## anonymous 5 years ago I'm confused on writing eigenvectors from eigenvalues. -3 4 0 -8 Matrix calcs to L1=3 L2=8. Matrix1 Matrix2 0 4 -11 4 0 -11 0 0 Right? I don't understand what to do next to get the eigenvector.

1. anonymous

peddani, I think the calculation for you characteristic polynomial may be off. If you let your matrix be A, then you're solving $\det(I \lambda -A)=0$to get your eigenvalues, which you can then use to find the vectors. When you form the polynomial, you should get$(\lambda + 3)(\lambda +8)=0$which means your eigenvalues will be $\lambda_1=-3$and$\lambda_2=-8$I'll tackle the eigenvectors next.

2. anonymous

Yes, I see I had the signs wrong now. Thanks. I'll be anxious to see the vectors when I get home. Thanks for the help.

3. anonymous

You find the eigenvectors by taking each of your eigenvalues and solve the system,$(I \lambda -A)x=0$For the first eigenvalue, -3, your matrix will be |0 4 ||x_1| =|0| |0 -5 ||x_2| |0| So the first equation you have to solve is $4x_1-5x_2=0$Since you have, in this system, *one* equation in *two* unknowns, an *infinite* number of solution is possible. So you may say, let$x_2=t \in \mathbb{R}$Then$4x_1-5t=0 \rightarrow x_1=\frac{5}{4}t$So your eigenvector for the eigenvalue lambda_1 = -3 is |x_1| = t | 1 | |x_2| |5/4| (this site doesn't really allow for decent matrix equation writing). It's just t multiplied by the column vector, (1.5/4). Note, you could have picked any kind of parametrization you wanted for x_1 (or x_2 first) since the equation linking the two forces the appropriate form of the parametrization of the remaining variable.

4. anonymous

Now you need to repeat the process for the second eigenvalue, lambda_2 = -8. The matrix equation you should get is |-5 4 ||x_1| = |0| | 0 0 ||x_2| |0| which gives you,$-5x_1+4x_2=0$Again, you have, for this system, one equation in two unknowns, so there are an infinite number of solutions. We may choose, say, $x_2=t$again (this isn't necessarily the same t as before). Then$-5x_1+4x_2=0 \rightarrow x_1=\frac{4}{5}t$and your eigenvector is |x_1| = t|4/5| |x_2| |1 |

5. anonymous

Note, the first eigenvector, |x_1| = t | 1 | |x_2| |5/4| has a typo. It should read, |x_1| = t |5/4| |x_2| |1 |

6. anonymous

Np. If this helps, you can thank me by becoming a fan :p