The determinant ${\det_n(A)}$ of an ${n \times n}$ matrix (with coefficients in an arbitrary field) obey many useful identities, starting of course with the fundamental multiplicativity ${\det_n(AB) = \det_n(A) \det_n(B)}$ for ${n \times n}$ matrices ${A,B}$. This multiplicativity can in turn be used to establish many further identities; in particular, as shown in this previous post, it implies the Schur determinant identity

$\displaystyle \det_{n+k}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \det_n(A) \det_k( D - C A^{-1} B ) \ \ \ \ \ (1)$

whenever ${A}$ is an invertible ${n \times n}$ matrix, ${B}$ is an ${n \times k}$ matrix, ${C}$ is a ${k \times n}$ matrix, and ${D}$ is a ${k \times k}$ matrix. The matrix ${D - CA^{-1} B}$ is known as the Schur complement of the block ${A}$.

I only recently discovered that this identity in turn immediately implies what I always found to be a somewhat curious identity, namely the Dodgson condensation identity (also known as the Desnanot-Jacobi identity)

$\displaystyle \det_n(M) \det_{n-2}(M^{1,n}_{1,n}) = \det_{n-1}( M^1_1 ) \det_{n-1}(M^n_n)$

$\displaystyle - \det_{n-1}(M^1_n) \det_{n-1}(M^n_1)$

for any ${n \geq 3}$ and ${n \times n}$ matrix ${M}$, where ${M^i_j}$ denotes the ${n-1 \times n-1}$ matrix formed from ${M}$ by removing the ${i^{th}}$ row and ${j^{th}}$ column, and similarly ${M^{i,i'}_{j,j'}}$ denotes the ${n-2 \times n-2}$ matrix formed from ${M}$ by removing the ${i^{th}}$ and ${(i')^{th}}$ rows and ${j^{th}}$ and ${(j')^{th}}$ columns. Thus for instance when ${n=3}$ we obtain

$\displaystyle \det_3 \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \cdot e$

$\displaystyle = \det_2 \begin{pmatrix} e & f \\ h & i \end{pmatrix} \cdot \det_2 \begin{pmatrix} a & b \\ d & e \end{pmatrix}$

$\displaystyle - \det_2 \begin{pmatrix} b & c \\ e & f \end{pmatrix} \cdot \det_2 \begin{pmatrix} d & e \\ g & h \end{pmatrix}$

for any scalars ${a,b,c,d,e,f,g,h,i}$. (Charles Dodgson, better known by his pen name Lewis Caroll, is of course also known for writing “Alice in Wonderland” and “Through the Looking Glass“.)

The derivation is not new; it is for instance noted explicitly in this paper of Brualdi and Schneider, though I do not know if this is the earliest place in the literature where it can be found. (EDIT: Apoorva Khare has pointed out to me that the original arguments of Dodgson can be interpreted as implicitly following this derivation.) I thought it is worth presenting the short derivation here, though.

Firstly, by swapping the first and ${(n-1)^{th}}$ rows, and similarly for the columns, it is easy to see that the Dodgson condensation identity is equivalent to the variant

$\displaystyle \det_n(M) \det_{n-2}(M^{n-1,n}_{n-1,n}) = \det_{n-1}( M^{n-1}_{n-1} ) \det_{n-1}(M^n_n) \ \ \ \ \ (2)$

$\displaystyle - \det_{n-1}(M^{n-1}_n) \det_{n-1}(M^n_{n-1}).$

Now write

$\displaystyle M = \begin{pmatrix} A & B_1 & B_2 \\ C_1 & d_{11} & d_{12} \\ C_2 & d_{21} & d_{22} \end{pmatrix}$

where ${A}$ is an ${n-2 \times n-2}$ matrix, ${B_1, B_2}$ are ${n-2 \times 1}$ column vectors, ${C_1, C_2}$ are ${1 \times n-2}$ row vectors, and ${d_{11}, d_{12}, d_{21}, d_{22}}$ are scalars. If ${A}$ is invertible, we may apply the Schur determinant identity repeatedly to conclude that

$\displaystyle \det_n(M) = \det_{n-2}(A) \det_2 \begin{pmatrix} d_{11} - C_1 A^{-1} B_1 & d_{12} - C_1 A^{-1} B_2 \\ d_{21} - C_2 A^{-1} B_1 & d_{22} - C_2 A^{-1} B_2 \end{pmatrix}$

