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Let $m\geq 0,n\geq 1$ be fixed integers. I want to compute the determinant of the $(2n+m)\times (2n+m)$ matrix given by $$a_{ij}=\left\{\begin{array}{cl}n+m & \text{if }i=j\leq 2n,\\ 2n& \text{if }i=j> 2n,\\ 0 & \text{if $i\neq j$ and $i,j\leq n$, $n< i, j\leq2n$ or $2n<i,j$,}\\ -1 &\text{otherwise.}\end{array}\right.$$ So, it's the sum of a matrix with diagonal blocks $0$ and every other block $-1$, and a diagonal matrix with the first $2n$ elements being $n+m$ and the rest $2n$.

Or just: $$\begin{bmatrix}D_n({n+m}) & -1 & -1\\ -1 & D_{n}(n+m) & -1\\ -1 & -1 & D_m({2n}) \end{bmatrix}$$

where $D_r(s)$ denotes the diagonal $r\times r$ matrix with diagonal $s$.

I'd like to use something like this, but here the blocks have different sizes and there are some rectangular ones, too.

For the context, I was trying to compute the number of spanning trees of the tripartite graph $K_{n,n,m}$ using the Matrix-Tree Theorem and ended up with that.

One detail is that in the problem I have to remove a row and a column corresponding to one vertex, which I choose to be the last one, but the resulting matrix is still with the same form. Also, for the case of $m=0$ ($m=1$ in the problem) this can be easily solved by this (or any) method since there are only 4 blocks, all of which have the same size, commute and are very simple.

Rodrigo de Azevedo
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LeviathanTheEsper
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2 Answers2

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If you're using the matrix tree theorem, then you've forgotten to delete one row and one column; the determinant of your current matrix is $0$, because the all-ones vector is in the null space.

But in general, we can compute the determinant of an $(m+n+p)\times (m+n+p)$ matrix of the form $$ \begin{bmatrix}aI_m & -1 & -1 \\ -1 & bI_n & -1 \\ -1 & -1 & cI_p\end{bmatrix} $$ by looking at the eigenvalues.

First of all, $a$, $b$, and $c$ are each eigenvalues with multiplicity $m-1,n-1,p-1$ respectively. (Or, if $a=b$ as in this case, we add the multiplicities.) To see this, just subtract the appropriate multiple of the identity matrix and look at the rank.

The remaining three eigenvalues come from the block structure. The block matrix acts on a vector of the form $(x,x,\dots,x, y,y,\dots, y,z,z,\dots,z)$ in exactly the same way that the $3\times 3$ matrix $$ \begin{bmatrix} a & -n & - p \\ -m & b & -p \\ -m & -n & c \end{bmatrix} $$ acts on the vector $(x,y,z)$. So just compute the determinant of this $3\times 3$ matrix, multiply by $a^{m-1}b^{n-1}c^{p-1}$, and you get the determinant of the original block matrix.

Misha Lavrov
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  • Thanks. Indeed I had to remove the row and column (At first I didn't think it was necessary since both matrices had the same form but they don't, and don't have the same determinant). – LeviathanTheEsper Oct 24 '18 at 18:57
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Your matrix is Laplacian. In particular, all row/column sums of your $A$ are zero. Hence $\det(A)=0$.

user1551
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  • Would the downvoter please explain the downvote? I'm usually fine if someone downvotes my answer, but in this case, I don't see why a simple and correct answer should be downvoted. – user1551 Oct 23 '18 at 15:46
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    It's a valid answer only because the question is stated imprecisely; as mentioned at the end of the question, we delete one row and column before taking the determinant, at which point the answer is no longer $0$. – Misha Lavrov Oct 24 '18 at 14:35
  • Thanks. Indeed, my matrix should have been that one but with a row and column removed. About the downvote, I don't really know. – LeviathanTheEsper Oct 24 '18 at 18:58