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Say $A=[e_1, e_2, e_3]$ is an ordered basis for $\mathbb R^3$, then is $B=[e_1-e_2, e_1+e_2, e_1+e_3]$ also an ordered basis?

Any hints on how I could go about proving this would be greatly appreciated.

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3 Answers 3

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Since you have three vectors and the dimension of $\mathbb{R}^3$ as a real vector space is $3$, it suffices to check that the vectors are linearly independent. A set of vectors $\{v_1, \ldots, v_n\}$ is linearly independent if the following condition is met: suppose there are $a_i \in \mathbb{R}$ such that $$a_1v_1 + \ldots + a_nv_n = 0$$ then all $a_i$ are zero.

Let us use this to look at your set of vectors. Let $a_1, a_2, a_3 \in \mathbb{R}$ such that $$a_1(e_1 - e_2) + a_2(e_1 + e_2) + a_3(a_1 + a_3) = 0.$$ Using distributivity, we find that this is equivalent to $$(a_1 + a_2 + a_3)e_1 + (-a_1 + a_2)e_2 + a_3e_3 = 0.$$ Since $e_1, e_2, e_3$ are linearly independent, we have that $$\begin{cases} 0 &= a_1 + a_2 + a_3 \\ 0 &= -a_1 + a_2 \\ 0 &= a_3 \end{cases}$$ which shows that $a_3 = 0$, $2a_2 = 0$ (by adding the first two equations, where we already deleted $a_3$) and $2a_1 = 0$ (by substracting the first two equations). Hence we have that $a_1 = a_2 = a_3 = 0$, so the vectors are linearly independent.

Note that this is a quite long explanation, so if you are comfortable with using determinants etc, this would be faster solution.

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Assume a linear combination of the vectors in $B$ sums to zero. Show that forces a linear combination of elements of $A$ that sums to zero.

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As $\det([e_1-e_2, e_1+e_2, e_1+e_3])=2$, $B$ is an ordered basist as well.

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