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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.
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0peddani, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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 \[4x_15x_2=0\]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\[4x_15t=0 \rightarrow x_1=\frac{5}{4}t\]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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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 = t4/5 x_2 1 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note, 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 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Np. If this helps, you can thank me by becoming a fan :p
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