How can I visualize independent and dependent set of vectors?

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.

(Source of Images)

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.

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 $${\rm dim}\bigl({\rm span}(a_1,\ldots, a_r)\bigr)=r\ .$$