![]() Here, I have a general matrix, a, b, c, d. What are its eigenvalues? What does that equation looks like for its two eigenvalues? So this will be a special case of this. Those same s1 and s2.īut now I move on to a general two by two matrix. So this equation has solutions e to the st when the matrix has the eigenvalues lambda equal s. So lambda for the matrix case is the same as s, s1 and s2 for the single second order equation. And of course you see that's exactly the same equation that we had for the exponent s. So there's my equation for the eigenvalues of a companion matrix. But it comes with a minus sign, so it's plus C. And the other part of the determinant is this product, minus C. And when I take the determinant of a two by two matrix, it's just that times that, which is minus lambda times minus lambda is lambda squared. What are the eigenvalues of that matrix? We just take the matrix, subtract lambda from the diagonal, and take the determinant. So I just want to go through the steps of finding its eigenvalues. So let me take that to be my matrix A, my companion matrix. But we really are going to be interested in all two by twos. So there is a two by two matrix that we're interested in. Two equations for two unknowns, y and y prime. And then, when I wrote the equation down- I won't repeat that- it led us to a two by two matrix. So now I have a vector unknown, y and y prime. And then I introduced y prime as a second unknown. ![]() So I started with one second order equation. So I started with the equation y double prime plus By prime plus Cy equals, say, 0. So we met this matrix when we had a second order equation. So can I begin with very easy two by two matrix, the kind that we met first, called a companion matrix. ![]() We shift by lambda times the identity to subtract that from the diagonal. So lambda is how much we shift the matrix to make the determinant 0. And now we have an equation for the eigenvalue lambda. If it was invertible, the only solution would be x equals 0. Now, when is that possible? That matrix can't be invertible. And now I see that this matrix times the vector gives me 0. So I just write the same equation this way. So I want to move that onto the left hand side. It just changes length by a factor lambda, which could be positive. In other words, when I multiply by A, that special vector x does not change direction. We're looking for a vector, x, and a number, lambda, the eigenvalue, so that Ax is lambda x. Good to separate out the two by two case from the later n by n eigenvalue problem.Īnd of course, let me remember the basic dogma of eigenvalues and eigenvectors. Two by two eigenvalues are the easiest to do, easiest to understand. This is a good time to do two by two matrices, their eigenvalues, and their stability.
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