# Trace of $3 \times 3$ matrix A and the polynomial $p(z)=det(z\cdot I_{3} - A)$.

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!

• The title is because, first I searched for any relationship between a matrix and it's trace, and I found it has something to do with eigenvalues, but that is a topic I haven't seen in my course of algebra, so I don't know how to find them or what are they. – Jose Fallas Rojas Nov 9 '18 at 1:44
• The $c_j$'s are by definition eigenvalues of $A$. I would write something like "The trace of a $3\times 3$ matrix and the polynomial $p(z)=\det(zI_3-A)$". – user587192 Nov 9 '18 at 2:34

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.