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?
Best Answer
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.