Convergence of a symmetric Matrix

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I am kind of stuck trying to write down a mathematical proof. Since i am a Computer Sience person, I know some maths terms more and some less. There are terms which i don't know about and thus can't form the equation the way i want in order to have the proof that i want.

Yes this is homework, but by writing it i hope to get more insight and hopefully answers that can help me solve it.

And here comes the proof question:

Given Matrix $M = \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}$ and vector $x_{0}=\begin{pmatrix} a \\ b \end{pmatrix}$. Show that the sequence of numbers $(x_{k} |k \geq 0)$ with $x_{0}=\begin{pmatrix} a \\ b \end{pmatrix}$ (where $b \neq 0$) does not converge.

What i have done so far:

I have written out the sequence of numbers.

$(x_{0}\,\,\, x_{1}\,\,\, x_{2}\,\,\, x_{3})$

Furthermore one can write this as:

$\left(x_{0}\,\,\,\frac{K*x_{0}}{K*x_{0}}\,\,\,\frac{K*\frac{K*x_{0}}{||K*x_{0}||}}{||K*\frac{K*x_{0}}{||K*x_{0}||}||}\,\,\,\frac{K*\frac{K*\frac{K*x_{0}}{||K*x_{0}||}}{||K*\frac{K*x_{0}}{||K*x_{0}||}||}}{K*\frac{K*\frac{K*x_{0}}{||K*x_{0}||}}{||K*\frac{K*x_{0}}{||K*x_{0}||}||}}\,\,\, ...\right)$

I do understand that this is some kind of recusrion happening here. As a next step i tried to factor out the $K$'s in each element of the sequence, so the sequence becomes:

$\left(x_{0}\,\,\,\frac{K*x_{0}}{K*x_{0}}\,\,\,\frac{K^{2}*\frac{x_{0}}{||K*x_{0}||}}{||K^{2}*\frac{x_{0}}{||K*x_{0}||}||}\,\,\,\frac{K^{3}*\frac{\frac{x_{0}}{||K*x_{0}||}}{||K^{2}*\frac{x_{0}}{||K*x_{0}||}||}}{K^{3}*\frac{\frac{x_{0}}{||K*x_{0}||}}{||K^{2}*\frac{x_{0}}{||K*x_{0}||}||}}\,\,\,\right)$

But now i am stuck because i dont know how to simplify the elements in the sequence further and end up with something that proofs the statement above.

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Presumably $x_k=Mx_{k-1}$. If this is the case, then you are overthinking. It's easy to see that $x_k=\pmatrix{a\\ (-1)^kb}$. Therefore, if $b\ne0$, the second entry of $x_k$ keeps oscillating between $b$ and $-b$ and it doesn't converge as $k\to\infty$.