W4 - Eigenvectors
How did we get the x and y values from the equations? Would someone break it down for me. Given there are no constants (and all values are coefficients of X or Y) I can’t solve for X or Y. Thank you!
Hi @shaheenb
To find the eigenvectors corresponding to the eigenvalues \lambda = 2 and \lambda = 3, we substitute these values into the equation (A - \lambda I)x = 0. For \lambda = 2, we get the system of equations 2x + y = 2x and 3y = 2y, simplifying to y = 0. Any x value works, so we choose x = 1 for simplicity, giving the eigenvector [1, 0]. For \lambda = 3, the system is 2x + y = 3x and 3y = 3y, simplifying to y = x. Again, we choose x = 1 for simplicity, giving the eigenvector [1, 1].
Hope it helps! Feel free to ask if you need further assistance.
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Right! Just to add one more level of explanation to Alireza’s solution, the point is that eigenvectors are not unique: there are infinitely many of them for any given eigenvalue. Any other vector that points is the same direction will have the same property. So if the vector (1,1) is an eigenvector for the eigenvalue \lambda = 3, then (2,2) or (3,3) will also be eigenvectors. So when you are solving the equations that you get, you can just choose one value and then let the other be determined from that plus the remaining equations.
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