# How to notate off-diagonal blocks for a tridiagonal block matrix

I have a block diagonal matrix that also has partitioned identity matrices in the off-diagonal blocks, like a tridiagonal block matrix. It looks something like $$\begin{bmatrix} \textbf{A} & I & \textbf{0} & \textbf{0}\\ I & \textbf{A} & I & \textbf{0} \\ \textbf{0} & I & \textbf{A} & I\\ \textbf{0} & \textbf{0} & I & \textbf{A} \end{bmatrix}$$ where both $$\textbf{A}$$ and $$I$$ are $$m\times m$$ matrices. In reality, the matrix is much bigger than what I've shown, but the pattern is the same as the one shown above. I know that I can rather elegantly write the diagonal part as $$I^{4\times 4} \otimes \textbf{A}$$ but I'd like to know if there is also a clean and easy notation to add the off diagonal matrices as well. I can't find a way outside of just constructing the matrix and showing it much like I have shown it here. I did find something on mathoverflow (here) that presents a block tridiagonal offhand, but quite frankly I don't understand how it's correct. Any insight would be appreciated.

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UPDATE: I've gotten this far with it, assuming that $$\textbf{A}$$ is tridiagonal WHERE the off diagonal elements are 1 AND the diagonal is a single value, i.e. $$A_{jj} = A_{ii}$$ for $$1\leq j < i \leq m$$, then the expression in the answer to the mathoverflow link is almost correct, except the final identity matrix being subtracted needs to be multiplied by the diagonal element of the original tridiagonal matrix. Referencing that post and slightly fixing the expression gives $$\textit{A} = J_m \otimes I_n + I_m \otimes J_n - J_{11} I_{mn}$$ So for my example the matrix can be written as $$\textbf{A}^{m\times m} \otimes I^{4\times4} + I^{m\times m} \otimes \textbf{A}^{4\times4} - \textbf{A}_{11}I^{4m\times4m}$$ Again, this only works given those particular restrictions on $$\textbf{A}$$. Sorry about the conflicting notation, I'm trying to keep my notation consistent with what I originally wrote, but also keep the notation of the mathoverflow post the same when referencing their expression. If anyone has any insight on how this can be extended to an arbitrary tridiagonal for $$\textbf{A}$$ or even just an arbitrary matrix, please let me know.

• $I^{4\times 4} \otimes \textbf{A} + \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 &1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \otimes I^{m\times m}$ Commented Nov 6, 2021 at 20:52
• Though this absolutely works, having to construct that off-diagonal matrix you're using to notate what the block tridiagonal is, is what I'm trying to avoid doing. Is there an inline way to talk about that kind of matrix of arbitrary size? Commented Nov 6, 2021 at 23:40
• Denote $D_{n}$ to be the off-diagonal matrix of size $n\times n$. Then you can write $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 &1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$ as $(D_{4} + D_{4}^T)$, where $T$ denotes transposition. Also, the powers of $D_{n}$ give you all the other levels of diagonals (with $D_{n}^{n} =0$). Commented Nov 7, 2021 at 9:07
• Just found out they're called "shift matrices". en.wikipedia.org/wiki/Shift_matrix You can use these as "building blocks". Commented Nov 7, 2021 at 9:21
• This works perfectly, I can write $I^{4\times4} \otimes \textbf{A} + \left( L_{4} + U_{4} \right) \otimes I^{m\times m}$. If you put this in all this in an answer, I'd gladly accept it. Commented Nov 7, 2021 at 18:03

Let $$L_{4}$$ and $$U_{4}$$ be the $$4\times4$$ upper and lower Shift Matrices. Then you can write the matrix as: $$I^{4\times4} \otimes \textbf{A} + \left( L_{4} + U_{4} \right) \otimes I^{m\times m}$$