Why can't a set of four vectors in $\mathbb{R}^3$ be linearly independent? Why can't a set of four vectors in $\mathbb{R}^3$  be linearly independent?
I know that if the determinant of the vectors together is not $0$ then the vectors are linearly independent. But this is not relevant to this case of a non square matrix.
Is the answer "because there can only ever be 3 pivots when reduced" a good answer?
 A: This is a consequence of the Steinitz exchange lemma, which says that for a vector space $ V $, if $ L $ is a linearly independent set and $ S $ is a spanning set, then we have $ |L| \leq |S| $. Since we may take $ S $ to be a basis of $ V $, this implies that we must have $ |L| \leq \dim V $.
A: Let's create new names for the unit vectors of $\mathbb{R}^3$: $v_1=(1,0,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$
Consider the basis $\{v_1,v_2,v_3\}$. I will proceed with a proof by contradiction. Assume that there exists a set of $4$ linearly independent vectors in $\mathbb{R}^3$, $\{w_1, w_2, w_3, w_4\}$. 
We can then represent $w_1$ as a linear combination the elements in the set $\{v_1,v_2,v_3\}$.
$$w_1=a_1\cdot v_1+a_2\cdot v_2+a_3\cdot v_3$$
Since $w_1\neq(0,0,0)$, we know that $a_i\neq 0$ for some $i$. After reordering $v_1$, $v_2$, and $v_3$, we can assume that $a_1\neq0$ Then,
$$v_1=\frac{w_1}{a_1}-\frac{a_2}{a_1}\cdot v_2-\frac{a_3}{a_1}\cdot v_3$$
Since we can represent $v_1$ as a linear combination of the elements in $\{w_1, v_2, v_3\}$, the subset $\{w_1, v_2, v_3\}$ generates $\mathbb{R}^3$. Now we can represent $w_2$ as a linear combination of the elements in $\{w_1, v_2, v_3\}$.
$$w_2=b_1\cdot w_1+c_2\cdot v_2+c_3\cdot v_3$$
We know that some $c_i\neq0$ since if both $c_1$ and $c_2$ was equal to $0$, $w_1$ and $w_2$ would be linearly dependent (which contradicts our initial assumption). After reordering, we can assume that $c_2\neq 0$ So, 
$$v_2=\frac{w_2}{c_2}-\frac{b_1}{c_2}w_1-\frac{c_3}{c_2}v_3$$
We now see that the subset $\{w_1, w_2, v_3\}$ generates $\mathbb{R}^3$. Repeating this process one more time, we can see that the subset $\{w_1, w_2, w_3\}$ generates $\mathbb{R}^3$. So, we can represent $w_4$ as a linear combination of the elements in the set $\{w_1, w_2, w_3\}$, proving that they must be linearly dependent.
Source: Linear Algebra, Third Edition, by Serge Lang
A: The dimension of $R^3$ is 3 i.e its basis has 3 vectors. So additional vectors would be just linear combination of 3 vectors. Any set of 3 L. I vectors will be a basis. If you say 4 vectors are linearly independent in R^3 then it would mean they will be part of basis.  Hence dimension of R^3 will become 4 which is not so
