# Does similarity of matrices preserve sum of principal minors?

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.

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.
• 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..? Jan 19 at 18:08