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anybody wants to help me with eigenvectors?
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vectors are the result of the values ....
values are the result of the vectors ...
question number 3.. the eigen value is.. \[\lambda1= 4; \lambda2= -1; \lambda3= -2\]
can we gordon jauss to get vectors?
yes
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lol, yes amistre we can, but seems like overkill
:)
(A-Li)x = eugene vector ... maybe gonna have to look it up again
(A-Li)v = 0 where v is the eugene vector
Fool thinks it's overkill to do Gauss-Jordan, but I don't see another way offhand, so looking at\[(\lambda_1 I-A)=\]|dw:1330287299340:dw|which reduces to|dw:1330287361068:dw|and so we get\[x_1=-\frac12 t\]\[x_2=0\]\[x_3=t, t\in\mathbb R\]
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