Consider $A \in \mathcal{M}^{3{\times}3}(\mathbb{C}) $ where $$ A = \left[ \begin{matrix} a_{1 1} & a_{12}& a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{matrix} \right] $$
- Show that $p(z)= \det(z\cdot I_{3} - A)$ is a third degree monic polynomial.
- If $p(z)= (z- c_{1})(z-c_{2})(z-c_{3})$ where $c_{j} \in \mathbb{C}$, then:
$$ \operatorname{trace}(A) = c_{1} + c_{2} + c_{3},\quad \text{and}\quad \det(A)=c_{1}\cdot c_{2} \cdot c_{3} $$
The first part is just calculating the determinant, so I've done it.
To prove that $\det(A)=c_{1}\cdot c_{2} \cdot c_{3}$ I have to just plug in $z=0$ in the first part and computing $\det(-A)$. I see that $$ \det(-A)= -\det(A)$$ which is exactly what I need, because $$ p(0)=-c_{1}\cdot c_{2} \cdot c_{3},$$ hence $$ -\det(A)= -c_{1}\cdot c_{2} \cdot c_{3}. $$
I'm having trouble proving that $$\operatorname{trace}(A) = c_{1} + c_{2} + c_{3}.$$
So can anyone help me? Thanks!
If you write out explicitly $b_0,b_2,b_3$ in $$ p(z):=\det(zI_3-A) =b_3z^3+b_2z^2+b_1z+b_0 $$ then part II simply follows from Vieta's formula. In your post, you have actually figured out what are $b_0$ and $b_3$. All you need now is finding out $b_2$ which can be done by observation of the determinant of the $3\times 3$ matrix.