Yeah, last time we have seen the following fact, right. Okay, the X = K * T is a solution. Okay, is nontrivial solution. Of system, X′ = A * X, right. λ is on Eigen value of A and K. Which is, non zero vector Y is a corresponding Eigen vector, okay? E. Okay, we have seen this fact, right. Okay, so depending on the nature of the Eigen value and the corresponding Eigen vector, the type of the solution for this system will be different, right. So, we'd better to distinguish several cases like a foster. Let's think about the, okay, the case one, okay? A is an N by N matrix and its corresponding characteristic equation is a polynomial equation of degree N. So, its characteristic equation may have, okay, N different solution, okay? In particular, I'm assuming that, okay, A has n distinctively a Eigen values, right. And distinct real Eigen values, right. Say λ1 and the λ2 through the λ of the n, okay? And distinctively are Eigen values. And by the meaning of the Eigen value, right. Each Eigen value should have an Eigen vector, should have a corresponding Eigen vector, okay? So with Corresponding Eigen vectors, say, K1, K2 and this on through the case of Ks of n, okay? Respectively, okay? Now there was a K1, is an Eigen vector corresponding to Eigen value λ1. K2 is an Eigen vector corresponding to Eigen value λ2. Case of N is an Eigen vector corresponding to Eigen value λ7, right. Can we see the, leave alone that, okay? Such Eigen vectors corresponding to distinct Eigen values, right. There I am assuming that all those λ1, λ2 and λ7, they're all different, right. Okay, then we know that then, right. Those an Eigen vectors KJ They are linearly independent, right. Linearly independent N vectors, okay? So that If we make N by N matrix using these as N column vectors, right. So make N biometrics formed by those column vectors, okay? This is an N by N vector, right. And biometrics, right. Consider this determinant. This is not equal to zero, right. That is a well known factor from the linear independence, right. Okay, now we make a conclusion okay? So, okay, we'd better to call it as a theorem. Assume, okay. Assume that The coefficient matrix of this system, A, has N distinctive rear Eigen values, say, λ1, λ2 to λ7. Case of N with corresponding I N vectors K1, K2 case of N, okay. Then what, okay? Then we have N solutions. XJ is given by the KJ * E to the λ of J of t. We know each one of them, okay? Each one of those XJs, okay. Is a solution of this homogeneous system already, right. Because λ z an Eigen value of the coefficient matrix A and the KJ is Eigen vector, okay? So, okay, this is a non zero vector, right. Because this is a non zero vector, okay, so this is a non zero solution, right. So non-trivial solution of this one, okay. Furthermore, we can say that, okay. These N solutions, right. They're linear independent. Linearly independent solutions of X′ = A(X). Where on the whole reel line from minus infinity to infinity, okay. Because of this system of equation, okay, the unknown vector, X, has N components. Now, we have non-linear independent solution vector, okay? In other words, so these N solutions they form, right. In other words, right. These solutions X J = K J @ λ(T), where J is moving from one to the end. They are forming, so called, fundamental set of solutions, okay? Fundamental set of solutions of this problem, okay? Now we can naturally conclude that, okay. What is the general solution of this original problem, okay. Once we know the fundamental set of solutions, then General solution of the given problem system of equation is a linear combination of all those members, right? Okay, that's the general solution. Okay. General solution means okay. Set of all possible solutions. Okay, so we can conclude, right? We can conclude that if this is true, if there is a distinct values okay then. Okay, then the linear independent solution so that this is a fundamental set of solutions. And finally, okay, we can conclude that the given problem, x a promise equal to eight times of x has a general solution Which is given by the summation from j equal to one to the c sub j times xj that is equal to kj e to the lambda, jft, right? This is a general solution. Okay, right? This is the kind of the best case that we can have, right? Okay. Let's think about a simple example of this title. Okay, simple example of this type. I'm considering the following the homogeneous linear system with constantly coefficients, right? X prime of t is equal to 2x minus 7y. Y prime is equal to 5x plus 10y plus 4z. Z prime is equal to 5y plus 2z. Okay, that's first order linear system of equations homogeneous with constant coefficients, right? Okay, this is the equivalent to okay, x prime is equal to what is x capital x is equal to it has a three components xyz, right? Okay, I'm using this notation and then let's read out the coefficient matrix. Okay. First you have a 2, -7 and 0. Second you have a 5, 10 and 4. Finally you have a 0, 5 and 2, right? Okay. In matrix form, the system of equations you can rewrite it in this first way. Okay. The coefficient matrix A is a three by three matrix down there, right? Okay. We'd like to define the general solution of this linear homogeneous system of equations, right? What we have to do? Okay. The rule said first, we need to find all Eigen values of this matrix, right? This is not so good looking, right? Here's a 5 and here's a 2, right? Okay, too close. Okay. To find the Eigen values of A, right? We have to to think about the following things, right? 0 is equal to determinant of A minus 3I. Okay. Three by three identity matrix, right? What is this one? This is okay. What I'm saying? That's three, but I'm sorry, but A minus lambda times I3. Okay, determinant of A minus lambda times a three, right? What is A minus lambda times I3? That is 2 minus lambda minus 7 and 0. 