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Given linear functionals $\varphi_2 = ( 2 , -2\alpha, 1)$ and $\varphi_1 = (2,1 -\alpha, -1)$ how do I find kernel of $\varphi_1 - 2\varphi_2$? ($\alpha$ is some parameter)

Is it so simple as $ker(\varphi_1-2\varphi_2)$? (Of course the kernel will be dependent on parameter $\alpha$)

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How about simply computing $\phi_1-2\phi_2$ and then computing its kernel? –  Rasmus Jan 8 '13 at 9:22
    
I don't understand the second question. Please clarify. –  Rasmus Jan 8 '13 at 9:23
    
The second question is exactly what you've said in the first comment - computing the combination and then it's kernel. I'm asking if this is legit. –  NumberFour Jan 8 '13 at 9:26
    
@NumberFour: Yes. You're asked to find the kernel of $\phi_1 - 2\phi_2$, which is $\ker(\phi_1 - 2\phi_2)$. Where's the confusion? –  Clive Newstead Jan 8 '13 at 9:28
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As you want to find the kernel of $\phi_1-2\phi_2$, yes, you should simply find $\ker(\phi_1-2\phi_2)$. As this is really a reformulation, Rasmus was confused how you could be confused about the first formulation when you seem to have no problem with the second ... –  Hagen von Eitzen Jan 8 '13 at 9:28
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1 Answer 1

up vote 1 down vote accepted

$x \in ker (\phi_1 - 2 \phi_2) \iff (\phi_1 - 2 \phi_2) (x) = 0$.

And $(\phi_1 - 2 \phi_2)(x) = (-2x_1,(3\alpha +1)x_2,-3x_3)=(0,0,0)$, so you can solve the rest.

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