Determine if the set of vectors are linearly independent or linearly dependent $$u=(1,1,1,3)$$
$$v=(1,2,1,3)$$
$$w=(1,2,3,2)$$
I need help understanding the method of how to do solve this type of problem. I understand that the concept is just to find out if the constants $k_n$ in $k_1u+k_2v+k_3w=0$ all equal zero or not.  
Linear independence : this means that $k_1$, $k_2$, and $k_3$ are all equal to zero and that these are the only values that will make the overall equation equal to zero.
Linear dependence : this means that at least one of the $k$ values is not equal to zero.
This is how I tried solving this:


*

*Construct an augmented matrix $A$ out of the vectors.


$$A = \left[\begin{array}{ccc|c}1 & 1 & 1 & 0\\1 & 2 & 2 & 0\\ 1 & 1 & 3 & 0 \\ 3 & 3 & 2 & 0\end{array}\right]$$


*Use row operations as much as possible to get to row-echelon form


$$A = \left[\begin{array}{ccc|c}1 & 1 & 1 & 0\\0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$$


*We now have the following:


$$a_1+a_2+a_3=0$$
$$a_2+a_3=0$$
$$a_3=0$$
This is where I'm confused. I don't know what these $a$ values are supposed to represent. They can't be the vector values because we had those then changed them. Are they the $k$ values I'm looking for?
I would conclude that this system of equations is linearly independent but that's assuming I even understand what the $a_n$ values are...
 A: First, note that augmenting with $0$'s is superfluous, as row operations won't change a zero-column.
It would be useful to find the reduced row-echelon form of $A$. In this case
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right]
$$
This tells us that any scalars $a_1,a_2,a_3$ satisfying
$$
a_1 u+a_2 v+a_3 w=\vec 0
$$
must also satisfy
$$
a_1=a_2=a_3=0
$$
This is exactly the statement that $u,v,w$ are linearly independent!
A: To answer where those $a$'s come from, and what they represent, when you construct an augmented matrix like you did, what you really have is, 
$$[A|0] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 2 & 2 & 0 \\ 1 & 1 & 3 & 0 \\ 3 & 3 & 2 & 0 \\ \end{array} \right].$$
Which is equivalent to saying,
$$A\vec{x} = \vec{0},$$
or,
$$\left[\begin{array}{ccc} 1 & 1 & 1  \\ 1 & 2 & 2  \\ 1 & 1 & 3  \\ 3 & 3 & 2  \\ \end{array} \right] \left[\begin{array}{c} a_{1}  \\ a_{2}  \\ a_{3} \end{array} \right] = \left[\begin{array}{c} 0  \\ 0  \\ 0 \end{array} \right],  $$
which describes the system,
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{1} + 2\cdot a_{2} + 2\cdot a_{3} &= 0\\
1\cdot a_{1} + 1\cdot a_{2} + 3\cdot a_{3} &= 0\\
3\cdot a_{1} + 3\cdot a_{2} + 2\cdot a_{3} &= 0\\
\end{align}$$
But once row reduced,
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 0\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 0\cdot a_{2} + 0\cdot a_{3} &= 0.\\
\end{align}$$
Which is just 
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{3} &= 0.\\
\end{align}$$
But as Brian mentioned, continuing to RREF form we have
$$\begin{align}
1 \cdot a_{1} &= 0\\
1 \cdot a_{2} &= 0\\
1 \cdot a_{3} &= 0.
\end{align}$$
Thus the only solution to your original equation is the vector,
$$ \left[ \begin{array}{c} a_{1}  \\ a_{2}  \\ a_{3} \end{array} \right] = \left[ \begin{array}{c} 0  \\ 0  \\ 0 \end{array} \right]$$
Which is the definition of linear independence.
