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I am new to linear algebra; someone on Quora posted this "shortcut" method to find the characteristic equation of a $3\times 3$ matrix. Though they demonstrated it through an example, here's what it might look like more generally.

The characteristic equation of a matrix $A=(a_{ij})_{3\times 3}=\begin{pmatrix}a_{11}&a_{12}& a_{13}\\a_{21}&a_{22}& a_{23}\\ a_{31} & a_{32} & a_{33}\end{pmatrix}\tag*{}$ is given by: $$\lambda^3-\DeclareMathOperator{\Tr}{Tr}\Tr_1(A)\cdot\lambda^2+\Tr_2(A)\cdot\lambda-\det(A)=0$$ where

  • $\Tr_1(A)$ is the regular trace i.e., sum of elements along the diagonal of $A$. $\Tr_1(A)= a_{11}+a_{22}+a_{33}\tag*{}$

  • $\Tr_2(A)$ is the sum of principal minors of $A$ i.e., sum of minors along the diagonal elements. $\Tr_2(A)=\begin{vmatrix}a_{22}& a_{23}\\ a_{32} & a_{33}\end{vmatrix}+\begin{vmatrix}a_{11}& a_{13}\\ a_{31} & a_{33}\end{vmatrix}+\begin{vmatrix}a_{11}& a_{12}\\ a_{21} & a_{22}\end{vmatrix}\tag*{}$

Is this true? If it is so, it would mean that similar matrices have same $\Tr_2$, isn't it?

I have learned how to prove that similarity preserves trace and determinant using Vieta's formula and expansion of $|A-\lambda I|$. Can this fact about $\Tr_2$ also be proven in this way?

I don't know if this is a common knowledge or if there's more general form of this result, however, as a newbie, I find this result (if true) amazing.

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Yes. For an $n\times n$ matrix $A$, let $A^{(k)}$ denote the $k$-th compound power of $A$ consisting of the $k\times k$ minor determinants of $A$. This matrix is $\binom{n}{k}\times\binom{n}{k}$ with entries indexed by sequences $\mu,\nu$ indicating the rows and columns of a minor. Then $\mathop{\mathrm{Tr}}_k(A)=\mathop{\mathrm{tr}}(A^{(k)})$, so in particular $\mathop{\mathrm{Tr}}_1(A)=\mathop{\mathrm{tr}}(A)$ and $\mathop{\mathrm{Tr}}_n(A)=\det(A)$.

By the Cauchy-Binet theorem, for any $n\times n$ matrix $B$ we have $(AB)^{(k)}=A^{(k)}B^{(k)}$, and it's easy to see that $I^{(k)}=I$ where $I$ denotes the identity matrices of appropriate dimensions. So by the cyclic property of the trace, if $B$ is invertible we have $$\textstyle\mathop{\mathrm{Tr}}_k(BAB^{-1})=\mathop{\mathrm{tr}}\bigl[(BAB^{-1})^{(k)}\bigr]=\mathop{\mathrm{tr}}\bigl[B^{(k)}A^{(k)}(B^{-1})^{(k)}\bigr]=\mathop{\mathrm{tr}}(A^{(k)})=\mathop{\mathrm{Tr}}_k(A)$$

Since (up to sign) these are the characteristic coefficients of $A$, this aligns with the fact that similar matrices have the same characteristic polynomial.

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  • $\begingroup$ I wonder, is this how computer algebra systems compute eigenvalues i.e., by computing coefficients of the characteristic equation or is just expanding $\det(A-xI)$ more efficient..? $\endgroup$ Jan 19 at 18:08
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    $\begingroup$ @Nothingspecial I'm not an expert on CAS but I highly doubt they'd use either of those methods due to inefficiency and/or instability of computing determinants. That said, there are algorithms like Leverrier's which use a relationship between the traces of compound and composite powers to compute characteristic coefficients. $\endgroup$
    – blargoner
    Jan 19 at 18:45

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