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Wikipedia gives an identity for blockwise inversion, assuming the appropriate inverses exist:

$$\begin{bmatrix}\mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \end{bmatrix}$$

Is there a corresponding general formula for the matrix of cofactors:

$$\operatorname{adj}^T\begin{bmatrix}\mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix}$$ for when $\mathbf{A}^{-1}$ or $(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})^{-1}$ do not necessarily exist?

Is there at least one for bordered matrices, i.e. $\mathbf{D} \in \mathbb{C}^{1\times 1}$?

Christian Chapman
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  • For the adjugate of a bordered matrix, see http://math.stackexchange.com/questions/1514408/matrix-determinant-lemma-with-adjugate-matrix?rq=1 . I don't know if there is a formula in the general case, though. – darij grinberg Mar 30 '17 at 20:19
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    Ah! The link I posted proves something different from what I remembered it does. Instead, take a look at Theorem 3.1.3 in V. Prasolov, Problems and Theorems in Linear Algebra. This gives a formula for the determinant of a bordered matrix. To get its adjugate, you'll have to apply this formula to each minor. (At least, most of the minors will be bordered matrices. A few will not; those you can do using Cramer's rule.) – darij grinberg Mar 30 '17 at 20:54

1 Answers1

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$\newcommand{\inv}[1]{{#1}^{-1}} \newcommand{\adj}{\operatorname{adj}} $

Let $M = \begin{pmatrix} A_{m\times m} & B_{m \times n} \\ C_{n \times m} & D_{n \times n}\end{pmatrix}$ then the following always holds $$ \begin{pmatrix} (\det{D})^{m-1} I_m & 0 \\ 0 & (\det(A))^{n-1}I_n \end{pmatrix} \adj M = \begin{pmatrix}\adj(A\det(D)-B\adj(D)C) & 0 \\ 0 & \adj(D\det(A) - C\adj(A)B)\end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\ -C\adj A & \det(A)I_n\end{pmatrix}.\tag{0}$$

Proof: Assume, initially , that $M,A,D$ are invertible. This also implies the invertibility of $F = D - C\inv{A}B$ and $G=A-B\inv{D}C.$

We have $\det(M) = \det(A) \det(F) = \det(D)\det(G).$

Under these assumptions $$M^{-1} = \begin{pmatrix} \inv{G} & 0 \\ 0 & \inv{F} \end{pmatrix}\begin{pmatrix} I_m & -B\inv{D}\\ -C\inv{A} & I_n\end{pmatrix}$$ and $\adj M = \det(M) \inv{M} = \begin{pmatrix} \det(G)\inv{G} & 0 \\ 0 & \det(F) \inv{F} \end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\det(D)\inv{D}\\ -C\det(A)\inv{A} & \det(A)I_n\end{pmatrix}.$

So, $\adj M = \begin{pmatrix} \adj{G} & 0 \\ 0 & \adj(F) \end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\ -C\adj A & \det(A)I_n\end{pmatrix}. \tag{1}$

Now $G = \dfrac{A\det(D) - B\adj(D)C}{\det(D)},$ so $ \adj(G) = \left(\dfrac{1}{\det(D)}\right)^{m-1}\adj(A\det(D)-B\adj(D)C)$ and similarly $\adj(F) = \left(\dfrac{1}{\det(A)}\right)^{n-1}\adj(D\det(A) - C\adj(A)B). \tag{2}$

Plugging (2) into (1) we get $$ \begin{pmatrix} (\det{D})^{m-1} I_m & 0 \\ 0 & (\det(A))^{n-1}I_n \end{pmatrix} \adj M = \begin{pmatrix}\adj(A\det(D)-B\adj(D)C) & 0 \\ 0 & \adj(D\det(A) - C\adj(A)B)\end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\ -C\adj A & \det(A)I_n\end{pmatrix}.$$

In the space of $n \times n$ real (or complex) matrices, the invertible matrices are dense. Given any $M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ we will find a sequence of matrices $M_n = \begin{pmatrix} A_n & B_n \\ C_n & D_n \end{pmatrix}$ such that $M_n,A_n,D_n$ are invertible and $\lim_{n\to\infty}M_n = M.$ Since (0) holds for $M_n,A_n,B_n,C_n$ and since the LHS and RHS are continuous functions we can let $n \to \infty$ and get the result for all $M$.

Arin Chaudhuri
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