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OMAR91
 3 years 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 ???
OMAR91
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1your right, the i is there ... forgot how to subtract for a sec

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

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

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

amistre64
 3 years 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)
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