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OMAR91
I am trying to find e-values and e-vectors of the matrix: [-2 3 -2 2] I found that ʎ= √2 i Then when trying to find the e-vector I was stuck at V= [-2-√2 i 3 2- 2-√2i] How do I find V ???
[-2-x 3 -2 2-x] (-2-x)(2-x) -3(-2) -2(2-x)-x(2-x) +6 -4+2x -2x +x^2 +6 x^2 - 2 = 0; x = +- sqrt2 right?
your right, the i is there ... forgot how to subtract for a sec
you do an rref now right?
or: (A-L)x = 0 (-2-√2 i)x + 3y = 0 2x+(2-√2i)y = 0 we can setup a system of equations to solve for x and y
As I took in class, I should find a value to multiply it by the first row so the first element in the 2nd row becomes 0. Then it is in row-echelon form and the rest is easy...
Sorry I mean after multiplying something by the 1st row AND ADDING to the 2nd row, the 1st element of the 2nd row should become 0
we get something like: 1 -1+i/sqrt2 0 0 x = (1+i/sqrt2)y
i take it you are not good at row reductions?
I'm not good at math! :(
im gonna use a instead of isqrt2, in either case its simply an unknown amount -2-a 3 -2 2-a swap rows, not needed but its viable -2 2-a -2-a 3 divide row 1 by -2 1 -1+a/2 -2-a 3 multiply row 1 by 2+a and add the results to row 2 2+a (-1+a/2)(2+a) -2-a 3 ----------------------------- 0 (-1+a/2)(2+a) + 3 -1(2+a)+(a/2(2+a) + 3 -2 -a + a + a^2/2 + 3 a^2/2 + 1 now we might want to address that a = -isqrt2 (-i)^2 2/2 + 1 (-i)^2 + 1 -1 + 1 = 0 these leaves as as: 1 -1+a/2 0 0 therefore: x = y(1-a/2)