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Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$

I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?

Thanks.

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4 Answers 4

up vote 6 down vote accepted

Hint: Your third and fourth vectors are just scaled versions of the first two vectors.

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I'm ashamed to say that I neglected to even look for something as simple as that. I thought I'd have to use some elaborate proof sketch showing that it works for any pair of vectors. Thanks! –  agent154 May 16 '13 at 14:37
    
Sure, don't worry, we all overlook things once in a while ... :) –  Matt L. May 16 '13 at 14:39

A linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is a vector that can be expressed as $$a\pmatrix{1\\\tfrac{3}{2}\\0}+b\pmatrix{0\\3\\6}$$ for some real numbers $a$ and $b$.

A linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$ is a vector that can be expressed as $$c\pmatrix{2\\3\\0}+d\pmatrix{0\\1\\2}$$ for some real numbers $c$ and $d$.

But $$2\pmatrix{1\\\tfrac{3}{2}\\0}=\pmatrix{2\\3\\0}\qquad \tfrac{1}{3}\pmatrix{0\\3\\6}=\pmatrix{0\\1\\2}.$$

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Suppose $x=\alpha u_1 + \beta u_2$. Then $x=(\frac{\alpha}{2})u_1'+(3\beta)u_2'$, where $u_1, u_2$ are your first two vectors and $u_1', u_2'$ are your second two vectors.

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every combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ will be as $c_1\pmatrix{1\\\frac{3}{2}\\0}+c_2\pmatrix{0\\3\\6}=\frac{c_1}{2}\pmatrix{2\\3\\0}+3c_3\pmatrix{0\\1\\2} $

so :

any linear combination of$ \pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $ \pmatrix{2\\3\\0}$and$\pmatrix{0\\1\\2}$

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