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.)
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.
The row space is invariant under elementary row operations, so for #51 it makes sense to take the vectors as the rows in a matrix, and bring to reduced row-echelon form.
I would have used rows for #52, also, but since the reduced form turns out to be the identity matrix, you can conclude that in each case the vectors span ${\bf R}^3$.
The computation you've done, I'm not sure it proves anything, other than that the two spaces both have dimension 2.
Best Answer
You are right. Eigen values do not change with the choice of the basis. I believe the standard basis is specified so that the reader can recognize the vector space in which the problem is set in. Also, standard basis makes the calculations bit easier.