In recent days, I need to estimate the 1-norm (or $\infty$-norm) of the inverse of the following lower Toeplitz-like triangular matrix, i.e., \begin{equation} C = \begin{bmatrix} 1 &\\ -2 &\frac{3}{2} \\ \frac{1}{2}&-2 &\frac{3}{2} \\ &\ddots & \ddots &\ddots\\ & &\frac{1}{2}&-2&\frac{3}{2} \end{bmatrix}\in\mathbb{R}^{N\times N}. \end{equation} After I use the MATLAB to compute the inverse of the above matrix with different size $N$, it seems that we have

  1. $\|C^{-1}\|_1\leq \frac{3N}{2}$;

  2. $\|C^{-1}\|_{\infty} = N$.

where $\|C\|_1 = \max\limits_{1\leq j\leq N}\sum\limits^{N}_{i=1}|C_{ij}|$ and $\|C\|_{\infty} = \max\limits_{1\leq i\leq N}\sum\limits^{N}_{j=1}|C_{ij}|$.

Can we prove these conclusions?

  • $\begingroup$ Many thanks for your comments, I will add the definition of the matrix norm. $\endgroup$ – Hsien-Ming Ku Oct 30 '19 at 13:33

You have the following two closed form formulas: $$ \|C^{-1}\|_1=\sum_{i=1}^N \frac{(3^i-1)/2}{3^{i-1}}, $$ and $$ \|C^{-1}\|_\infty=\frac{(3^N-1)/2}{3^{N-1}}+\sum_{i=2}^N \frac{3^{N-i+1}-1}{3^{N-i+1}}. $$

  • $\begingroup$ many thanks for your answer, I got it according your reminder. $\endgroup$ – Hsien-Ming Ku Nov 5 '19 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.