## help with #1 |
ElainePanZH |
---|---|

Can someone please tell me how to get two eigenvectors when λ=-1? And what does D look like when one of the eigenvalues has a multiplicity greater than 1? Thanks. |

alanvce: June 3, 2015, 12:09 a.m. |
---|

To get eigenvector, Row reduce and you will end up with something like 1 1 1 and zeros for the rest of the rows, this means that V1+V2+V3=0 which means V1=-V2-V3 so let V2 be say a variable t and V3 a variable z. Then you have, V1= -t-z where t and z can be any real number, but zero since by definition 0 is not an eigenvector, (if you choose zero that gives you 000, see why) so basically since you have multiplicity 2 for eigenvalue -1, just choose two linearly independent vectors. For example, if you choose 1 for t and 0 for z you get V = (-1 1 0) so you have one now for the other one lets say choose 1 for z and zero for t so you have W= (-1 0 1) |

jlumxajaha: June 3, 2015, 9:09 p.m. |
---|

When lambda = -1, you just solve for it and you'll get 2 free variables. Those two variables give you two different vectors. When an eigenvalue has a multiplicity greater than 1, then the value is repeated. So suppose you had eigenvalues -1 and 2 where -1 had multiplicity of 2. Then D has values -1, -1, and 2 in diagonal with 0's everywhere else. It doesn't matter what order the eigenvalues are in. |