Finding Kernel of $2\times2$ matrix

Question

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

Answer 1

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.

Answer 2

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 $\left\{\left(\begin{matrix}\phi t\\t\end{matrix}\right):t\in \Bbb R\right\}$.

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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.