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What's a quick way to prove the following fact about adjugates of an invertible matrix $A$ and its inverse?

Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then

$$ |\det A[I,J]| = | (\det A)^{-1} \det A^{-1}[I^c,J^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$.

I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(

Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.

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Wow - I did not know that algebra proofs could have a "sexist bias". I am too curious to let it pass --- what do you mean exactly? – Federico Poloni Feb 8 '12 at 10:19
I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word "boy" has a gender neutral meaning back then? – John Jiang Feb 8 '12 at 19:39
Shouldn't the I and J be reversed in the right hand side? This is very easy to check in case of a 2x2 matrix. – Martijn Feb 20 '14 at 13:10
up vote 10 down vote accepted

The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$ M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix}, $$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.

You may verify the following identity ("magic wand Schur complement formula") $$ \begin{bmatrix}A & B\\\\ C & D\end{bmatrix} = \begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix} \begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix} \begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1} $$ By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).

Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.

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I guess for any thesis involving minors the Schur complement formula would be among the first things to try. – John Jiang Feb 8 '12 at 10:30

This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.

Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$ On the one hand, we have (Schur C.F) $$\det A=\det W\cdot\det(Z-YW^{-1}X).$$ Finally, we have $$A=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result.

These formula are obtained by factorizing $A$ into $LU$.

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Your answer is equally valid. Thanks! – John Jiang Feb 8 '12 at 10:27
Shouldn't the formula block under "Finally, we have" have the inverse of A on the left hand side? – Martijn Feb 20 '14 at 12:14

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