$\displaystyle \det_{n-2} (M^{n-1,n}_{n-1,n}) = \det_{n-2}(A)$

$\displaystyle \det_{n-1}( M^{n-1}_{n-1} ) = \det_{n-2}(A) (d_{22} - C_2 A^{-1} B_2 )$

$\displaystyle \det_{n-1}( M^{n-1}_{n} ) = \det_{n-2}(A) (d_{21} - C_2 A^{-1} B_1 )$

$\displaystyle \det_{n-1}( M^{n}_{n-1} ) = \det_{n-2}(A) (d_{12} - C_1 A^{-1} B_2 )$

$\displaystyle \det_{n-1}( M^{n}_{n} ) = \det_{n-2}(A) (d_{11} - C_1 A^{-1} B_1 )$

and the claim (2) then follows by a brief calculation (and the explicit form ${\det_2 \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc}$ of the ${2 \times 2}$ determinant). To remove the requirement that ${A}$ be invertible, one can use a limiting argument, noting that one can work without loss of generality in an algebraically closed field, and in such a field, the set of invertible matrices is dense in the Zariski topology. (In the case when the scalars are reals or complexes, one can just use density in the ordinary topology instead if desired.)

The same argument gives the more general determinant identity of Sylvester

$\displaystyle \det_n(M) \det_{n-k}(M^S_S)^{k-1} = \det_k \left( \det_{n-k+1}(M^{S \backslash \{i\}}_{S \backslash \{j\}}) \right)_{i,j \in S}$

whenever ${n > k \geq 1}$, ${S}$ is a ${k}$-element subset of ${\{1,\dots,n\}}$, and ${M^S_{S'}}$ denotes the matrix formed from ${M}$ by removing the rows associated to ${S}$ and the columns associated to ${S'}$. (The Dodgson condensation identity is basically the ${k=2}$ case of this identity.)

A closely related proof of (2) proceeds by elementary row and column operations. Observe that if one adds some multiple of one of the first ${n-2}$ rows of ${M}$ to one of the last two rows of ${M}$, then the left and right sides of (2) do not change. If the minor ${A}$ is invertible, this allows one to reduce to the case where the components ${C_1,C_2}$ of the matrix vanish. Similarly, using elementary column operations instead of row operations we may assume that ${B_1,B_2}$ vanish. All matrices involved are now block-diagonal and the identity follows from a routine computation.

The latter approach can also prove the cute identity

$\displaystyle \det_2 \begin{pmatrix} \det_n( X_1, Y_1, A ) & \det_n( X_1, Y_2, A ) \\ \det_n(X_2, Y_1, A) & \det_n(X_2,Y_2, A) \end{pmatrix} = \det_n( X_1,X_2,A) \det_n(Y_1,Y_2,A)$

for any ${n \geq 2}$, any ${n \times 1}$ column vectors ${X_1,X_2,Y_1,Y_2}$, and any ${n \times n-2}$ matrix ${A}$, which can for instance be found in page 7 of this text of Karlin. Observe that both sides of this identity are unchanged if one adds some multiple of any column of ${A}$ to one of ${X_1,X_2,Y_1,Y_2}$; for generic ${A}$, this allows one to reduce ${X_1,X_2,Y_1,Y_2}$ to have only the first two entries allowed to be non-zero, at which point the determinants split into ${2 \times 2}$ and ${n -2 \times n-2}$ determinants and we can reduce to the ${n=2}$ case (eliminating the role of ${A}$). One can now either proceed by a direct computation, or by observing that the left-hand side is quartilinear in ${X_1,X_2,Y_1,Y_2}$ and antisymmetric in ${X_1,X_2}$ and ${Y_1,Y_2}$ which forces it to be a scalar multiple of ${\det_2(X_1,X_2) \det_2(Y_1,Y_2)}$, at which point one can test the identity at a single point (e.g. ${X_1=Y_1 = e_1}$ and ${X_2=Y_2=e_2}$ for the standard basis ${e_1,e_2}$) to conclude the argument. (One can also derive this identity from the Sylvester determinant identity but I think the calculations are a little messier if one goes by that route. Conversely, one can recover the Dodgson condensation identity from Karlin’s identity by setting ${X_1=e_1}$, ${X_2=e_2}$ (for instance) and then permuting some rows and columns.)