, you can see that, okay, in particular, okay, for any j, okay, for any j, K sub j 1, okay, okay? When k is equal to 1, A minus lambda 1I times the k sub j comma 1 is equal to 0 vector. And that means that this must be equal to d, okay? K1, okay? Okay, Eigen vector corresponding to this Eigen value lambda sub 1, okay? That's the starting step. From this, inductively you can solve, okay, you must solve the system of equations, okay? It looks rather complicate but, okay, right? Pattern will be understood by looking at the concrete example, okay, okay?