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I came across a question in my linear algebra textbook and it said: "Given $x_1 = (1, 1, 1)^T$ and $x_2 = (3, -1, 4)^T$: Do $x_1$ and $x_2$ span $\Bbb R^3$? Explain."

I'm pretty sure that the answer is no (I thought you needed n vectors to span $\Bbb R^n$) but I'm not sure how to show that this is the case, if anyone could help it would be greatly appreciated.

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A perfectly good proof in any specific case such as this is to take (almost) any vector and show that it is not a linear combination of your vectors.

For the more general proof that you need three vectors, you have to point out that a set of three equations in two variables generally does not have a solution.

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Hint : Can you find $(\lambda,\mu)\in \mathbb{R}^2$ such that $\lambda x_1+\mu x_2 = (4,1,5)^T$ ?

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Given a basis of a vector space, the dimension is defined to be exactly the number of vectors in the basis. Since we know that $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$ span $\mathbb{R}^3$, hence the dimension is 3. So it can't be spanned by two vectors, otherwise the dimension would be 2.

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$(1,0,0)$, $(0,1,0)$, $(0,0,1)$ and $(1,1,1)$ also span $\mathbb{R}^3$. That doesn't imply that the dimension of $\mathbb{R}^3$ is $4$. – posilon Jun 14 '15 at 12:36

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