I am looking for an expression that gives the determinant of a matrix of the form

\begin{bmatrix} A & B & 0 & \dots & 0 & C \\ B & A & B & & 0 & 0 \\ 0 & B & A & \ddots & 0 & \vdots \\ 0 & & & \ddots & A & B \\ C & 0 & \dots & \dots & B & A \\ \end{bmatrix}

Just to clarify. The above matrix is a block tridiagonal matrix with "extra" block entries in the "corners" of the matrix. All block entries are of the same size. They are all square matrices of the same size.

My first port of call was to recursively apply the block formula given in the following link under the heading "Block Matrices".


The expression gets quite messy. I was wondering if there was a paper or well known formula among mathematicians which would make easy work of such a task.

Each of the blocks $A,B,C$ are tridiagonal, toeplitz matrices which are invertable.

EDIT: attempting to implement the suggested solution given by Federico

I can write the above matrix as $$\mathcal{A} = \bar{\mathcal{A}} + UWV^T \in \mathbb{R}^{ml \times ml},$$ where $$U = [e_l \otimes\mathbb{I}_m,e_1 \otimes\mathbb{I}_m], W = diag(C,C), V = [e_1 \otimes\mathbb{I}_m,e_l \otimes\mathbb{I}_m]$$

and $$ \bar{\mathcal{A}} = \begin{bmatrix} A & B & & & & \\ B & A & B & & & \\ & B & A & \ddots & & \\ & & & \ddots & A & B \\ & & & & B & A \\ \end{bmatrix} $$

This allows me to say

$$\det\left(\bar{\mathcal{A}} + {UWV}^T\right) = \det (W^{-1} + V^T\bar{\mathcal{A}}^{-1}U)\det{W}\det \bar{\mathcal{A}}$$

The problem for me now is to somehow find an expression for the determinant of $\bar{\mathcal{A}}$ and the determinant of $W^{-1} + V^T\bar{\mathcal{A}}^{-1}U$.

  • 5
    $\begingroup$ do you know what happens when the blocks have dimension one? $\endgroup$ – Martin Rubey Nov 5 '18 at 19:23
  • 2
    $\begingroup$ @MartinRubey In that case it's a circulant matrix plus two rank-$1$ updates, which can be handled by Sherman-Morrison and the matrix determinant lemma. To the OP, can you give us any information about the blocks? $\endgroup$ – MTyson Nov 5 '18 at 22:36
  • $\begingroup$ The blocks are invertiable, they are tridiagonal and teoeplitz. This is a lot of extra information, I should have included it! I will update the post accordingly. $\endgroup$ – ZingyMcGhee Nov 6 '18 at 9:54

Just a sketch of an idea that seems to work:

  • You can get rid of the corrections $C$ using the matrix determinant lemma (or, better, replace them with $B$, which makes the matrix block circulant).
  • Once you have made those corrections, you can change basis using the generalized Schur decomposition and reduce to a case in which $A$ and $B$ are upper triangular.
  • After you have done this reduction, you can reorder the entries to obtain a block triangular matrix, in which every diagonal block is tridiagonal Toeplitz or circulant. Those matrices have explicit formulas for their determinants (and, actually, even for their eigenvalues).
  • $\begingroup$ Thank you for your help thus far. Would it be possible to expand on the second and the third point a little? I am working through your solution and have made edit to show my progress. It is the case that if $A = QSZ^{*}$ then $S$ is diagonal if $A$ is tridiagonal? Also I would have to find the matrices $Q,S,Z$ explicitly if I was to continue with this method? ( the same goes for B) $\endgroup$ – ZingyMcGhee Nov 6 '18 at 14:21
  • $\begingroup$ @ZingyMcGhee Sorry - that was a typo; the Schur decomposition takes you to a case in which $A$ and $B$ are upper triangular, not diagonal. Editing it immediately. $\endgroup$ – Federico Poloni Nov 6 '18 at 14:58
  • $\begingroup$ Also: yes, I am afraid that to compute the determinant, you would have to find Q and Z. Many computational libraries/programming environments let you do that (Numpy/Scipy, Matlab...). Or, at least, you would have to find the diagonal entries of the two triangular factors. Not very practical with pen and paper, I am afraid. $\endgroup$ – Federico Poloni Nov 6 '18 at 15:02
  • $\begingroup$ I was looking for some expressions that I can use in a proof... Would you happen to know if expressions exist for the block entries of the inverse of a matrix of the form $\bar{\mathcal{A}}$ and the determinant of $\bar{\mathcal{A}}$? $\endgroup$ – ZingyMcGhee Nov 6 '18 at 15:13
  • $\begingroup$ For the determinant of $\bar{\mathcal{A}} $ there is formula in this paper actually arxiv.org/pdf/0712.0681.pdf $\endgroup$ – ZingyMcGhee Nov 6 '18 at 15:20

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