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Is the following proof correct?

Theorem. If $v_1,v_2,v_3,v_4$ is a linearly independent list then $$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$ is also a linearly independent list.

Proof. Assume that $v_1,v_@,v_3,v_4$ is a linearly independent list, Consider now the following equation. $$0=0(v_1-v_2)+0(v_2-v_3)+0(v_3-v_4)+0v_4\tag{1}$$ Let $a_1,a_2,a_3$ and $a_4$ be arbitrary scalars in $\mathbf{F}$ and assume that the following equation holds $$0=a_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+a_4v_4\tag{2}$$ After some algebraic manipulation we arrive at the following equation. $$0=a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+(a_4-a_3)v_4\tag{3}$$ Since the list $v_1,v_2,v_3,v_4$ is linearly independent it follows that given any vector in $span(v_1,v_2,v_3,v_4)$ the choice of scalars is unique and since $$0=0v_1+0v_2+0v_3+0v_4\tag{4}$$ It follows that all the scalars in $(3)$ must be $0$, consequently the only way to produce the $0$ vector as a linear combination of the vectors in the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is that indicated in $(1)$.

$\blacksquare$

Here $\mathbf{F}$ is either $\mathbb{C}$ or $\mathbb{R}$.

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    $\begingroup$ Sounds ok - would just like to point out 'Consider now the following equation.' sounds a bit strange for $(1)$. Perhaps remove that line and $(1)$ and say something along the lines 'by inspection, $(2)$ holds when all the scalars are equal to $0$' at the end of your proof $\endgroup$ – Shuri2060 Jul 25 '17 at 17:31
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    $\begingroup$ Nitpicking, but you made a small typo just before your first labeled equation, where you write "Assume $v_1,v_{@},v_3,v_4$ is a linearly independent list." Did you mean $v_2$ not $v_{@}$ ? $\endgroup$ – Vivek Kaushik Jul 25 '17 at 17:37
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Equation (1) should be omitted: it's an obvious fact that has no consequence on the rest.

The final argument is too fast: from linear independence of $v_1,v_2,v_3,v_4$ you deduce \begin{cases} a_1=0 \\ a_2-a_1=0 \\ a_3-a_2=0 \\ a_4-a_3=0 \end{cases} and, from this, $a_1=a_2=a_3=a_4=0$. This should be mentioned, although easy.

A different approach is to consider the coordinates of the new vectors with respect to the original ones and so the matrix \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{bmatrix} A standard Gaussian elimination leads to the reduced row echelon form \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} which proves that the coordinate vectors are linearly independent and so also the vectors are.

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  • $\begingroup$ So your point is that i should show how each of the scalars $a_1,a_2,a_3$ and $a_4$ is 0. $\endgroup$ – Atif Farooq Jul 25 '17 at 17:59
  • $\begingroup$ The original matrix is lower triangular with $1$'s in the diagonal, hence invertible. $\endgroup$ – lhf Jul 25 '17 at 18:49
  • $\begingroup$ @AtifFarooq Yes, that's the point; notwithstanding it's easy, it should appear in the proof. $\endgroup$ – egreg Jul 25 '17 at 19:27
  • $\begingroup$ @lhf That's true, but Gaussian elimination would give more information in case the vectors aren't linearly independent. $\endgroup$ – egreg Jul 25 '17 at 19:27
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It's valid. A note:

Before $(2)$ when you say "...and assume that the following equation holds..." it would be better to perhaps phrase this as "We wish to solve ... for $a_j$." Saying the former means that you're assuming solutions exist - but you're not sure! This is a tiny point, and is perhaps reflected in my writing style.

We may also streamline a bit by saying the following after $(3)$.

As $\{v_1, v_2, v_3, v_4\}$ is a linearly independent set, we must have that $a_1 = 0, \ a_2-a_1=0, \ a_3 - a_2 = 0,$ and $a_4-a_3=0.$ Back substitution yields that each $a_j = 0$, and so the original set is also a linearly independent set. $\square$

Again, these are just small suggestions, but nonetheless your proof is correct.

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