Finding the fewest vectors of a set S that still has the same span as S by inspection My book says by inspection which seems to imply without too much work. I can work out the problem with much difficultly by writing it out and solving the set.
S is the set with the vectors:
\begin{bmatrix}
      4         \\[0.3em]
       3
     \end{bmatrix}
\begin{bmatrix}
      -2           \\[0.3em]
       5
     \end{bmatrix}\begin{bmatrix}
      2           \\[0.3em]
       1
     \end{bmatrix}
I know that I can plug all of these into a matrix and solve it but that isn't what the book wants, there should be something obvious here and I am not seeing it. A top voted answer on yahoo answers just states that the first two are not scalar multiples of eachother so that the last one is redundant which doesn't make sense to be because it could be true that the third one is not in the span of the other two. How do I show that two of these are a linear combination or something similar without doing the work to prove it?
 A: The vectors span a subspace.  The subspace is in $\mathbb{R}^2$, so it is either 0,1, or 2 dimensional.  The first two vectors are linearly independent, so they already span a 2 dimensional set.  The only 2 dimensional subspace of $\mathbb{R}^2$ is all of $\mathbb{R}^2$, so the first two vectors already span $\mathbb{R}^2$.  The third vector is in $\mathbb{R}^2$ so it must be in the span of the first two.
A: 
How do I show that two of these are a linear combination or something similar without doing the work to prove it?

You will have to do some work to show that only two vectors suffice to span $\mathbb{R}^{2}$. You have to show that two out of the three vectors are linearly independent, so the third is necessarily a linear combination of the other two. To make this explicit, for $\gamma,\delta \in \mathbb{R}$, show that 
$\gamma$$\left[\begin{matrix}4\\3\end{matrix}\right] +\delta\left[\begin{matrix}-2\\5\end{matrix}\right]=\left[\begin{matrix}0\\0\end{matrix}\right] \Longrightarrow \gamma=\delta =0$ 
Assuming that you've proved that the first two vectors are linearly independent, included below shows how the third vector is a linear combination of the first two. 
\begin{align*}
&4\alpha -2\beta = 2\\
&3\alpha+ 5\beta = 1  \\
\end{align*}
$\Longrightarrow $
$$
\alpha - 7\beta = 1 \Leftrightarrow \alpha = 1+7\beta
$$
$\Longrightarrow$
$$
3\alpha + 5\beta \Leftrightarrow 3+26\beta = 1 \Leftrightarrow \beta = -\frac{1}{13}\Longrightarrow \alpha = \frac{6}{13}
$$
Therefore:
 $\alpha$$\left[\begin{matrix}4\\3\end{matrix}\right] $ + $\beta$$\left[\begin{matrix}-2\\5\end{matrix}\right] =\left[\begin{matrix}\tfrac{24}{13}\\\tfrac{18}{13}\end{matrix}\right]+\left[\begin{matrix}\frac{2}{13}\\-\frac{5}{13}\end{matrix}\right]=\left[\begin{matrix}\frac{26}{13}\\\frac{13}{13}\end{matrix}\right]=\left[\begin{matrix}2\\1\end{matrix}\right] $ 
So the third vector is in the span of the first two. (But you know this apriori if you show that the first two vectors are linearly independent)
A: The vectors are 2D. You only need two linearly independent vectors to span all of two-dimensional space. So, all we have to do is observe that some two of the three are linearly independent. In 2D, two vectors are linearly independent iff they have different directions (they are not parallel). In fact, all three of the given vectors have different directions. So any two of them are linearly dependent and will therefore span all of $\mathbb{R}^2$.
