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after i plug in the value, i have the matrix |dw:1334524320293:dw|
multiplied 1st row by 2 and add to second row, multiply 1st row by -2 and add to 3rd row... |dw:1334524445998:dw|
multiply second row by -1/2 and add to 3rd row... |dw:1334524531744:dw|
so now i have these equations v1+v2+2v3=0 -2v2+2v3=0 -6v3=0
i'm not sure what the vector is, if v3=0, then -2v2+2(0)=0 so v2 must be 0 also and then v1+0+0=0 would mean v1 is zero too, right?
but that's what you wanted |A-I(Lambda)| =0
just put any value, for example v3=a and find the other
well for other eigenvalues, i have vectors with nonzero values...ok, if i plug in 3for v3, the last equation wouldn't make sense. and the answer isn't 0 vector. the vectors i have for my other values are correct
you should get a singular matrix the last row should be 0
-2v2+2a=0 v2=a v1+v2+2v3=0 v1+a+2a=0 v1 = -3a eigenvector (1,1,3)
it doesn't have to be all 0
@phi is right the matrix should be singular. So you made a mistake somewhere
remember you are solving det(A- lambda*I)= 0
i was taught the bottom left triangle has to be 0s.
so it would be the bottom left corner, and the values right above and to the right
the determinant has to be 0
i've checked and checked to see if i made a mistake...i will look again
what is the original matrix?
i think your eigenvalue is wrong
ugh idiot i did do it wrong. value is right, i will rework this
just trying to help