I'm having a hard time proving or finding a proof for the following result. It should follow from an application of the Laplace expansion.

Let $n\in\mathbb{N}$, $[n]=\{1,\dots,n\}$, and $A\in\mathbb{R}^{n\times n}$. Then, $$\det\left( I_{n}+A\right) =\sum\limits_{G\subseteq\left[ n\right] } \det\left( A_{G}\right).$$ where $A_{G}$ is the matrix $A$ with all columns and rows not in $G$ removed.

For instance, here is the proof for $n=3$:

$\det\left( I_{n}+A\right) =\det\left( \begin{array} [c]{ccc} 1+a_{1,1} & a_{1,2} & a_{1,3}\\ a_{2,1} & 1+a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & 1+a_{3,3} \end{array} \right) $

$\;\;\;\;=\left( 1+a_{1,1}\right) \left( 1+a_{2,2}\right) \left( 1+a_{3,3} \right) +a_{1,2}a_{2,3}a_{3,1}+a_{1,3}a_{2,1}a_{3,2}$

$\;\;\;\;\;\;\;\; -\left( 1+a_{1,1}\right) a_{2,3}a_{3,2}-a_{1,2}a_{2,1}\left( 1+a_{3,3}\right) -a_{1,3}\left( 1+a_{2,2}\right) a_{3,1}$


$\;\;\;\;\;\;\;\; -a_{2,3}a_{3,2}+a_{1,1}a_{2,2}a_{3,3}-a_{1,1}a_{2,3}a_{3,2}-a_{1,2} a_{2,1}a_{3,3}$

$\;\;\;\;\;\;\;\; +a_{1,2}a_{3,1}a_{2,3}+a_{2,1}a_{1,3}a_{3,2}-a_{1,3}a_{2,2}a_{3,1} +a_{1,1}+a_{2,2}+a_{3,3}+1$

$\;\;\;\;=\underbrace{a_{1,1}a_{2,2}a_{3,3}+a_{1,2}a_{2,3}a_{3,1}+a_{1,3} a_{2,1}a_{3,2}-a_{1,1}a_{2,3}a_{3,2}-a_{1,2}a_{2,1}a_{3,3}-a_{1,3} a_{2,2}a_{3,1}}_{=\det A=\det\left( A_{\left\{ 1,2,3\right\} }\right) }$

$\;\;\;\;\;\;\;\;+\underbrace{a_{1,1}a_{2,2}-a_{1,2}a_{2,1}}_{=\det\left( A_{\left\{ 1,2\right\} }\right) }+\underbrace{a_{1,1} a_{3,3}-a_{1,3}a_{3,1}}_{=\det\left( A_{\left\{ 1,3\right\} }\right) }+\underbrace{a_{2,2}a_{3,3}-a_{2,3}a_{3,2}}_{=\det\left( A_{\left\{ 2,3\right\} }\right) }$

$\;\;\;\;\;\;\;\;+\underbrace{a_{1,1}} _{=\det\left( A_{\left\{ 1\right\} }\right) }+\underbrace{a_{2,2}}_{=\det\left( A_{\left\{ 2\right\} }\right) }+\underbrace{a_{3,3}}_{=\det\left( A_{\left\{ 3\right\} }\right) }+\underbrace{1}_{=\det\left( A_{\varnothing}\right) }$

$\;\;\;\;=\sum\limits_{G\subseteq\left[ n\right] }\det\left( A_{G}\right) $.

  • $\begingroup$ Do you know the formula for the coefficients of the characteristic polynomial in terms of principal minors? If so, you can just set $\lambda=-1$ in that formula. $\endgroup$ – Hans Lundmark May 3 '16 at 15:27

You should recall the formula $$ \chi_A(X) \overset{def}= \det(X\, \mathrm{id}_n - A) = \sum_{i=0}^n [(-1)^k \mathrm{tr}(\Lambda^k A)] X^{n-k}, $$ which, if you have never seen it proved, I have written down here (Your question holds for a general commutative unital ring $R$, but if you only care about a vector space over a field or the real numbers, feel comfortable to assume $R$ is that field of the real numbers.) Then it is an exercise to show that $$ \mathrm{tr}(\Lambda^k A) = \underset{|G|=k}{\sum_{G \subseteq [n]}} \det(A_G). $$ If you plug in $X=-1$, the left hand side becomes $(-1)^n \det(I_n+A)$ and the right-hand side becomes $(-1)^n \sum_{G \subseteq [n]} \det(A_G)$.

Hint for the exercise : the map $\Lambda^k A : \Lambda^k R^n \to \Lambda^k R^n$ is defined by $v_1 \wedge \cdots \wedge v_k \mapsto A v_1 \wedge \cdots \wedge A v_k$. Express it in coordinates over the standard basis (i.e. the one in which $A$ is written with coefficients) and deduce that the expression of $A$ in coordinates uses the minors of $A$ as its coefficients. Then sum up the diagonal terms.

Hope that helps,


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.