# Lipschitz constant of a function of matrix

The function is given by $$f(X) = (AX^{-1}A^\top + B)^{-1}$$ where $$X$$, $$A$$, and $$B$$ are $$n \times n$$ positive definite matrices.

I'm trying to find the Lipschitz constant such that $$\| f(X)-f(Y) \| \leq L \|X-Y\|$$ where $$X \geq 0$$ and $$Y \geq 0$$. Motivated by Lemma 3.1 in Nonlinear Systems(H. Khalil, 3rd Ed.), I tried to find the derivative of $$f(X)$$ (i.e. $$\| \frac{ \partial f(X)}{\partial X} \|$$) but it's not easy to find the derivative of a function of a matrix over a matrix.

How can I find the Lipschitz constant? or Please let me know if there exists the way to calculate the derivative of a function of a matrix.

• I understand that $A$ (and then the others matrices as well) is not assumed to be symmetric. Then what do you mean here by "$A$ is positive definite matrix" (and $X\ge0$) ? Is it $v^\top A v >0$ for nonzero vectors $v$? Or $A$ has positive real eigenvalues? Commented Jun 18, 2019 at 11:52
• it's not easy to find the derivative of a function of a matrix over a matrix Actually it's quite easy. All you need here is the formula $(X+H)^{-1} = X^{-1} - X^{-1}HX^{-1} + O(\|H\|^2)$, which follows from the Neumann series. Applying it twice you should get your derivative. Commented Jun 18, 2019 at 13:15

I assume $$X\ge0$$ means $$u^\top X u\ge0$$, and that $$B$$ is definite positive $$\inf_{\|u\|=1} u^\top B\, u:=\beta>0.$$ I also assume matrix norms are the Euclidean operator norms.

Compute the differential by the chain rule, as suggested in comments by F.Poloni: $$Df(X)H=(AX^{-1}A^{\top}+B)^{-1}AX^{-1}\cdot H\cdot X^{-1}A^{\top}(AX^{-1}A^{\top}+B)^{-1}$$ $$=(A^{\top}+XA^{-1}B)^{-1}\cdot H \cdot(A+BA^{-\top}X)^{-1}=$$ $$=\big(B^\top+Y\big)^{-1}B^\top A^{-\top}\cdot H\cdot A^{-1}B^\top\big(B^\top+Z)^{-1},$$

where $$Y:=B^{\top}A^{-\top} X A^{-1}B\ge0$$ and $$Z:=BA^{-\top}XA^{-1}B^\top\ge0$$, conjugated to $$X\ge0$$. Thus for any unit norm vector $$u\in\mathbb{R}^n$$ $$\big\|\big(B^\top+Y\big)u\big\|\ge u^\top\big(B^\top+Y\big)u\ge u^\top B u \ge\beta$$ and $$\big\|\big(B^\top+Z)u\big\|\ge u^\top\big(B^\top+Z)u \ge u^\top B u \ge\beta.$$ Hence $$\big\|\big(B^\top+Y\big)^{-1}\big\|\le \beta^{-1}$$ and $$\big\|\big(B^\top+Z)^{-1}\big\|\le \beta^{-1}.$$ Therefore $$\|Df\|_\infty\le\|B\|^2\|A^{-1}\|^2\beta^{-2}$$ which is also a Lipschitz constant for $$f$$, since its domain is convex, $$\{X\ge0\}$$.

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Rmk. The above bounds on the $$L_2$$ operator norms (or others matrix norms) could be improved, but not up to $$\big\|\big(A^{\top}+XA^{-1}B\big)^{-1}\big\|\le\big\|A^{-1}\big\|,$$ even for symmetric definite positive matrices. Take e.g. $$n=2$$ and $$A=:I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad X:=\begin{bmatrix} 5/2 & 1 \\ 1 & 1/2 \end{bmatrix}\quad B:=\begin{bmatrix} 1 & -1/2 \\ -1/2 & 1/2 \end{bmatrix}$$

Then $$\big(A^{\top}+XA^{-1}B\big)^{-1}=\big(I+XB\big)^{-1}= \begin{bmatrix} 4/15 & 4/15\\ -4/15 & 16/15 \end{bmatrix}$$ whose maximum singular value is $$\displaystyle{2\over 15}\sqrt{29}+{2\over 5}>1=\|A^{-1}\|$$. The same holds for the Frobenius, and other common entry-wise norms (due to the coefficient $$16/15>1$$).

Let $$h=Y-X$$. Using the first order expansion of the matrix inverse, $$f(X+h)=(A(X^{-1}-X^{-1}hX^{-1})A^\top+B)^{-1}+O(\|h\|^2)$$ Now let $$Z=AX^{-1}A^\top+B$$ and let $$g = AX^{-1}hX^{-1}A^\top$$. Then $$f(X+h)=(Z-g)^{-1}+O(\|h\|^2)=Z^{-1}+Z^{-1}gZ^{-1}+O(\|h\|^2).$$ Since $$Z^{-1}=f(X)$$, it follows that $$f(X+h)-f(X)=Z^{-1}AX^{-1}hX^{-1}A^\top Z^{-1}+O(\|h\|^2).$$ Using $$Z^{-1}AX^{-1}=(XA^{-1}Z)^{-1}=(A^T+XA^{-1}B)^{-1}$$, we obtain that $$\|Z^{-1}AX^{-1}\|\leq \|A^{-1}\|.$$ Similarly, $$\|X^{-1}A^\top Z^{-1}\|\leq \|A^{-1}\|$$. Consequently, $$\|f(X+h)-f(X)\|\leq \|A^{-1}\|^2\|h\|+O(\|h\|^2),$$ yielding a Lipschitz constant of $$L=\|A^{-1}\|^2$$.

• Yet it is not clear why $\|Z^{-1}AX^{-1}\|\le\|A^{-1}\|$(here I had to introduce a factor); could you explain it? Commented Jun 19, 2019 at 8:47