by Jose Fallas Rojas
Last Updated November 09, 2018 01:20 AM - source

Consider $A \in \mathcal{M}^{3{\times}3}(\mathbb{C}) $
$$ A =
\left[
\begin{matrix}
a_{1 1} & a_{12}& a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{matrix}
\right]
$$

$\bullet$ Show that $p(z)= \det(z\cdot I_{3} - A)$ is a third degree monic polynomial.

$\bullet$ 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} \ $ and $\ \det(A)=c_{1}\cdot c_{2} \cdot c_{3}$

The first part is just calculate 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.

November 09, 2018 01:19 AM

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