Editing Cauchy–Binet formula (section)

== A simple proof ==

The following simple proof relies on two facts that can be proven in several different ways:<ref>{{cite book |last1=Tao |first1=Terence |author-link=Terence Tao |title=Topics in random matrix theory |date=2012 |publisher=American Mathematical Society |url=https://terrytao.files.wordpress.com/2011/08/matrix-book.pdf |location=Providence, RI |isbn=978-0-8218-7430-1 |doi=10.1090/gsm/132 |series=Graduate Studies in Mathematics |volume=132 |page=253}}</ref>

# For any <math>1 \leq k\leq n</math> the coefficient of <math>z^{n-k}</math> in the polynomial <math>\det(zI_n+X)</math> is the sum of the <math>k\times k</math> principal minors of <math>X</math>. 
# If <math>m \leq n</math> and <math>A</math> is an <math>m\times n</math> matrix and <math>B</math> an <math>n\times m</math> matrix, then
::: <math>\det(zI_n+BA)=z^{n-m}\det(zI_m+AB)</math>.

Now, if we compare the coefficient of <math>z^{n-m}</math> in the equation <math>\det(zI_n+BA)=z^{n-m}\det(zI_m+AB)</math>, the left hand side will give the sum of the principal minors of <math>BA</math> while the right hand side will give the constant term of <math>\det(zI_m+AB)</math>, which is simply <math>\det(AB)</math>, which is what the Cauchy–Binet formula states, i.e.

: <math>
\begin{align}
&\det(AB)= \sum_{S\in\tbinom{[n]}m} \det((BA)_{S,S})=\sum_{S\in\tbinom{[n]}m} \det(B_{S,[m]}) \det(A_{[m],S}) \\[5pt]
= {} & \sum_{S\in\tbinom{[n]}m}\det(A_{[m],S}) \det(B_{S,[m]}).
\end{align}
</math>