Eigenvalues of block matrices with upper triangular blocks

This past week, Rujun Jiang and I posted a preprint on almost simultaneously diagonalizable quadratic forms (and other related ideas). At some point in the future, I will try write a short expository blog post about the main contributions of this work. Today, I’m simply going to present a neat factI don’t claim that this is new in any way and imagine this fact is known somewhere. On the other hand, I was personally unable to find a reference for it and would appreciate if anyone had pointers for me! about the eigenvalues of block matrices with upper triangular blocks that I needed/learned/proved while working on this project.


First, let’s define the sets of matrices we care about.

Definition 1. A matrix TCn×mT\in{\mathbb{C}}^{n \times m} is upper triangular if mj+i>min(n,m)    Ti,j=0.\begin{aligned} m -j + i > \min(n,m) \implies T_{i,j} = 0.\end{aligned}

Pictorially, an upper triangular matrix looks like (0000)or(0000)\begin{aligned} \begin{pmatrix} * & \cdots & *\\ & \ddots & \vdots\\ & & *\\ 0 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & 0 \end{pmatrix}\quad\text{or}\quad \begin{pmatrix} 0 & \cdots & 0 & * & \cdots & *\\ \vdots & \ddots & \vdots & & \ddots & \vdots\\ 0 & \cdots & 0 & & & * \end{pmatrix}\end{aligned} when nmn\geq m and nmn\leq m respectively.

Suppose (n1,,nk)(n_1,\dots,n_k) is such that pnp=n\sum_p n_p = n. Given TCn×nT\in{\mathbb{C}}^{n\times n}, we will let [T]p,q[T]_{p,q} denote the (p,q)(p,q)th block of TT so that [T]p,qCnp×nq[T]_{p,q} \in{\mathbb{C}}^{n_p\times n_q}.

Proposition 1. Let (n1,,nk)(n_1,\dots,n_k) such that pnp=n\sum_p n_p = n and TCn×nT\in{\mathbb{C}}^{n\times n} such that [T]p,q[T]_{p,q} is upper triangular for all p,q[k]p,q\in[k]. Then, the determinant of TT depends only on the diagonal entries of blocks [T]p,q[T]_{p,q} where np=nqn_p = n_q.

Pictorially (for the case (n1,n2,n3)=(2,2,3)(n_1,n_2,n_3) = (2,2,3)), this proposition says that the determinant of a matrix with support (@@@@@@@@@@@)\begin{aligned} \left(\begin{array} {cc|cc|ccc} @ & * & @ & * & & * & *\\ & @ & & @ & & & *\\\hline @ & * & @ & * & & * & *\\ & @ & & @ & & & *\\\hline * & * & * & * & @ & * & *\\ & * & & * & & @ & *\\ & & & & & & @ \end{array}\right)\end{aligned}

depends only on the entries in the @@ positions.

At a high level, this proposition states that the determinant (whence also the characteristic polynomial and eigenvalues) of a block matrix with upper triangular blocks depends on only a very small number of entries of the block matrix.


I will now present a really neat proof of this fact using a proof strategy that Kevin Pratt suggested.

Proof. We will show the strongerThis stronger statement in fact implies that the permanent, or in fact any generalized matrix function, of such a matrix depends only on a small number of entries. statement: For any permutation σSn\sigma \in S_n, if iTi,σ(i)\prod_i T_{i,\sigma(i)} is nonzero, then for every ii, we have that (i,σ(i))(i,\sigma(i)) is a diagonal entry of some block [T]p,q[T]_{p,q} where np=nqn_p = n_q.

First assign a weight wiw_i to each i[n]i\in[n]: Given i[n]i\in[n], let pp denote the block containing ii and let rr denote position of ii within the ppth block. Set wirnp/2.\begin{aligned} w_i \coloneqq r - n_p/2.\end{aligned} Next for each i,j[n]i,j\in[n], assign the weight wi,jwjwiw_{i,j}\coloneqq w_j - w_i.

