# $\det(I+A)$= sum of all principal minors of $A$

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 $n=3$:

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

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

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

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

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

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

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

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

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

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

• 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. – 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.