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**Linear Systems: Complex Roots | MIT 18.03SC Differential Equations, Fall 2011**. Following along are instructions in the video below:

1 00:00:00,000 –> 00:00:05,772 PROFESSOR: Welcome to this recitation. In this recitation, were going to look at linear systems with complex roots. So the system were examining is the one given as x dot equals minus 3x minus 2, and y dot equals 5x minus y. And youre asked to use the matrix methods to solve this system. So why dont you take a pause here and try to solve this problem? And Ill be right back. 12 00:00:28,390 –> 00:00:37,460 Welcome back. So the first step is to write this system in matrix form. 15 00:00:42,520 –> 00:00:45,510 So we introduced a vector [x, y], the matrix multiplying column vector [x, y] again. The coefficients are going to be minus 3, minus 2, 5, minus 1. 19 00:01:02,440 –> 00:01:04,959 So the first step in solving this system is to find the eigenvalues of the matrix A. So the eigenvalues of matrix A are basically the solutions to this following determinant equal to 0, minus 3 minus lambda, minus 2, 5, minus 1 minus lambda, determinant equals to 0. Here, the lambda are the unknown eigenvalues. And to get this determinant, were basically multiplying these two terms, minus minus 2 dot 5, which gives us a plus sign. So here, were going to have lambda squared. 3*lambda plus 1*lambda gives us 4*lambda. And 3 plus 10 gives us 13. 33 00:02:05,450 –> 00:02:10,410 So this second-order polynomial in lambda will give us the two eigenvalues for this matrix. So lets examine the discriminant. So we have b squared minus 4a*b. And this gives us minus 36. So the discriminant is negative. And that tells us that we are going to have two complex roots, which is the title of the recitation. And these two complex roots are going to be complex conjugate of each other.

So the formula gives us plus or minus i of 6 for the root of the discriminant. Here, we have minus 4 over 2. So the two roots are basically minus 2 plus or minus i*3 or 3i. 49 00:03:06,110 –> 00:03:08,920 So these are our two roots. So now, lets focus on one of the roots to get the eigenvector associated with the eigenvalue. 53 00:03:18,370 –> 00:03:21,490 So lets focus on the positive one, for example. 55 00:03:24,395 –> 00:03:27,680 And we could do all the following again– AUDIENCE: Minus 2. PROFESSOR: Minus 2. Thank you. We can do all the following calculation that Im going to go do now for the complex conjugate, and I will explain at the end how that basically not change the result. So for this eigenvalue, we need to compute now the eigenvector. So to do that, we basically have to use minus 3 minus– Im just going to write the system here, let you see what Im doing. 67 00:03:57,810 –> 00:04:02,060 And we are solving this system. So where does this system come from? It comes from the fact that were looking for an eigenvector, v_plus, that is defined as a*v_plus equals lambda_plus v_plus. And you can then bring everything on the left-hand side, a minus lambda_i applied to v_plus gives us the zero vector. So thats what we have here. The unknowns are a_1, a_2, and were going to try to solve for this. So if we plug in now for the value of lambda_plus that we have, we have minus 3 plus 2, which gives us minus 1. And then, we have a minus 3i and minus 2. And for the second line, second entry of this matrix you have 5, minus 1 minus minus 2. So we have 2 minus 1, which is

1. And then, we have minus 3i. 86 00:05:00,890 –> 00:05:04,770 [a 1, a 2] equals to [0, 0]. So here, you can check for yourself that these two equations given by the first line and the second line are actually the same. And so basically, to get a_1 and a_2, it is sufficient to just solve, for example, the first one, where here, I just wrote minus 1 minus 3i multiplied by a_1 minus 2a_2 equals to 0. And I just brought the minus 2 on this side. So here, you can see that if we pick a_1 equals to 0– equals to plus 2, which would be our first entry, we can then cancel out these two and just have a_2 equals to minus 1 minus 3i. So this would be one eigenvector associated with this eigenvalue. We could have picked other ones. Theyre basically parallel to this one. 104 00:05:56,810 –> 00:06:00,290 So now what? So what we need to remember is the meaning of all of this. Seeking the eigenvalues and the eigenvectors is basically equivalent to seeking a solution in the form exponential lambda*t with the direction of the eigenvector associated with this eigenvalue. So now that we actually have this eigenvector and this eigenvalue, we can write down the solution. And Im just going to write the solution in x, which has entries basically x and y. And one way of writing it would be just to basically first start by writing what we have there. 117 00:06:37,870 –> 00:06:42,940 Im just going to spell it out. So we have this multiplied by 2 minus 3i. So what do we do with this? Well, we remember our earlier formula. So this is exponential minus 2t plus exponential 3i*t. So we can split the exponential 3i*t into a cosine and a

sine. And this, were going to also be able to split it into the complex part and the real part. And then, were going to combine the real part and the complex part. So lets do that. Exponential minus 2t multiplying, basically, cosine 3t plus i sine 3t for the entry 2 minus 1 minus 3i. So we have an i here and an i here. So things can be combined into a real part. So in the first entry here, what are we going to have? Were going to have cos 3t multiplying 2. Thats going to be in the real part. 137 00:07:53,960 –> 00:07:57,396 And Im going to keep some space. 139 00:07:58,770 –> 00:08:05,260 And another entry here at the second entry of this vector is going to give us cosine 3t multiplied by minus 1. Oops, here, it should be a 3t. Sorry. 144 00:08:18,336 –> 00:08:22,710 So minus cosine 3t. Now, where are we going to have another real part here? Its going to come from a multiplication of i sine 3t by 3i. So the two is together gives a minus 1. And we end up with a plus 3 sine 3t. So were done for the real part. Now lets focus on the imaginary part. What do we have? We have an i sine 3t multiplying a 2. 155 00:08:50,540 –> 00:08:56,280 And we have a minus 3i here multiplying cosine 3t. So we want to have a minus 3 cosine 3t, and finally, this minus 1 multiplying this sine 3t. 159 00:09:11,020 –> 00:09:16,200 So now, we did– AUDIENCE: [INAUDIBLE]. PROFESSOR: Oh yeah. Thank you. 2, from this operation. So now, we did split our solution into a real part and an imaginary part. So how can we write the general solution of the system? Well, we

for this linear system of equations, if we have a complex number that is a solution to the linear equation, then its real part and its imaginary part are also two independent solutions. So we can write the general solution of the system as a linear combination of the real part and the imaginary part. And I can just label this u_1 of t and u_2 of t here and vector. And we can then write the general solution in terms of any constant– that would be determined by the initial condition if we had one– exponential minus 2t along u_1 plus c_2 exponential minus 2t along vector u_2, which are also functions of t, just the difference from what we had before. So here, basically we seek for the eigenvalues values of the matrix. We looked for the eigenvector associated with the complex eigenvalue. We were able to write the full solution. And then, because of the linearity property, we were able to just then extract two linearly independent solutions, the real part of the solution we had and the imaginary part of the solution we had. So what I mentioned earlier was that we could do this whole calculation for the other eigenvalue with a minus. If you try to do it and trickle down your minus, you would see that basically you would just end up with minus signs here basically in front of the sines. And what you could do then is just simply absorb that minus sign for the general solution in c_1 and c_2. And basically, it gives you exactly the same form for the general solution. So you dont need to redo it for the second one. You would still end up with only two linearly independent solutions, not four. OK, so that ends this recitation. 209 00:11:47,020 –> 00:11:48,393

tags:

linear systems, complex roots

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