$$ \begin{pmatrix}3&2&5\\2&-3&6\\0&0&-3\end{pmatrix} $$ It's a $3 \times 3$ matrix
Q. The imaginary parts of the eigen values of the matrix are:
1) $(0,0,0)$
2) $(0,2,-2)$
3) $(3,-3,0)$
4) $(1,-1,0)$
I tried by solving the characterstic equation but I couldn't find any number that would satisfy it. I also thinking if we could take the derivative to see how many times its cutting the $x$ axis But I'm not sure how to compute the values.
The characteristic equation is $$ 0=\det\begin{pmatrix}3-x & 2 & 5 \\ 2& -3-x & 6\\0 & 0 & -3-x\end{pmatrix}. $$ Because the last row contains two zeros, it gives you a very nice equation: $$ 0=\det\begin{pmatrix}3-x & 2\\ 2& -3-x \end{pmatrix}\cdot(-3-x). $$ You see directly that $x=-3$ is an eigenvalue.
Next, we also check the other eigenvalues, which are given by the equation \begin{align} 0=&\det\begin{pmatrix}3-x & 2\\ 2& -3-x \end{pmatrix} \\=&(3-x)(-3-x)-4 \\=& x^2-9-4\\=&x^2-13. \end{align} Hence the equation $0=x^2-13$ gives $x=\pm\sqrt{13}$.
Therefore, we get the eigenvalues $$ -3,-\sqrt{13}\text{ and }\sqrt{13}. $$
Since all eigenvalues are real, their imaginary parts are ...
The correct answer is ...