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Given $V=\mathbb Q^3$, and a vector $v=(x_1,x_2,x_3) \in V$ such that $x_1\neq x_2$. Is there any basis $B$ in $V$ in which the coordinates of the vector $v$ are $(1,0,0)$?

a) No, because $v$ could be the null vector.

b) We can't answer the question.

c) Yes, because $v$ is not the null vector.

d) It depends on the value of $x_3$.

My attempt:

I don't know where to start. I'm lost, and I need at least a hint.

1 Answers 1

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Hint: Asking if the coordinates of $v$ are $(1,0,0)$ is the same as asking if there is a basis for $V$ consisting of $v$ together with two other vectors $w$ and $x$. Since $x_1\ne x_2$, you know that $v$ is nonzero (right?). What do you know about extending a set of vectors to a basis?

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    OK, but you have already solved the problem. It's the option 'c', right? It could be the canonical basis, for example? – 2017-02-07
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    Well, it could be the canonical basis if $(x_1,x_2,x_3) = (1,0,0)$. But in any case, if you extend it to a basis $\{v,w,x\}$, then clearly the coordinates of the three basis vectors are $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. – 2017-02-07