How would one go about doing this?
I tried creating a general matrix in the following way:
$$ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} $$
Now from here, my reasoning was to just try and find a general expression for the characteristic polynomial which is $det(A-\lambda I)$ but that turned out to be a dead end (at least for me) because I was not able to factor anything out and thus make it an eigenvalue that works regardless of the numbers in the matrix.
Any help on this would be very much appreciated.
Let $\phi(t) = \det (tI -A) = t^3 \det(I - {1 \over t} A)$.
Note that $\phi$ is a polynomial in $t$ and so is continuous.
Hence $\lim_{t \to -\infty} \phi(t)= -\infty$, $\lim_{t \to \infty} \phi(t)= \infty$ and so $\phi$ must take the value $0$ somewhere.
Alternatively:
Note that $\phi$ has 3 roots.
If $z \in \mathbb{C}$ then $\phi(\overline{z}) = \overline{\phi(z)}$, hence if $z$ is a root then $\overline{z}$ is also a root. In particular, if there is a non real root, then its conjugate is also another root. Hence the number of non real roots is even, and hence we can have either $0$ or $2$ non real roots. And so this means we have $1$ or $3$ real roots.