No.1. Carefully write out three times: the definition of a set T to be linearly independent, the definition for & set B to be a basis for a vector space S, the ...

Prove that if $S=\left\{v_{1}, v_{2}, \ldots, v_{r}\right\}$ is a linearly independent set of vectors, then so is every nonempty subset of $S$.

In this problem, we need to prove that if we want we to V three is linearly independent set of vectors, then so are the given subsets of this set. Now note that since we want me to meet three is linearly independent set of vectors. Thus this means that none of the vectors can be written as a linear combination of the other victims. So if we want cannot be written as a linear combination of V two and V three, then that would mean that we won cannot be written as a linear combination of just be to either and hence we want. We too will be linearly independent. Similarly, everyone will not be able to be written as a linear combination of V three, just be three, so it will not be a scale and multiple A. V three. And this we want V three is linearly independent. Similarly, since we want V two, V three, this is a linearly independent set of vectors. V two cannot be written as a linear combination of V one and V three, and thus we cannot be written as a combination, or rather as a scalar multiple, A. V three and V two, V three. This is also linearly independent vectors. Also, since this set is linearly independent, it means that we want we to envy three are all non zero vectors, and since they are non zero vectors, the set V one, which is a singleton set containing only a non zero vector, this will be linearly independent. Similarly, we do is be linearly independent, and the single done set containing only be three will also be linearly independent. Hence this, These six sets are all linearly independent sets of records.