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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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.

Mathematics
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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.
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.
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.

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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 |
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 |
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