Find the trace of a matrix without using eigenvalues.

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!

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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.

user587192
November 09, 2018 01:19 AM

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