Does $\{u_1, u_2, u_3, u_4\}$ spanning $\mathbb R^3$ mean that $\{u_1,u_2,u_3\}$ also does? Since it's a subset? Does $\{u_1, u_2, u_3, u_4\}$ spanning $\mathbb R^3$ mean that $\{u_1,u_2,u_3\}$ also does?  Since it's a subset? A little unclear about this...
 A: Take $u_2, u_3$ and $u_4$ that span $\mathbb R^3$, then take $u_1=u_2$. Now you have $\{u_1,u_2,u_3\}=\{u_2,u_3\}$ which does not span $\mathbb R^3$
A: A concrete example:
$$u_1=(1,0,0),u_2=(2,0,0),u_3=(0,1,0),u_4=(0,0,1)$$
$\{u_1,u_2,u_3,u_4\}$ clearlly spans $\mathbb{R}^3$.
On the other hand $u_4 \notin span\{u_1,u_2,u_3\}$, and therefore $\{u_1,u_2,u_3\}$ does not span $\mathbb{R}^3$.
A: In general: if $A$ spans $V$ and $A\subset B$, then $B$ spans $V$. The converse is obviously false, but if $B$ spans $V$ and $b\in B$ is a linear combination of elements of $B\backslash\{b\}$, then $B\backslash\{b\}$ spans $V$.
A: Less formally, but to get the idea in words : for a set to span a vector space is a sufficient condition. It means that you have enough "good" vectors in the set for it to span the vector space. If you add more vectors to your set, then you will still have your good vectors in the set, so it will still span the space. If you remove a vector from the set, however, you might remove one of the good vectors, and in that case you will no longer span the space.
Conversely, if you start with a set which does not span the space, it means that your set does not have enough good vectors in it. If you remove vectors, then certainly it will stil not have enough good vectors, so it will still not span the space. However, if you add vectors, then it is possible that the vectors you add are good vectors, and that the resulting set will span the space.
A: The following two statements are true:
(1) If $\{u_1, u_2, u_3\}$ span $V$, then (for any $u_4$) it follows that $\{u_1, u_2, u_3, u_4\}$ span $V$.
(2) If $\{u_1, u_2, u_3, u_4\}$ do not span $V$, then it follows that $\{u_1, u_2, u_3\}$ do not span $V$. And, similarly, of course, $\{u_2, u_3, u_4\}$ do not span $V$, either.
The following two statements are false:
(3) If $\{u_1, u_2, u_3, u_4\}$ span $V$, then it follows that $\{u_1, u_2, u_3\}$ span $V$.
(4) If $\{u_1, u_2, u_3\}$ do not span $V$, then (for any $u_4$) it follows that $\{u_1, u_2, u_3, u_4\}$ do not span $V$. 
But statement (2) doesn't tell us anything new -- it follows directly from statement (1). In fact, it's called the "contrapositive" of statement #1. Similarly, (4) doesn't tell us anything new, because it's just the contrapositive of (3).
That's the logic. Intuitively, a set of vectors needs to be "big" and "fat" in order to span a space. So, adding vectors increases your chances of spanning the space, and removing vectors reduces your chances.
A: Hint: what will happen if $u_1=0$?
A: It is not true that if $u_1, u_2, u_3, u_4$ span ${\mathbb R}^3$, then $u_1, u_2, u_3$ necessarily span ${\mathbb R}^3$ as well. Examples have been given in previous answers, e.g, $u_1 = (1,0,0)$, $u_2 = (2,0,0)$, $u_3 = (0,1,0)$, $u_4 = (0,0,1)$.
What is true, however, is that if $u_1, u_2, u_3, u_4$ span ${\mathbb R}^3$, then among those four vectors there are three vectors that do span ${\mathbb R}^3$. They might be $u_1, u_2, u_3$, but could also be $u_1, u_2, u_4$, or $u_2, u_3, u_4$.
This last claim is a special case of the following general statement: let $W$ be a vectorspace and let $A$ be a subset of $W$. Let $V \subseteq W$ be the span of $A$. Then there is a subset $B$ of $A$ that is a basis of $V$. (In your case $W = {\mathbb R}^3$, $A = \{u_1, u_2, u_3, u_4\}$, $V = {\mathbb R}^3$, and $B$ will consist of exactly $3$ elements because the dimension of $V$ is $3$).
