Spanning sets and vector spaces Let $V=\mathbb{R}^{3}$, find $S$ such that ${\rm Span}(S)=V$ and such that $S$ is linearly dependent.
For this question I have no idea where to start, how would I even think of such an $S$? I would prefer an explanation towards the solution rather than just the answer to this one. Thanks!
 A: Hint:
One set that spans $\mathbf{R}^3$ is the set made out of the unit vectors $\hat \imath,\hat\jmath$ and $\hat k$, which is $$\{(1,0,0),(0,1,0),(0,0,1)\},$$ since every element $(x,y,z)$ of $\mathbf{R}^3$ can be written as a linear combination using those vectors : $$(x,y,z)=x\hat\imath+y\hat \jmath+z\hat k=x(1,0,0)+y(0,1,0)+z(0,0,1).$$ Now use the fact that if some vector in a list of vectors in $V$ is a linear combination of other vectors in that same list, then the list is linearly dependent.
A: Note that for
the three  triplets
$\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)$,  $\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)
$ and
$
\left(\begin{array}{c}
0\\
0\\
1
\end{array}\right)
$,
you have for an arbitrary one like $\left(\begin{array}{c}
a\\
b\\
c
\end{array}\right)$
its expression as a linear combination:
$$\left(\begin{array}{c}
a\\
b\\
c
\end{array}\right)
=
a
\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)
+
b\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)
+
c\left(\begin{array}{c}
0\\
0\\
1
\end{array}\right)
$$
