Finding complex eigenvalues

Question

For the matrix \begin{pmatrix}1/2 & 1 & 3/4\\2/3 & 0 & 0\\0 & 1/3 & 0\end{pmatrix} Find the eigenvalues and corresponding eigenvectors. I did this with an online calculator and got that there were $2$ complex eigenvalues, but I am struggling to do the work by hand. I know that I have to get it to the characteristic polynomial with the determinant of $A -\lambda \cdot I$ but after that I am lost.

Answer 1

If $A=\begin{pmatrix}1/2 & 1 & 3/4\\2/3 & 0 & 0\\0 & 1/3 & 0\end{pmatrix}$, $|A-\lambda I|=det \begin {pmatrix}1/2-\lambda & 1 & 3/4\\2/3 & -\lambda & 0\\0 & 1/3 & -\lambda\end{pmatrix}=\frac{3}{4}\times\frac{2}{9}-0-\lambda((\frac{1}{2}-\lambda)(-\lambda)-\frac{2}{3})$.

Hence we have to solve, $-\lambda^3+\frac{1}{2}\lambda^2+\frac{2}{3}\lambda+\frac{1}{6}=0$.

Answer 2

let me consider the matrix $$A=\pmatrix{6&12&9\\8&0&0\\0&4&0}.$$ the eigenvalues are given by $$0=det\pmatrix{6-x&12&9\\8&-x&0\\0&4&-x}=-x\left(-x(6-x)-96)\right)-4(-72)=-x^3+6x^2+96x+288$$ has a real root at $x = 14.193354$ so the original matrix has an eigevvalue $1.18277$; not a nice number.