5, 10 minus lambda and 4. 0, 5, 2 minus lambda, right? And we compute this determinant, right? Okay. Can you remind how to compute the determinant of three by three matrix is, right? Okay. Here's the shortcut to compute it, right? Determinant of this three by three matrix is, right? For us to multiply this main diagana and trays. So you have a 2 minus lambda and the 10 minus lambda times 2 minus lambda plus this and that way. So okay. Plus minus 7 times 4 times 0 and plus multiply of these three factors, right? That is again zero because of this 0, right? 0 times 5 times 5 that is equal to zero. Okay then minus. Okay, sub diagonal direction. Okay, so multiply of two to three entries down there. There's 0 times 10 minus lambda times 0. That is 0 okay. Then minus okay again, 2 minus lambda times this times this 2, right? So that is what? Okay, 2 minus lambda times 4 times 5. Finally minus. Okay, these two numbers times 2 minus lambda. So minus minus 7, 5 times 5 and times 2 minus lambda. Okay, that must be equal to 0, right? That's the determinants, right? I'm not doing this simple computation but to simplify this equation you'll get minus times lambda minus 2 and the lambda minus 5 and the lambda minus 7. Okay, 10 must be equal to 0, right? This is the characteristic equation zero is equal to this one. That's a polynomial equation in lambda of degree three, right? Okay. The isotomy is minus 1 times lambda to the cubed plus and so on, right? Okay, this is the characteristic equation. Okay, what are the solutions? Okay, we have three distinct solutions, right? They are Eigen values. Okay, Eigen value lambda one, you can read it. There's 2 second Eigen value, that is 5. So the Eigen value that is 7, right? So we have three distinct Eigen values. Okay. Corresponding each Eigen value, there must be a nontrivial solution. Nontrivial factor satisfying. Okay, now If lambda is an Eigen value then you have a corresponding Eigen vector satisfying dissimulation. Okay, [COUGH] okay, so let's take lambda 1 is equal 2, right? The corresponding Eigen to find the corresponding Eigen vector, you've to solve the following equation. Okay, A minus 2 I3 minus 2 lambda A3 that is equal 2 okay minus 2 times of identity, that is 0 minus 7, 0, 5 -2 that is 8 and 4. 0 and 5 and 0, right. And unknown vector k, you can write it as a k1, k2 and k3, right?. We need this must be a 0 vector, right? Okay, that's the way to compute the corresponding Eigen vector. Okay. From the first one, okay, look at the product of this one. This is equal to okay, 0 times k1 one plus -7 times k2 plus 0 times k3. So the first Line gives you -7k2 = 0. From the second one 5k1+ k2, right, okay. 5 times the k1, 8 times the k2 + 4k3 = 0. Finally 0 times the k1 plus 5k2 plus zero is equal to 0. Right, okay, so this is simultaneous system of equation. It's the same as this one. Okay, from the first and the third, you will get simply the k2 = 0, right? Okay, if k2 is equal to 0, then from the second you will get 5k1 + 4k3 and that is equal to 0, right? So that this system of equation is eventually simplified into this two, right? What does that mean then? Okay, So k must be equal to the vector that I'm looking for, okay, is equal to k to 0 from this equation. K1 is arbitrary, then k3 is equal to- 5/4 times k1, right? That's the solution, right? That's the solution. Okay, This you can write it as a k1 times 1, 0 and -5/4, right? Okay, where the k1 is arbitrary, okay. This is the candidate for the corresponding Eigen vector, right? Because I'm calling this number 2 as a for * Eigen value lambda several one. Okay, so we do better to call it this one by case of 1. Okay, capital case of 1. Okay, as you can see from this expression, you can see that okay, corresponding to the Eigen value, lambda 1 is equal to there are infinitely many distinct Eigen vectors. For k1 is arbitrary but not 0 because we are looking for the nonzero vector. Okay, so that we can conclude that okay, take the k1 to be arbitrarily constant to be number 4, right? Then corresponding Eigen vector K1 is equal to 1, 0, -5. Okay, this is the corresponding Eigen vector. Okay, for Eigen value, lambda 1 is equal 2, right? Okay, To the same process for the other two Eigen values, right, okay. Similarly you can compute okay, let me keep it. I'm getting k1 is equal to now by taking k1 is equal to 4, we have 1, 0, -5, right? This is the corresponding Eigen vector. Okay, then for another Eigen values, lamda 2 is equal to 5 and lambda 3 is equal to 7. Okay, do the similar process then you will get the Eigen vectors are okay. K2 is equal to is easy. The computation minus -7, 3, 5 and responding to lambda 3 is equal to 7. Corresponding Eigen vector K sub 3 is equal to -7, 5 and 5. Okay, this corresponds to lambda 1 is equal 2, right? Look at the following, okay. Lambda 2 and 5 and 7, they're all different, then we know that corresponding Eigen vectors K1, K2 and K3, they are linearly independent. So if we make a solution xj is equal to Kj e to the lambda j of t, okay? Where j is moving from 1 to 3, they are linearly independent. In other words this is a fundamental set of solutions. Okay, so general solution is the what? General solution of the problem is arbitrary constants c1 times K1, 1, 0, -5 e to the 2t plus another arbitrary constant c2 times K2 -7, 3, 5 and e to the 5t. Plus another arbitrary constant c3 times K3 -7, 5, 5 and e to the 7 of t. Okay, that's the general solution. Okay, so we succeeded to solve this given system of equations completely. Right, okay, this is the general solution, okay. [MUSIC]
