## anonymous one year ago I am getting the incorrect answer for determining the eigenvectors of a matrix. I will post my work in the next post.

1. amistre64

eugene vectors are such that: Ax = Ix, or succinctly; (A-I)x = 0 and im sposing we start with the characteristic equation of A-I ... if memory serves

2. amistre64

curious on what your matrix is tho :)

3. anonymous

The matrix in question is $\left[\begin{matrix}1 & 2 \\ 2 & 3\end{matrix}\right]$ It's easy to work out the eigenvalues with the formula $\left| A - \lambda I \right| = 0$ They are $2+\sqrt{5}$ and $2-\sqrt{5}$ However, when I try to determine the eigenvectors, I get the wrong answer. Please note that the eigenvectors are unit vectors, so we can use $v _{1}^{2} + v_{2}^{2} = 1$ The correct eigenvectors are $\left[\begin{matrix}.5257 & .8507 \\ -.8506 & .5257\end{matrix}\right]$ Attached is my work which gives me the incorrect answer. I would like to understand what I'm doing wrong.

4. beginnersmind

Where do the values $$2+\sqrt{5} \text{ and } 2-\sqrt{5}$$ come from?

5. beginnersmind

|dw:1442189681242:dw| Sorry, too lazy to use LaTex. Solving the last line should give the value of lambda. Is this what you did?

6. anonymous

Yes, that is what I did, but my question is not about lambda. My question is about the eigenvectors (values of V1 and V2).

7. beginnersmind

Ok apparently that does give $$\lambda = 2 \pm \sqrt{5}$$ I'll go through the math in a second but I have a comment first. $v _{1}^{2} + v_{2}^{2} = 1$ You should almost never use this to express v_1 as $$\sqrt{1-v_2^2{}}$$. Instead you note that if v is an eigenvector then so is cv for any nonzero scalar c. So there is an eigenvector for which the first coordinate is 1.

8. beginnersmind

So you solve for the eigenvector by setting one of the coordinates to 1, then normalize at the end.

9. beginnersmind

So anyway, here's what I did. Assuming v2 = 1, v1 = 2/(1+sqrt(5)) Normalizing (v1,v2) gives (0.5257, 0.85065)

10. beginnersmind

This is for $$\lambda = 1+ \sqrt{5}$$. I don't really want to go through the calculation in detail, because it's kind of easy, but a pain to write out. The only trick you need to remember is that you can chose one component freely, because all the eigenvectors for a given eigenvalue lie on a line (except when that component is 0, of course).

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