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 one year ago
I am trying to find evalues and evectors of the matrix:
[2 3
2 2]
I found that ʎ= √2 i
Then when trying to find the evector I was stuck at
V= [2√2 i 3
2 2√2i]
How do I find V ???
 one year ago
I am trying to find evalues and evectors of the matrix: [2 3 2 2] I found that ʎ= √2 i Then when trying to find the evector I was stuck at V= [2√2 i 3 2 2√2i] How do I find V ???

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1[2x 3 2 2x] (2x)(2x) 3(2) 2(2x)x(2x) +6 4+2x 2x +x^2 +6 x^2  2 = 0; x = + sqrt2 right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1your right, the i is there ... forgot how to subtract for a sec

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you do an rref now right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1or: (AL)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

OMAR91
 one year ago
Best ResponseYou've already chosen the best response.0As 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 rowechelon form and the rest is easy...

OMAR91
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we get something like: 1 1+i/sqrt2 0 0 x = (1+i/sqrt2)y

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i take it you are not good at row reductions?

OMAR91
 one year ago
Best ResponseYou've already chosen the best response.0I'm not good at math! :(

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im gonna use a instead of isqrt2, in either case its simply an unknown amount 2a 3 2 2a swap rows, not needed but its viable 2 2a 2a 3 divide row 1 by 2 1 1+a/2 2a 3 multiply row 1 by 2+a and add the results to row 2 2+a (1+a/2)(2+a) 2a 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(1a/2)

OMAR91
 one year ago
Best ResponseYou've already chosen the best response.0AWSOME! Thanks a lot...
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