In my situation I have an $s\times k$ integer matrix $X$ of rank $s$ and an $k\times k$ positive diagonal matrix $D$, where $s\leq k$. Say I can bound $||X||_{\infty}$, and I know $D$ precisely. Can I bound the norm of $(XDX^{t})^{-1}$ from above? I tried looking at the determinant, which has a nice expression:
$det(XDX^{t})=\sum_{\sigma}d_{\sigma}\cdot m^{2}_{\sigma}$ where $\sigma$ is a choice of $s$ elements from $1,…,k$, $d_{\sigma}$ is the product of the appropriate elements of $D$ and $m_{\sigma}$ is the appropriate minor of $X$. This shows the determinant is not too small (?).
If anyone has any idea or any scheme to bound the inverses of such matrices, I would be very intersted and thankful!
Also, in my particular case $X=(Id_{s},X_0)$ the elements of $D$ are between greater than $1$,and $d_{1}>d_{2}>\dots>d_{k}$, if that helps.
Best Answer
I will make use of the euclidean norm for vectors and the operator norm for matrices.
Let $$X=\begin{bmatrix} I_s & X_0 \end{bmatrix} \quad D= \begin{bmatrix} D_s & 0 \\ 0 & D_r \end{bmatrix},$$ where $D_s$ denotes the $s\times s$ block of the matrix $D.$ Then $$XDX^t= D_s+X_0D_rX_0^t\ge D_s$$ Hence by the Cauchy-Schwarz inequality we have $$\|XDX^tv\|\,\|v\|\ge \langle XDX^t v,v\rangle \ge \langle D_sv,v\rangle\ge \min_{1\le j\le s} d_j\, \|v\|^2$$ Therefore for $d:=\displaystyle \min_{1\le j\le s}d_j$ we have $$\|XDX^tv\|\ge d\, \|v\|$$ The operator $XDX^t$ is invertible and $\|(XDX^t)^{-1}\|\le d^{-1}.$
Remark The entries $d_{s+1},d_{s+2},\ldots, d_k$ are not involved in the estimate.