Solving linear system using "reduced row echelon form" when augmented columns is "zero" I just learnt about Reduced Row Echelon Form and solved few rather large matrices. However I have no idea how to approach this rather simple system:
$$x+2y=0\\
3x+6y=0$$
Here is my 2x2 augmented matrix:
$$ A = \left[
      \begin{array}{cc|c}
        1&2&0\\
        3&6&0
      \end{array}
    \right]$$
Then I calculated Reduced Row Echelon Form by formula $R2 = 3R1 - R2$:
$$ RREF(A) = \left[
      \begin{array}{cc|c}
        1&2&0\\
        0&0&0
      \end{array}
    \right]$$
Now what? I can guess that I have infinite answers because I have zero at last column. However I am not sure about it.
P.S: I am actually trying to find eigenvector of $A=\begin{bmatrix}3&2\\3&8 \end{bmatrix}$. I found $\lambda = 2,9$. Here is my solution to find eigenvector for $\lambda=2$:
$$    \begin{align*}
        (A - \lambda I)\vec{V} &= 0\\
        (\begin{bmatrix}3&2\\3&8\end{bmatrix} - \begin{bmatrix}2&0\\0&2\end{bmatrix})\vec{V} &= 0\\
        \begin{bmatrix}1&2\\3&6\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\
        \begin{bmatrix}1x+2y\\3x+6y\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\
    \end{align*}
$$
Does this means that my matrix have infinite number of eigenvectors?
 A: First thing: you do not have infinite answers: this means $x=\infty$, $y=\infty$ which is nonsense.  What you have is an infinite number of answers: it is the number of answers that is infinite, not the answers themselves.  (Analogy: "I have a large number of friends on Facebook" is not the same as "I have large friends on Facebook".)
Second thing: whether or not there are an infinite number of answers has nothing to do with the right hand side.  It is because the left hand side includes a non-leading (non-pivot) column.  So you would let the variable corresponding to this column be a parameter, and then solve for the rest in terms of this parameter.
If the zeros on the RHS worry you, start off by solving the following system:
$$\left\{\eqalign{x+2y&=3\cr 3x+6y&=9\ .\cr}\right.$$
Then go back to your question - it is really exactly the same.
See if you can finish the problem.
Re: your PS.  If the matrix $A$ has an eigenvalue $\lambda$, then there are always an infinite number of eigenvectors corresponding to $\lambda$.
A: It is correct that you have an infinite number of solutions. The reason for this is because the equations are multiples of each other. E2 = 3E1
$$3(x+2y=0)$$
$$3x+6y=0$$
This is true for all systems of equations that are multiples of each other.
For example:
$$2x+4y=8$$
$$6x+12y=24$$
will also have an infinite number of solutions because they're multiples of each other.
A: it means that you have only one independent variable and $t(2,-1)$ where $t\in \mathbb R$ is your solution space.
