I wonder if there are any theorems which can help me to calculate an upper bound for the spectral norm of: $$\left\| \left[ I + \sum_{i=1}^{\overline{n}\in\mathbb{N}} \big( C_i - I\big)\right]^{-1}\right\|_2$$ with the identity matrix $I \in \mathbb{R}^{n\times n}$ and $C_i$ are triangular matrices arising from a Cholesky decompostion of $A_i = C_i C_i^T $. The matrices $A_i \in \mathbb{R}^{n\times n}$ are symmetric positive semidefinite.

My first idea was to use that if $A\leq B$ the following relation holds for s.p.d. matrices $$\big\|(A + B)^{-1}\big\|_2 \leq \big\| A^{-1}\big\|_2.$$

But the problem is that the matrices $C_i - I$ are negative definite if I'm right.

  • 1
    $\begingroup$ The title asks about inverting the sum of identity and triangular matrices, which is tractable. But the body instead asks about an upper bound on the spectral norm of such an inverse. I don't think the problem statement lends itself to much more than a construction of the (triangular) inverse of a triangular matrix. $\endgroup$
    – hardmath
    Nov 25, 2018 at 21:20
  • 4
    $\begingroup$ There's no upper bound here- Take a single term with $C_{1}=\epsilon I$ the 2-norm of $(I+(C_{1}-I))^{-1}$ is $1/\epsilon$, which goes to infinity as $\epsilon \rightarrow 0$. $\endgroup$ Nov 26, 2018 at 3:25
  • $\begingroup$ But whats the reason for $\epsilon \to 0$ since I am not talking about $C_1 = 0$? $\endgroup$
    – user29088
    Nov 26, 2018 at 11:26
  • $\begingroup$ So are you asking about an upper bound for the spectral norm in terms of the entries of the $C_i$? What do you hope to have on the right hand side of the bound? $\endgroup$ Nov 26, 2018 at 14:13
  • 2
    $\begingroup$ You can modify my earlier example to use $C_{1}=(1/2)I$ and $C_{2}=(1/2)I+\epsilon I$. $\endgroup$ Nov 26, 2018 at 18:41

1 Answer 1


Answer assembled from @BrianBorchers comments.

There is no upper bound. Example: Take a single-term $C_1=\epsilon I$. Then $$ ||(I+(C_1-I))^{-1}||_2=\frac{1}{\epsilon}\overset{\epsilon\to0}{\longrightarrow}\infty $$

This example can be modified to avoid the case of $C_1$ being close to 0-matrix. With $C_1=\frac{1}{2}I$ and $C_2=\frac{1}{2}I+\epsilon I$, the situation is identical.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.