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This one is quite straightforward, but I just want to make sure that my reasoning is clear.

I have following proposition:

Proposition. If $S = \{\mathbf{v_{1}},\mathbf{v_{2}}...,\mathbf{v_{n}}\}$ is linearly independent then any subset $T = \{\mathbf{v_{1}},\mathbf{v_{2}}...,\mathbf{v_{m}}\}$, where $m < n$, is also linearly independent.

My attempt:

We prove proposition by contrapositive.

Suppose $T$ is linearly dependent. We have

$$\tag1 k_{1}\mathbf{v_{1}} + k_{2}\mathbf{v_{2}}\cdots k_{j}\mathbf{v_{j}} ... k_{m}\mathbf{v_{m}} = \bf O $$

Where there is at least one scalar, call it $k_{j}$, such that $k_{j} = a$ ($a ≠ 0$) and all other scalars are zero.

Since $T$ is the subset of $S$, the linear combination of vectors in $S$ is:

$$\bigl(k_{1}\mathbf{v_{1}} + k_{2}\mathbf{v_{2}}\cdots k_{j}\mathbf{v_{j}} ... k_{m}\mathbf{v_{m}}\bigr) + k_{m+1}\mathbf{v_{m+1}}\cdots +k_n\mathbf{v_{n}} = \bf O$$

Let $k_{j} = a $, and set all other scalars for zero:

$$\underbrace{\bigl(0\cdot\mathbf{v_{1}} + 0\cdot\mathbf{v_{2}}\cdots a \cdot \mathbf{v_{j}} ... 0\cdot\mathbf{v_{m}}\bigr)}_{\mathbf{= O} \text{ by } (1)} + \underbrace{0\cdot\mathbf{v_{m+1}}\cdots +0\cdot\mathbf{v_{n}}}_{\mathbf{= O} \text{ because all scalars = 0}}= \bf O$$

We can see that linear combination of $S$ equals to zero but we have at least one non-zero scalar, which implies that $S$ is not linearly independent, which is a contradiction. Therefore, if $S$ is linearly independent, arbitrary subset $T$ must be linearly independent as well. $\Box$

Is it correct?

Stokolos Ilya
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    I think you should not say that $k_j=a$ and all the other scalars are zero, and you don't need this anyway. You can/need only say that there is at least one non-zero scalar. The dependence for $T$ becomes a dependence for $S$ as you have outlined. – Mark Bennet Sep 15 '19 at 07:56

2 Answers2

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Proof looks good, but scalars $k_{1},...,k_{j-1},k_{j+1},...,k_{m}$ may not be all zero. But still you will get set $ \{ v_{1},...,v_{n} \}$ as linear dependent, due to false assumption. Also you can not mix both proof by contradiction and proof by contrapositive in general.

Manoj Kumar
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I don't see anything wrong with your proof. Just be careful with the claim that you get a contradiction. The contrapositive of a statement is logically equivalent to the statement itself, so you don't get any contradiction whatsoever when proving a contrapositive.

If you were to use a proof by contradiction, you would start off by assuming that $S$ is linearly independent but $T$ is not, and show that it leads to some impossibility.

jl00
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  • Should I just delete "which is a contradiction" part then? – Stokolos Ilya Sep 15 '19 at 07:20
  • Yes, it is enough just to say that you're proving the contrapositive at the beginning of the proof. – jl00 Sep 15 '19 at 07:21
  • To be honest, I'd thought proofs by contradiction and by contrapositive were the same thing, i.e, if you want to prove $P \implies Q$, then you assume that $\lnot Q$ and show that $P$ is impossible (contradiction), or in other words, show that $\lnot P$ (contrapositive). Just the wording a bit different. – Stokolos Ilya Sep 15 '19 at 07:22
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    @Nelver This may help you – Manoj Kumar Sep 15 '19 at 07:29
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    In fact they are not the same, since we have $P \implies Q \iff \neg Q \implies \neg P$ (the contrapositive), whereas in a proof by contradiction, you assume $\neg (P \implies Q) \iff \neg (\neg P \lor Q) \iff \neg \neg P \land \neg Q \iff P \land \neg Q$ – jl00 Sep 15 '19 at 07:29