Finding eigenvectors
For each eigenvalue lambda, solve (A - lambda I) v = 0. Any non-zero solution is an eigenvector, and any scalar multiple of it is too.
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Analogy
Once you know the stretch factor, you search for the direction that gets that stretch.
Example
A - 2I = [[2,1],[2,1]].2x + y = 0, so y = -2x.Take x = 1: v = (1, -2).Check: A v = (2,-4) = 2v.
In practice
Gives the actual directions, which you need to build the diagonal form.
Practice
Flashcards
- Find an eigenvector of [[4,1],[2,3]] for lambda = 2.(1, -2).
Interview questions
- Find the eigenvalues and eigenvectors of A = [[4,1],[2,3]].
- Characteristic equation L^2 - 7L + 10 = 0
- Eigenvalues 5 and 2
- For 5: (A - 5I)v = 0 gives v = (1,1)
- For 2: gives v = (1,-2)
- Check using trace 7 and determinant 10
Questions people ask
Why are the rows of A minus lambda I dependent?
Because the determinant is zero by design. The two equations in the example are the same line, 2x plus y equals 0, repeated. That leaves one free variable, which is why you get a whole line of eigenvectors, not a single vector. That is why eigenvectors are described as directions or lines.
Is an eigenvector unique?
No. Any non-zero multiple of an eigenvector is also an eigenvector for the same eigenvalue. For lambda equals 2, both (1,-2) and (2,-4) work. We usually pick the simplest one, or scale it to length 1 so that comparing answers is easier.