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I have a large matrix in the following form: $M=\begin{bmatrix} A_{11}& {A}_{12}&...&A_{1N}\\A_{21}& {A}_{22}&...&A_{2N}\\...&...&...&...\\A_{N1}& {A}_{N2}&...&A_{NN} \end{bmatrix}$,

where $A_{ij}=A_{ji}$, and each $A_{ij}$ is a diagonal block matrix of the form: $A_{ij}=\begin{bmatrix} a_{11}& 0&...&0\\0& {a}_{22}&...&0\\...&...&...&...\\0& 0&...&a_{NN} \end{bmatrix}$.

All the diagonal entries of $A_{ij}$ are positive, and thus each $A_{ij}$ is a positive definite.

Now, I want to check if the matrix $M$ is positive definite. I think with its special form, there should be a practical way to check its definiteness. I have try the followings:

Diagonal dominant: The matrix $M$ also has an interesting characteristic. Let $m_{ij}$ be the entries of $M$. The sum of the off-diagonal entries on each row is exactly $(N-1)$ times the diagonal entries. So if $N>3$ the matrix is not diagonal dominant.

Cholesky decomposition: I have tried to check the square roots but the number of elements inside the square root increases as $i,j$ increase so it becomes impractical.

So, I want to ask if there is any efficient way to check if $M$ is positive definite. Thank you.

Cong
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1 Answers1

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I assume that $M$ is an $N \times N$ block matrix and that each block is $N \times N$, as you have indicated (but not stated explicitly) in your post.

Any matrix with diagonal blocks (assuming the blocks have the same size) can be converted to a block-diagonal matrix. In particular, suppose that $a_{ijk}$ denotes the $k$th diagonal entry of the block $A_{ij}$, so that $$ A_{ij} = \pmatrix{a_{ij1} \\ & \ddots \\ && a_{ijN}}. $$ There exists a permutation matrix $P$ such that $$ PMP^T = \pmatrix{B_1\\ & \ddots \\ && B_N}, $$ where $$ B_k = \pmatrix{ a_{11k} & \cdots & a_{1Nk}\\ \vdots & \ddots & \vdots \\ a_{N1k} & \cdots & a_{NNk}}. $$ It follows that $M$ is positive definite if and only if every $N \times N$ matrix $B_k$ is positive definite.

Where an ordinary Cholesky decomposition on the $N^2 \times N^2$ matrix would have complexity $O((N^2)^3) = O(N^6)$, attempting a separate Cholesky decomposition of each $B_k$ has complexity $N \cdot O(N^3) = O(N^4)$.

If you're interested in what the matrix $P$ looks like, it can be written as $$ P = \sum_{i,j = 1}^N (e_{i} \otimes e_j)(e_j \otimes e_i)^T $$ where $e_i$ denotes the $i$th canonical basis vector (the $i$th column of the identity matrix), and $\otimes$ denotes the Kronecker product.

Ben Grossmann
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