# General inversion for complex block Toeplitz matrices

I have been looking at inversion methods of block Toeplitz matrices, and found the paper by Akaike for real block Toeplitz matrices. Is there any good reference to look at inversion of complex-values, block Toeplitz matrices? In particular, I would be interested to know the case of the following: $$\begin{eqnarray} A = \left(\begin{matrix} a & 0 & b\\ b & a & 0 \\ 0 & b &a \end{matrix}\right) \end{eqnarray}$$ where $$a$$ are $$n\times n$$ complex matrices and $$b$$ is real, and I want to know $$A^{-1}$$.

EDIT: A more general case, which is related to the above question. Suppose we are given the following $$(N\times m)\times(N\times m)$$ block matrix, where each block with dimension $$m\times m$$ so that: $$\begin{eqnarray} A = \left(\begin{matrix} a & 0 & ... & 0\\ b & a & ... &0 \\ \vdots & b &\ddots &\vdots\\ 0 & ...&... & a \end{matrix}\right) \end{eqnarray}$$ that is, we can write $$A=aI + bP$$, where $$P$$ is the lower-diagonal ladder operator (that is, with $$I_{m\times m}$$ matrices along the lower-diagonal); so the matrix above is a block, lower-bidiagonal matrix. Then we can look for the inverse as: $$\begin{eqnarray} (aI +bP)(c_{0}I+\sum_{k=1}^{Q}c_{k}P^{k})=I \end{eqnarray}$$ How do we determine the coefficients $$c_{k}$$ is this general case?

• This is a circulant matrix, and there is a nice general formula for its inverse (which is also circulant). math.stackexchange.com/questions/1563907/… Apr 8, 2021 at 13:36
• @Z See also the wiki page on circulant matrices Apr 8, 2021 at 13:44
• Thanks for your comments and links Apr 8, 2021 at 13:47
• I don't think this is useful enough to be an answer, but I think it's interesting. Your matrix can be written in the form $A = a I + b P$, where $I$ is the identity matrix and $$P = \pmatrix{0&0&1\\1&0&0\\0&1&0}.$$ Because $A = f(P)$ for some polynomial $f$, it must be the case that $A^{-1} = g(P)$ for some other polynomial $g$. Moreover, by the Cayley Hamilton theorem, every matrix of the form $g(P)$ can be written in the form $c_0 I + c_1 P + c_2 P^2$. In particular, the matrix $P$ satisfies $P^3 = I$. Apr 8, 2021 at 14:00
• With that established, finding the inverse amounts to solving the system $$(aI + bP)(c_0 I + c_1 P + c_2 P^2) = I \implies\\ ac_0 I + (bc_0 + ac_1)P + (bc_1 + ac_2)P^2 + bc_2 P^3 = I \implies\\ (ac_0 + bc_2)I + (bc_0 + ac_1)P + (bc_1 + ac_2)P^2 = 1 \cdot I + 0 \cdot P + 0 \cdot P^2 \implies\\ \pmatrix{a & 0 & b\\b&a&0\\0&b&a}\pmatrix{c_0\\c_1\\c_2} = \pmatrix{1\\0\\0}$$ In other words: if $(c_0,c_1,c_2)$ is the first column of $A^{-1}$, then $A^{-1} = c_0 I + c_1 P + c_2 P^2$. Apr 8, 2021 at 14:00

As I clarify below (and in a comment on your question), it is sufficient to solve the system $$A \pmatrix{c_1\\c_2\\c_3} = \pmatrix{1\\0\\0}.$$ For your particular matrix, one nice approach to answering this question is to use Cramer's rule. We have $$c_i = \frac{\det(A_{i})}{\det(A)},$$ where $$A_i$$ is the matrix obtained by replacing the $$i$$th column with $$(1,0,0)$$. In fact, each $$\det(A_i)$$ is a cofactor corresponding to an entry of the first row, i.e. it is a $$2 \times 2$$ determinant, and because of the zero-entries of $$A$$ the $$3 \times 3$$ determinant can be quickly computed in this case using the rule of Sarrus. We find that $$\det(A) = a^3 + b^3, \\ \det(A_1) = a^2, \quad \det(A_2) = -ab, \quad \det(A_3) = b^2.$$ Putting it all together, we find that $$A^{-1} = \frac{a^2}{a^3 + b^3} I - \frac{ab}{a^3 + b^3} P + \frac{b^2}{a^3 + b^3}P^2 \\ =\frac 1{a^3 + b^3}\pmatrix{a^2 & b^2 & -ab\\ -ab & a^2 & b^2\\ b^2 & -ab & a^2}.$$
Your matrix can be written in the form $$A = a I + b P$$, where $$I$$ is the identity matrix and $$P = \pmatrix{0&0&1\\1&0&0\\0&1&0}.$$ Because $$A = f(P)$$ for some polynomial $$f$$, it must be the case that $$A^{-1} = g(P)$$ for some other polynomial $$g$$. Moreover, by the Cayley Hamilton theorem, every matrix of the form $$g(P)$$ can be written in the form $$c_0 I + c_1 P + c_2 P^2$$. In particular, the matrix $$P$$ satisfies $$P^3 = I$$.
With that established, finding the inverse amounts to solving the system $$(aI + bP)(c_0 I + c_1 P + c_2 P^2) = I \implies\\ ac_0 I + (bc_0 + ac_1)P + (bc_1 + ac_2)P^2 + bc_2 P^3 = I \implies\\ (ac_0 + bc_2)I + (bc_0 + ac_1)P + (bc_1 + ac_2)P^2 = 1 \cdot I + 0 \cdot P + 0 \cdot P^2 \implies\\ \pmatrix{a & 0 & b\\b&a&0\\0&b&a}\pmatrix{c_0\\c_1\\c_2} = \pmatrix{1\\0\\0}$$ In other words: if $$(c_0,c_1,c_2)$$ is the first column of $$A^{-1}$$, then $$A^{-1} = c_0 I + c_1 P + c_2 P^2$$.