Equivalently, if i,ji,j corresponds to p,q,r,sp,q,r,s, then set wi,j(sr)+(np+nq)/2nq.\begin{aligned} w_{i,j} \coloneqq (s- r) + (n_p + n_q)/2 - n_q.\end{aligned}

Next, as TT has upper triangular blocks, we also have that Ti,j0    (sr)+min(np,nq)nq0.\begin{aligned} T_{i,j}\neq 0 \implies (s-r) + \min(n_p,n_q) - n_q\geq 0.\end{aligned}

We deduce that if Ti,j0T_{i,j}\neq 0, then wi,j0w_{i,j}\geq 0. Furthermore, if Ti,j0T_{i,j}\neq 0 and wi,j=0w_{i,j} = 0, then it must be the case that np=nqn_p = n_q and s=rs = r.

Finally, note that for any permutation σSn\sigma\in S_n, we have iwi,σ(i)=iwσ(i)iwi=0.\begin{aligned} \sum_{i} w_{i,\sigma(i)} = \sum_i w_{\sigma(i)} - \sum_i w_i = 0.\end{aligned}

This concludes the proof as then if iTi,σ(i)\prod_i T_{i,\sigma(i)} is nonzero, it must be the case that for all i[n]i\in[n], we have Ti,σ(i)0T_{i,\sigma(i)}\neq 0 and wi,σ(i)=0w_{i,\sigma(i)} = 0. ◻


Intuitively, this proof looks at a matrix with upper triangular blocks and compares its support with some carefully constructed weighting of its entries. For (n1,n2,n3)=(2,2,3)(n_1,n_2,n_3) = (2,2,3), this looks like (),(01010.50.51.510101.50.50.501010.50.51.510101.50.50.50.51.50.51.50120.50.50.51.51011.50.51.50.5210).\begin{gathered} \left(\begin{array} {cc|cc|ccc} * & * & * & * & & * & *\\ & * & & * & & & *\\\hline * & * & * & * & & * & *\\ & * & & * & & & *\\\hline * & * & * & * & * & * & *\\ & * & & * & & * & *\\ & & & & & & * \end{array}\right),\\ \left(\begin{array} {cc|cc|ccc} 0 & 1 & 0 & 1 & -0.5 & 0.5 & 1.5\\ -1 & 0 & -1 & 0 & -1.5 & -0.5 & 0.5\\\hline 0 & 1 & 0 & 1 & -0.5 & 0.5 & 1.5\\ -1 & 0 & -1 & 0 & -1.5 & -0.5 & 0.5\\\hline 0.5 & 1.5 & 0.5 & 1.5 & 0 & 1 & 2\\ -0.5& 0.5 & -0.5 & 1.5 & -1 & 0 & 1\\ -1.5 & -0.5 & -1.5 & -0.5 & -2 & -1 & 0 \end{array}\right).\end{gathered} The proof then observes that: The support of TT corresponds exactly to the entries with nonnegative weight. Additionally, the intersection of the support of TT with the entries of weight zero is exactly the set of diagonal entries of blocks [T]p,q[T]_{p,q} where np=nqn_p = n_q.


In the setting of our paper, we have that the blocks of TT are furthermore Toeplitz, i.e., their entries are constant along each diagonal. Then using the above proposition (along with other more elementary facts), we are then able to make crazy-looking statements such as (ababcdcdeee)=σ(ababcdcdeee)=σ(abcde),\begin{aligned} \begin{pmatrix} a & * & b & * & & * & *\\ & a & & b & & & *\\ c & * & d & * & & * & *\\ & c & & d & & & *\\ * & * & * & * & e & * & *\\ & * & & * & & e & *\\ & & & & & & e\\ \end{pmatrix} &\stackrel{\sigma}{=} \begin{pmatrix} a & & b & & & & \\ & a & & b & & & \\ c & & d & & & & \\ & c & & d & & & \\ & & & & e & & \\ & & & & & e & \\ & & & & & & e\\ \end{pmatrix}\\ &\stackrel{\sigma}{=} \begin{pmatrix} a & b&\\ c & d&\\ &&e \end{pmatrix},\end{aligned} where =σ\stackrel{\sigma}{=} indicates that the matrices have the same spectrum.

Last updated Feb 28, 2021