# Characteristic polynomial of the block matrix

I wonder what is the characteristic polynomial of $$\begin{bmatrix} 0 & A_1 & 0 & ... & 0\\ 0 & 0 & A_2 & ... & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & ... & A_{m-1}\\ A_m & 0 & 0 & ... & 0\\ \end{bmatrix}$$ where $$A_1,..., A_m$$ are square matrices with different sizes. The regular block matrix determinant formula does not apply here since they require diagonal blocks to be square, but here it is not. I would appreciate it so much if anyone could help.

I think imposing different sizes is too general, you (probably) won't find a generic formula.

If we set $$m$$ the size of each matrix (provided they are the same), by redefining a sort of linear algebra on non commutative rings (study of $$\mathcal{M}_m(k)$$-modules) you can get $$\chi_M = \det \left(X^n I_m + (-1)^{m(n-1) + n} A_n A_1 \cdots A_{n-1}\right) = \det \left(X^n I_m - (-1)^{(m+1)(n-1)} A_n A_1 \cdots A_{n-1}\right)$$ Which gives us $$\chi_M = \chi_{(-1)^{(m+1)(n-1)}A_n A_1 \cdots A_{n-1}} (X^n) = \chi_{(-1)^{(m+1)(n-1)}A_1 \cdots A_n} (X^n)$$ (just do as if the block matrix were coefficient (non commutative) and replace $$-1$$ by $$(-1)^m$$ in each of your operation, then apply set again, this time the normal way. I can't do much better (maybe check my result with two matrices).

Otherwise I have a way to calculate the determinant without the technique above

The way to do it is to exchange rows or columns, for this, I denote: $$f(A_n,A_1,\dots,A_{n-1}) := \det \begin{pmatrix} &A_1 \\ &&\ddots\\ &&&A_{n-1}\\ A_n \end{pmatrix}$$ with $$m$$ the size of each square matrix and $$mn$$ the total size of this block matrix.

By inverting the successively rows $$1$$ and $$mn-m+1$$, $$2$$ and $$mn - n+2$$, $$\dots$$, $$m$$ and $$mn$$ we get: $$f(A_n,A_1,\dots,A_{n-1}) := (-1)^m \det \begin{pmatrix} A_n \\ &&A_2\\ &&& \ddots\\ &&&&A_{n-1}\\ &A_1 \end{pmatrix}\\ = (-1)^m \det (A_n) \det \begin{pmatrix} &A_2 \\ && \ddots\\ &&&A_{n-1}\\ A_1 \end{pmatrix} = (-1)^m \det(A_n) f(A_1,A_2,\dots ,A_{n-1}).$$ By iterating we have $$f(A_{n}, A_1,\dots , A_{n-1}) = (-1)^{km}\det(A_n) \det(A_1) \dots \det(A_{k-1}) f(A_k,\dots,A_{n-1}) \\ =(-1)^{(n-1)m}\det(A_n) \det(A_1) \dots \det(A_{n-2}) f(A_{n-1}) \\= (-1)^{(n-1)m} \det(A_1) \dots \det(A_n).$$

The problem may have nice a solution but I can't give much more, it's very tricky. You can always try to study its eigenvalues and think of it as a linear morphism instead of a matrix?

I (very) probably have made some sign errors but it is not relevant for the study of the spectrum...

I would add that it is no surprise to find the product of the $$A_i$$'s in this one, If you divide your vector space in $$V_1 \oplus \cdots \oplus V_n$$ with $$V_i$$ of dimension $$i$$, you can see that the $$A_i$$ induce linear functions $$f_i: V_{i+1} \longrightarrow V_i$$ with $$V_{n+1} = V_1$$. It looks like a cyclic "base", if you have seen Frobenius decomposition you are probably familiar with it, so if you take $$x \in V_2$$ you will have to apply $$A$$ $$n$$ times to it to get back to $$V_2$$, and it will be like you have applied $$f_1$$, $$f_2$$ etc.

• If you want details for the first part, just ask! Commented Jun 14, 2023 at 8:45
• Thank you so much for your reply! But in my research I have to consider different size of blocks, since I'm considering, from structural perspective, the relationship between perron root of a nonnegative irreducible matrix and its primitive components (i.e., cycles). It is not practical to make those cycles have the same shape... Commented Jun 14, 2023 at 14:54
• Haha good luck!! Commented Jun 14, 2023 at 14:54
• Thank you anyway, Jules! Commented Jun 14, 2023 at 15:10