Prove that if the set of vectors is linearly independent, then the arbitrary subset will be linearly independent as well.

Question

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. $\Box$

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Is it correct?

Answer 1

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.

Answer 2

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.