MathOverflow Asked by Laurent Lessard on November 29, 2021
Suppose we are given a block $2times 2$ matrix that is positive-definite, and let’s suppose for simplicity that the blocks along the main diagonal are the identity. So
$$
begin{bmatrix} I & X \ X^T & I end{bmatrix} > 0
$$
It turns out that any such matrix admits a factorization of the following form:
$$
begin{bmatrix}I & X \
X^T & Iend{bmatrix} =
begin{bmatrix}M_1 & N_1 & 0\
M_2 & 0 & N_2end{bmatrix}
begin{bmatrix}M_1 & N_1 & 0\
M_2 & 0 & N_2end{bmatrix}^T
$$
such that the $N_i$ are square and invertible, and the $M_i$ are tall and full-rank. Here is one possible construction. Let $X = U_1 Sigma U_2^T$ be a compact SVD, so $Sigma > 0$ is diagonal, and the $U_i$ have orthonormal columns. Let $bar U_i$ be the orthogonal completion of $U_i$, so that the matrix $begin{bmatrix}U_i & bar U_iend{bmatrix}$ is orthogonal. Now define the matrices as follows: $M_i = U_iSigma^{1/2}$, and $$N_i = begin{bmatrix} U_i(I-Sigma)^{1/2} & bar U_iend{bmatrix}$$ These matrices are always well-defined because we must have $I – XX^T > 0$, and thus $0 < Sigma < I$ in order to guarantee that the $2times 2$ block matrix is positive.
Two questions. First, does this factorization have a name? I came up with it, but I figure I’m probably not the first to do so. Second, can one generalize this factorization to the $3times 3$ case or greater. For example, how does one factor the following matrix?
$$
begin{bmatrix}I & X & Y\ X^T & I & Z \ Y^T & Z^T & Iend{bmatrix} > 0
$$
Some motivation for the factorization in the $2times 2$ case: Suppose the matrix is in fact the joint covariance matrix for two Gaussian random zero-mean vectors $w_1$ and $w_2$. Then using the factorization above, we can write:
begin{align*}
w_1 &= M_1 e_c + N_1 e_1 \
w_2 &= M_2 e_c + N_2 e_2
end{align*}
where $e_c$, $e_1$, and $e_2$ are mutually independent white Gaussian noise. We can think of $e_c$ as the "common" noise driving $w_1$ and $w_2$. It’s like an innovations decomposition, but symmetric in the sense that we are not required to choose which variable is written in terms of the other.
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