Finding Eigenvectors from Eigenvalues
Date: 08/24/2005 at 01:15:17 From: Matthew Subject: finding eigenvector from eigenvalue Given the matrix | 0 0 -2| A = | 0 -2 0| |-2 0 3| find the eigenvalues, and then the corresponding eigenvectors. I can find the eigenvalues, but my eigenvectors turn out to be just zero, and I thought the whole point of an eigenvector was that it didn't equal 0? I have solved the determinant of A-lI to get eigenvalues of -1, 4, and -2, and I checked that they are correct by checking that the product of those eigenvalues equals the determinant of A (-1*4*-2 = 8 = detA). However, plugging the values in to find the eigenvectors I get something like (for l = -1): |-1 0 -2| |x| |0| | 0 -3 0| |y| = |0| |-2 0 2| |z| |0| which follows through to give me x = 0, y = 0, z = 0. I am confused, I thought there were supposed to be an infinite number of eigenvectors for each eigenvalue, or at least they are not supposed to be zero. Pointing out what I am doing wrong would be most appreciated.
Date: 08/24/2005 at 08:19:37 From: Doctor George Subject: Re: finding eigenvector from eigenvalue Hi Matthew, Thanks for writing to Doctor Math. Let's work out an eigenvector for the eigenvalue of -2. We want to find the column vector (x,y,z) such that _ _ _ _ _ _ | 0 0 -2| | x | | x | | 0 -2 0| | y | = -2 | y | |-2 0 3| | z | | z | - - - - - - If we multiply the first row by the vector we get -2z = -2x z = x (1) If we multiply the second row by the vector we get -2y = -2y (2) If we multiply the third row by the vector we get -2x + 3z = -2z -2x = -5z (3) Equations (1) and (3) require that x = 0 and z = 0. Equation (2) is true for all y, so we can pick y = 1 for convenience. Now we have an eigenvector for the eigenvalue of -2. _ _ | 0 | | 1 | | 0 | - - Since the eigenvector is only unique to within a scale factor, we will always find them by making convenient choices for one of the values, and then solving for the others. Does that make sense? See if you can work out the other eigenvectors now. Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/
Date: 08/24/2005 at 09:04:16 From: Matthew Subject: Thank you (finding eigenvector from eigenvalue) Wow, thank you so much. That problem was driving me crazy. I have worked out all my eigenvectors now! :-)
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