Are these vectors independent or dependent? I have this problem:

Determine whether the column vectors 
  $$
\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\, , 
\begin{bmatrix}0\\0\\2\\1\end{bmatrix}\,,
\begin{bmatrix}0\\3\\2\\1\end{bmatrix}\,,
\begin{bmatrix}4\\3\\2\\1\end{bmatrix}
$$
   are independent or dependent, no explanation necessary.

Right away, I thought it was dependent; just by looking at it, we can easily see the reduced form is in a staircase form, which would mean that it was dependent.
I.e, divide every row by the coefficient leading, then perform row reductions.
I asked Wolfram, and it said it was independent.
I do not quite understand what I have done incorrectly. Perhaps I am looking at the problem the wrong way; maybe they were talking about rows, not columns!
 A: I think you are confusing the words dependent and independent. If you set these vectors as rows (or columns) of a matrix and row reduce the matrix, you will indeed find that the matrix is in "staircase form", as you call it. This means that the vectors are linearly independent, not linearly dependent. 
A: Suppose there is a linear combination $$a(0,0,0,1)+b(0,0,2,1)+c(0,3,2,1)+d(4,3,2,1) \\
=(4d,3(c+d),2(b+c+d),a+b+c+d)=(0,0,0,0)$$
So we have $d=0$ (since $4d=0$), so that $c=0$ (since $3c+3d=3c=0$), so that $b=0$ (since $2(b+c+d)=2b=0$), and finally $a=0$ (since $a+b+c+d=a=0$).
Thus, we see that the set of vectors is independent.
A: Call the vectors $a,b,c,d$ then:
$a=[0,0,0,1]$
$\frac 12 (b-a)=[0,0,1,0]$
$\frac 13 (c-b)=[0,1,0,0]$
$\frac 14 (d-c)=[1,0,0,0]$
So your four vectors form a basis for the space, since they generate the vectors of the standard basis, and are therefore independent (and also form a spanning set).
A: Form a $4\times 4$ matrix with the four vectors, starting from the rightmost one:
$$
\begin{bmatrix}
4 & 0 & 0 & 0 \\
3 & 3 & 0 & 0 \\
2 & 2 & 2 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}
$$
This is a lower triangular matrix with no zero diagonal elements, so it's invertible and hence has rank $4$.
The row echelon form (staircase form) is really easy to find, isn't it?
Then…
