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The problem is:

Suppose $c\mathbf{u} = \mathbf{0}$ for some nonzero scalar $c$. Show that $\mathbf{u}=\mathbf{0}$.

Using the axioms for a vector space, I figured I could use $c(d\mathbf{u}) = (cd)\mathbf{u}$ first, and multiplying both sides with the inverse of $c$: $$\begin{align} c\mathbf{u}=\mathbf{0} \\ c^{-1}(c\mathbf{u})=\mathbf{0} c^{-1} \\ (c^{-1}c)\mathbf{u}=\mathbf{0} \\ 1\mathbf{u} = \mathbf{0} \end{align}$$

And then use the axiom that $1 \mathbf{u} = \mathbf{u}$ to show that $\mathbf{u} = \mathbf{0}$.

Is this somewhat correct? Is there a better, or more precise way of doing this? Thank you!

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1  
Yes, it looks perfectly fine. –  user9413 Sep 4 '11 at 13:49
    
@Chandrasekhar: Thanks! –  Jodles Sep 4 '11 at 13:50
3  
And to be nitpicky, I don't know if it's acceptable to everyone to write $\mathbf 0 c^{-1}$. The scalar is usually written to the left of the vector. –  Srivatsan Sep 4 '11 at 15:05
    
@Srivatsan: Good to know, thanks! –  Jodles Sep 4 '11 at 15:57

1 Answer 1

up vote 2 down vote accepted

Depending on the definition of a vector space, you may need to explain why $c^{-1}0=0$ (which is used on the RHS of the third line of your proof). In many (or most?) textbook definitions of a vector space, that "the scalar multiple of a zero vector is the zero vector itself" is not part of the definition, but a consequence of the axioms.

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Good point, thanks! –  Jodles Sep 4 '11 at 15:58

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