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Let $A$ be a $3\times 3$ matrix over reals. Then its characteristic polynomial $\det(xI-A)$ is of the form $x^3+a_2x^2+a_1x+a_0$. It is well known that $$-a_2=\mbox{trace}(A) \mbox{ and } -a_0=\det(A).$$ Note that these constants are expressed as functions of $A$ without referring to eigenvalues of $A$.

Q. What is interpretation of $a_1$ in terms of $A$ without considering its eigenvalues?


This could be trivial, but I have never seen it.

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If your matrix is: $$A=\begin{bmatrix} A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}$$ Then your $a_1$ is: $$a_1=\det\left(\begin{bmatrix} A_{22} & A_{23}\\A_{32} & A_{33}\end{bmatrix}\right)+\det\left(\begin{bmatrix} A_{11} & A_{13}\\A_{31} & A_{33}\end{bmatrix}\right)+\det\left(\begin{bmatrix} A_{11} & A_{12}\\A_{21} & A_{22}\end{bmatrix}\right)$$ And I think it's called $a_1=\text{Tr}(\text{Adj}(A))$.
Calculating it is similar to calculating the determinant: You are calculating the sub-determinant( I'm not sure about the name) for $A_{11}$, $A_{22}$ and $A_{33}$, and you are summing them.

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For invertible $A$, we have Adj$(A)=det(A)\cdot A^{-1}$ so the trace of the Adj$(A)$ is $det(A)\cdot (\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})=x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3=a_1 $ where $x_i$ are the eigenvalues. Since it holds for invertible $A$, it holds for all matrices.

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If$$A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix},$$then the answer is\begin{multline}-a_{12} a_{21} + a_{11} a_{22} - a_{13} a_{31} - a_{23} a_{32} + a_{11} a_{33} + a_{22} a_{33}=\\=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}+\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}+\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}\end{multline}

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  • $\begingroup$ Isn't it positive? $\endgroup$ – Botond Mar 6 '18 at 13:11
  • $\begingroup$ @Botond I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Mar 6 '18 at 13:13

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