- #1
Roodles01
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In my example I have matrix A = (1 2)
. . . . . . . . . . . . . . . . . . . . . . (3 2)
Finding the eigenvalue through the method I understand & can get the result
i.e.
k = 4 & -1
I suspect my algebra is the shaky link, here, but to find the eigenline I find a bit more of a challenge.
OK I start by substituting the eigenvalue into the eigenvector equation;
Ax = kx
giving
(1 2) (x) = 4 (x)
(3 2) (y)...(y)
which gives rise to the following simultaneous equations
x + 2y = 4x
3x + 2y = 4y
Now the bit I don't get . . .
How do these both reduce to 3x - 2y = 0
I'm sure it's simple & I can't see the wood for the trees, but this is stupidly defeating me.. . . . Grrr!
Please could someone point me in the right direction.
. . . . . . . . . . . . . . . . . . . . . . (3 2)
Finding the eigenvalue through the method I understand & can get the result
i.e.
k = 4 & -1
I suspect my algebra is the shaky link, here, but to find the eigenline I find a bit more of a challenge.
OK I start by substituting the eigenvalue into the eigenvector equation;
Ax = kx
giving
(1 2) (x) = 4 (x)
(3 2) (y)...(y)
which gives rise to the following simultaneous equations
x + 2y = 4x
3x + 2y = 4y
Now the bit I don't get . . .
How do these both reduce to 3x - 2y = 0
I'm sure it's simple & I can't see the wood for the trees, but this is stupidly defeating me.. . . . Grrr!
Please could someone point me in the right direction.
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