18.03 LA.5: Eigenvalues and Eigenvectors [1] Eigenvectors and Eigenvalues [2] Observations about Eigenvalues [3] Complete Solution to system of ODEs [4] Computing Eigenvectors [5] Computing Eigenvalues [1] Eigenvectors and Eigenvalues Example from Differential Equations Consider the system of first order, linear ODEs. = 5y1 + 2y2 dt dy1 dy2 = 2y1 + 5y2 dt We can write this using the companion matrix form: y y2 = [5 2] 2 5 y2 y1 . Note that this matrix is symmetric. Using notation from linear algebra, we can write this even more succinctly as y' = Ay. This is a coupled equation, and we want to uncouple it. Method of Optimism We've seen that solutions to linear ODEs have the form ert. So we will look for solutions 1
y1 = edta y2 = e1tb Writing in vector notation: y = ett a b = extx Here A is the eigenvalue and x is the eigenvector. To find a solution of this form, we simply plug in this solution into the equation y' = Ay: deltx = de1+x dt Aetx = ext Ax If there is a solution of this form, it satisfies this equation Xe^tx = ext Ax. Note that because est is never zero, we can cancel it from both sides of this equation, and we end up with the central equation for eigenvalues and eigenvectors: \x = Ax Definitions · A nonzero vector x is an eigenvector if there is a number À such that Ax = \x. · The scalar value À is called the eigenvalue. Note that it is always true that A0 = \ . 0 for any A. This is why we make the distinction than an eigenvector must be a nonzero vector, and an eigenvalue must correspond to a nonzero vector. However, the scalar value \ can be any real or complex number, including 0. 2
This is a subtle equation. Both \ and x are unknown. This isn't exactly a linear problem. There are more unknowns. What is this equation saying? It says that we are looking for a vector x such that x and Ax point in the same direction. But the length can change, the length is scaled by A. Note that this isn't true for most vectors. Typically Ax does not point in the same direction as x. Example If X = 0, our central equation becomes Ax = 0x = 0. The eigenvector x corresponding to the eigenvalue 0 is a vector in the nullspace! Example Let's find the eigenvalues and eigenvectors of our matrix from our system of ODEs. That is, we want to find x and \ such that [5 2] 2 5 [?] ? = 1? By inspection, we can see that [5 2] 1 2 5 1 = 7 1 1] . We have found the eigenvector x1 = 17 1 corresponding to the eigenvalue À1 = 7. So a solution to a differential equation looks like 1 y = et 1 Check that this