How can I visualize independent and dependent set of vectors? Can someone help me visualize those concepts? It will also help me understand it better.
Thanks :) 
 A: Let's work in $\mathbb{R}^2$. You can visualize vectors as arrows, and add them by laying the arrows head-to-tail.

That's what we're going to do to visualize linearly dependent and linearly independent vectors.
Here's a collection of vectors:
$$
\begin{aligned}
A &= (2, 0)\\
B &= (1, -1)\\
C &= (1, 1)
\end{aligned}
$$
If you drew each of these vectors as arrows on the same coordinate grid, it would look like (a better drawn version of) this.
        |    C
        |   /
        |  /
        | /
--------o--------A
        | \
        |  \
        |   \
        |    B

(Note that the vector $A$ coincides overlaps with the $x$-axis).
Are these vectors linearly independent? No, they're dependent, since $A = B + C$. And if you laid $B$'s tail at the head of $C$, you would get $A$.
        |    C
        |   / \
        |  /   \
        | /     \
--------o--------A
        |
        |  
        |   
        |    

You can think about linear independence as efficiency -- do you really need everything you've got, or are some of your tools extraneous?
Suppose someone asked you to build a bridge with the vectors $A, B, C$. Now they ask you to build the same bridge with just $B$ and $C$. Can you do it?
Sure, since whenever you would have used an $A$ you could just use one copy of $B$ plus one copy of $C$.
(In fact you could build the same bridge with any two of them, since you can also make $C$ out of $A$ and $B$, by using $A - B = C$.)
Even if they gave you a shorter copy of $C$, like $(\frac{1}{2}, \frac{1}{2})$, you're allowed to scale vector by a constant before you use them, so you could double it and still make $A$. This is where linear combinations come in.
A linearly independent set is one where none of the elements can be made with a combination of the others, even with scaling allowed.
A: The following is pretty intuitive:
A list $(a_1,a_2,\ldots,a_r)$ of vectors $a_k\in V$ is linearly independent iff none of the $a_k$, $1\leq k\leq r$, is a linear combination of its predecessors in the list. This implies that for each $k$ one has  $${\rm dim}\bigl({\rm span}(a_1,\ldots, a_k)\bigr)={\rm dim}\bigl({\rm span}(a_1,\ldots, a_{k-1})\bigr)+1\ ,$$  so that
$${\rm dim}\bigl({\rm span}(a_1,\ldots, a_r)\bigr)=r\ .$$
A: A set of linearly independent vectors: pick any vector in this set and you cannot write it as a linear combination of the other vectors in this set. For example suppose $\{x_1,x_2,x_3\}$ is a set of linearly independent vectors. Let's say you want to write $x_2$ as a linear combination of $x_1$ and $x_3$ i.e. $x_2 = a\cdot x_1 + b\cdot x_3$ with $ a,b \in \mathbb{R}$. Since the vectors are linearly independent, this is impossible. A more concrete example (illustrating the exact same thing) is done by taking $x_1 = \sin x , x_2 = \tan x , x_3 = \cos x$
A set of linearly dependent vectors: again pick any vector in this set. Then we can write it as a linear combination of the other vectors in the set. Here an example could be $\{x,2,x+2\}$ we see that 
$x = 1\cdot(x+2) -1 \cdot x$
$2 = 0\cdot x + 1\cdot 2$
$x+2 = 1\cdot x + 1\cdot 2$  
A: Two parallel vectors are linearly dependent. Three vectors lying in a plane are linearly dependent. Four vectors lying in the same three-dimensional hyperplane are linearly dependent.
In n-dimensional space, you can find at most n linearly independent vectors.
Think of the vectors as rods with which you want to span up a tent: one rod gives you just a line, two rods give you a face. you need a third rod outside of that plane (linearly independent) to span up a volume. Any additional rods cannot span into a fourth dimension, so four rods in three dimensions must be linearly dependent.
A: Euclidean vectors are the easiest to visualize.  Let's go through this in steps.  We'll skip a set of only one vector until the end and start with a set of two vectors.
If a set of two Euclidean vectors is linearly dependent then there exists a line containing all both vectors and the origin.  If a set of two Euclidean vectors is linearly independent then there does not exist a line which contains both vectors and the origin.

If a set of three Euclidean vectors is linearly dependent then there exists a plane containing all three vectors and the origin.  If a set of three Euclidean vectors is linearly independent then there does not exist any plane containing all three and the origin.

Likewise if a set of $n$ Euclidean vectors is linearly dependent then there exists an $(n-1)$-dimensional subspace containing all $n$ vectors and the origin.  If a set of $n$ Euclidean vectors is linearly independent then there does not exist any $(n-1)$-dimensional subspace containing all $n$ vectors and the origin.  This is a bit harder to visualize for $n\gt 3$.
Now let's handle the case of a set of only $1$ vector.  It should be true that a set of one Euclidean vector is linearly dependent if and only if there exists a $0$-dimensional subspace containing that vector and the origin.  But what is a zero-dimensional subspace?  It's a point.  So if a zero-dimensional subspace is a point, then the only zero-dimensional subspace containing the origin must be the origin.  So we see that a set of $1$ vector is linearly dependent if and only if that one vector is the zero vector.

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