[Math] Finite superset of linearly dependent set is linearly dependent

Question

Larson Edwards Falvo – Elementary Linear Algebra

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1. Does 58 follow from contrapositive of 57?

2. I'm not sure I follow the proof but based on previous question Subsets of finite sets of linearly independent vectors are linearly independent I think this is what it means:

WOLOG say $S_1 = \{u, u_1, …, u_n\} \subseteq \{v, u, u_1, …, u_n\} = S_2$

WOLOG say we have in $S_1$ that $$u = c_1u_1 + … + c_nu_n$$

Then we have in $S_2$ that

$$u = c_1u_1 + … + c_nu_n + 0v$$

Is that right?

Answer 1

There are several slightly different approaches possible.

One is the characterization of a linearly dependent set as one where one of the elements is a linear combination of (some of) the others.

So, if $S_1\subseteq S_2$ and $S_1$ is linearly dependent, then one of the elements $v\in S_1$ can be written as $$ v=c_1u_1+\dots+c_nu_n $$ with $u_i\in S_1$, $u_i\ne v$ ($i=1,2,\dots,n$). Then also $S_2$ clearly satisfies the same property, so it is linearly dependent.

(Note that the hypothesis that $S_1$ is nonempty is redundant, as the empty set is linearly independent.)

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Another approach: you have proved the statement in exercise 57. This implies the more general statement that

A set of vectors is linearly independent if and only if each of its subsets is linearly independent.

One direction is exercise 57, the other direction is trivial, because a set is a subset of itself.

Thus exercise 58 directly follows.