Given the matrix $$A = \left(\begin{array}{ccc} 1&-0.85&0\\ 1.7&-1&0\\ 0 & 0.85 & 4 \end{array}\right)\in\mathbb{C}^{3\times3}$$
I am now looking for the eigenvalues of $A$. Calculating the chararacteristical polynom via $\operatorname{det}\left(A-\lambda\mathcal{I}_3\right)$ it gives me $$ -\lambda^3+4\lambda^2 -0.445\lambda+1.78 \stackrel{!}{=}0$$
One solution is trivial ($\lambda=4$), so because of $$-\lambda^3+4\lambda^2 -0.445\lambda+1.78 = (x-4)\cdot \left(-x^2-0.445\right)$$
Thus I just have to solve $\left(-x^2-0.445\right)=0$ which is of course $$\lambda = \pm i\cdot\sqrt{0.445} \approx \pm i\cdot 0.667$$ In particular: The real part of those $\lambda$ is exactly $0$
Are my results correct? I am asking, because Wolfram Alpha gives me $$\lambda \approx -5.55\times10^{-17}\pm i\cdot 0.667$$ In particular: The real part of $\lambda$ is not equal to zero, but very small.
So, is this caused by rounding errors of wolfram alphas calculation only? But where could this real part come from? Or am I missing something in my solution that is causing a small real part of $\lambda$?
Thanks in advance
Your results are correct. The issue is that Wolfram Alpha will use inaccurate arithmetic if you specify numbers in the dotted decimal format $x.y$, like you do with the entry $0.85$.
If you instead write $85/100$ you will get the result you want.