Finding Kernel of $2\times2$ matrix I seem to have forgotten how to find a kernel and it's frustrating me. I want to solve
$$
        \begin{pmatrix}
        1-\phi & 1 \\
        1 & -\phi \\
        \end{pmatrix}
 \begin{pmatrix}
        x \\
        y \\
        \end{pmatrix}
 =  \begin{pmatrix}
        0 \\
        0 \\
        \end{pmatrix}
$$
($\phi = \frac{1 + \sqrt{5}}{2}$ if that's relevant) this of course just amounts to finding the Kernel. However, I keep getting a trivial solution and I'm not sure where I'm going wrong. 
$$ \left[
    \begin{array}{cc|c}
      1-\phi&1&0\\
      1&-\phi&0
    \end{array}
\right] \implies \left[
    \begin{array}{cc|c}
      1&\frac1{1-\phi}&0\\
      1&-\phi&0
    \end{array}
\right] 
 \implies \left[
    \begin{array}{cc|c}
      1&\frac1{1-\phi}&0\\
      0&-\phi-\frac1{1-\phi}&0
    \end{array}
\right]  $$
Adding a certain multiple of the second row to the first would then result in 
$$\left[
    \begin{array}{cc|c}
      1&0&0\\
      0&1&0
    \end{array}
\right]  $$
Clearly, there is a gap in my knowledge somewhere
 A: If you are confused by the method you have been taught, why not go back to the very meaning of the equation you have? 
$$\left(\begin{matrix}1-\phi&1\\1&-\phi\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=0\\\implies\\(1-\phi)x+y=0\\x-\phi y=0$$
Then solving this system of equations by first substituting the second equation into the first, $$(1-\phi)\phi y+y=0\implies\text{we can set } y=t\text{ (*explanation below*)}\\\implies x=\phi t\text{ (by substitution into the second equation)}\tag 1$$
So the kernel is just $\left\{\left(\begin{matrix}\phi t\\t\end{matrix}\right):t\in \Bbb R\right\}$.

Edit
As mentioned in the comments, it is probably worth mentioning why $(1-\phi)\phi=-1$. You probably know this from how you obtained $\phi$, but here is another way of checking: $$(1-\phi)\phi=\left(1-\frac{1+\sqrt5}2\right)\cdot\frac{1+\sqrt5}2=-\frac{1-\sqrt5}{2}\cdot\frac{1+\sqrt5}2=-\frac{1^2-\sqrt5^2}{2^2}=\frac{-4}4=-1$$
This is what leaves LHS of equation $(1)$ as an equation which says $0=0$, and so we can choose any $y$ we want.
A: Let us recall what $\phi$ is. Namely, the positive solution to $x^2 = x + 1$ (you can verify this if you'd like). Thus we have
$$
\begin{align*}
\phi^2 &= \phi + 1 \\
\phi &= 1 + \phi^{-1} \text{ (dividing by $\phi$)} \\
\phi - 1 &= \phi^{-1} \\
\frac{1}{\phi - 1} &= \phi \\
\frac{1}{1 - \phi} &= -\phi
\end{align*}
$$
and there are a bunch more identities you can write down. What's important is to recognize that
$$ -\phi - \frac{1}{1 - \phi} = 0 \text{ and } \frac{1}{1 - \phi} = - \phi. $$
Thus in your row reduction, you have
$$ \left[
    \begin{array}{cc|c}
      1 & -\phi & 0\\
      0 & 0 & 0
    \end{array} \right] $$
and you can figure out the kernel from there.
