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I came across the following problem:

Let ${v_{1},v_{2},v_{3},v_{4}}$ be a basis of $\Bbb R^{4}$ and $v=a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}+a_{4}v_{4}$ where $a_{i}\in \Bbb R,i=1,2,3,4.$ Then ${v_{1}-v,v_{2}-v,v_{3}-v,v_{4}-v}$ is a basis of $\Bbb R^4$ if and only if

(a) $a_{1}=a_{2}=a_{3}=a_{4},$

(b) $a_{1}a_{2}a_{3}a_{4}=-1,$

(c) $a_{1}+a_{2}+a_{3}+a_{4}\neq1,$

(d) $a_{1}+a_{2}+a_{3}+a_{4}\neq0.$

My attempts: Since ${v_{1}-v,v_{2}-v,v_{3}-v,v_{4}-v}$ is a basis of $\Bbb R^4$ we can express $v=a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}+a_{4}v_{4}=a_{1}(v_{1}-v)+a_{2}(v_{2}-v)+a_{3}(v_{3}-v)+a_{4}(v_{4}-v)$ (here, I am not sure whether same scalars ${a_{1},a_{2},a_{3},a_{4}}$ can be used) and hence calculating we get $v( a_{1}+a_{2}+a_{3}+a_{4})=0$ and so $a_{1}+a_{2}+a_{3}+a_{4}=0 $ for $v\neq0.$ Am I right? If not, where did I go wrong? Please help. Thanks in advance for your time.

3 Answers 3

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You can't use the same scalars to get $v$ as a linear combination of the $v_i-v$ vectors.

Note that since any basis for $\Bbb R^4$ has $4$ vectors in it, then you need only determine what conditions the $a_i$ must satisfy for the $v_i-v$ to be linearly independent. We begin by supposing that $\begin{align}0 &= b_1(v_1-v)+b_2(v_2-v)+b_3(v_3-v)+b_4(v_4-v)\\ &= b_1v_1+b_2v_2+b_3v_3+b_4v_4-(b_1+b_2+b_3+b_4)v\end{align}$ for some $b_1,b_2,b_3,b_4\in\Bbb R$. To conclude that the $v_i-v$ form a basis, we'll need to show that the $b_i$ are necessarily all $0$. Using the definition of $v$ to rewrite everything, we have $\begin{align} 0= & \bigl(b_1-(b_1+b_2+b_3+b_4)a_1\bigr)v_1+\bigl(b_2-(b_1+b_2+b_3+b_4)a_2\bigr)v_2\\ & +\bigl(b_3-(b_1+b_2+b_3+b_4)a_3\bigr)v_3+\bigl(b_4-(b_1+b_2+b_3+b_4)a_4\bigr)v_4,\end{align}$ and since the $v_i$ comprise a basis, then each $b_i-(b_1+b_2+b_3+b_4)a_i=0$. Adding all the $b_i-(b_1+b_2+b_3+b_4)a_i=0$ equations together gives us $(b_1+b_2+b_3+b_4)-(b_1+b_2+b_3+b_4)(a_1+a_2+a_3+a_4)=0.\tag{#}$ If we happen to know that $a_1+a_2+a_3+a_4\neq 1$, then dividing both sides of $(\#)$ by the non-zero number $1-(a_1+a_2+a_3+a_4)$ yields $b_1+b_2+b_3+b_4=0,$ and so $0=b_i-(b_1+b_2+b_3+b_4)a_i=b_i$ for each $i$, as desired. This shows that if option (c) holds, then the $v_i-v$ comprise a basis for $\Bbb R^4$.

I'll leave the other direction to you.

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    @CameronBuie thank you sir. I am also editing my question.2012-12-09
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I think maybe there is a mistake in your calculation: the same vector always has different coordinate when we changed the basis, so you can not assume the same scalars can be used. We can just let the determinant of matrix the coordinate be non-zero (the calculation is a little bit tedious). The answer is $a_{1}+a_{2}+a_{3}+a_{4}\ne 1$

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    I also thought that.But i could not reach any one of the given options by using different scalars.2012-12-09
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The vectors $v_1-v,v_2-v,v_3-v,v_4-v$ constitute a basis for $\mathbb R^4$ iff they are linearly independent iff $\lambda_1(v_1-v)+\lambda_2(v_2-v)+\lambda_3(v_3-v)+\lambda_4(v_4-v)=0 \Longrightarrow \lambda_1=\lambda_2=\lambda_3=\lambda_4=0$

But $\lambda_1(v_1-v)+\lambda_2(v_2-v)+\lambda_3(v_3-v)+\lambda_4(v_4-v)=0$ is equivalent to the system (since we know that $v_1,v_2,v_3,v_4$ is a basis) $\lambda_1(1-a_1)-\lambda_2a_1-\lambda_3a_1-\lambda_4a_1=0\\ -\lambda_1a_2+\lambda_2(1-a_2)-\lambda_3a_2-\lambda_4a_2=0\\ -\lambda_1a_3-\lambda_2a_3+\lambda_3(1-a_3)-\lambda_4a_3=0\\ -\lambda_1a_4-\lambda_2a_4-\lambda_3a_4+\lambda_4(1-a_4)=0$ with determinant $1-(a_1+a_2+a_3+a_4)$. This system has a unique solution with respect to $\lambda_i$ iff $a_1+a_2+a_3+a_4\neq1.$

Therefore the correct answer is $a_1+a_2+a_3+a_4\neq1$.

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    thank you sir for clarification.2012-12-09