Question

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 ???

Answer 1

[-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?

Answer 2

your right, the i is there ... forgot how to subtract for a sec

Answer 3

you do an rref now right?

Answer 4

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

Answer 5

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...

Answer 6

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

Answer 7

we get something like:

1 -1+i/sqrt2
0 0

x = (1+i/sqrt2)y

Answer 8

How is that?

Answer 9

i take it you are not good at row reductions?

Answer 10

I'm not good at math! :(

Answer 11

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)

Answer 12

AWSOME! Thanks a lot...