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Let the tridiagonal matrix ${\bf C} \in {\Bbb R}^{n \times n}$ and the diagonal matrix ${\bf D} \in {\Bbb R}^{n \times n}$ be

\begin{equation}\label{matrix-C,D} {\bf C} := \begin{bmatrix} 0 & 1 & & & &\\ -1 & 0 & 1 & & &\\ &-1 & 0 & 1 & &\\ & & \ddots&\ddots&\ddots\\ & & &-1 & 0 & 1\\ & & & &-1 & 1 \end{bmatrix}, \qquad {\bf D} :=\begin{bmatrix} 1& & & & &\\ & 1& & & &\\ && 1& & &\\ &&&\ddots&&\\ & & & & 1&\\ & & & & &\frac{1}{2} \end{bmatrix} \end{equation}

I want to prove the following lemma: for $\lambda>0$, $\left\lVert ({\bf C} + 2\lambda {\bf D})^{-1} \right\rVert_2\leq n$.

Can you please give me some tips? I wonder whether the upper bound $n$ in this lemma is sharp? is there any sharp bound for matrix $({\bf C} + 2\lambda {\bf D})^{-1}$?

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  • $\begingroup$ Because this matrix $(C+2\lambda D)^{-1}$ is a tri-diagonal matrix, we have a concrete formula for the entries of this matrix [link](). By analysis the entries of this matrix and using the inequality $$\lVert A\rVert_2^2\leq \lVert A\rVert_1 \lVert A\rVert_\infty$$, I proof this lemma. But I think there are a sharp upper bound for this matrix, I wonder how can I get this shap bound $\endgroup$ Commented Feb 17 at 2:21
  • $\begingroup$ entries of the inverse of a tridiagonal matrix : link $\endgroup$ Commented Feb 17 at 2:33
  • $\begingroup$ Thank you very much. I have modified the question according to your guidance. Matrix C is not a Toeplitz matrix because the last element in the last row is not 0. Similarly, matrix D is also not a Toeplitz matrix. I now think that perhaps I can use the Sherman-Morrison-Woodbury formula to form each element of the matrix and prove it through the relationships between the 2-norm, 1-norm, and infinity norm. $\endgroup$ Commented Feb 18 at 6:42
  • $\begingroup$ Right. I missed the southeast corner. In any case, both are nearly Toeplitz. You can write ${\bf C} = {\bf T} + {\bf e}_n {\bf e}_n^\top$, where $\bf T$ is Toeplitz, and ${\bf D} = {\bf I}_n - \frac12 {\bf e}_n {\bf e}_n^\top$, where ${\bf I}_n$ is also Toeplitz. Eventually, you should obtain a rank-$1$ update to a Toeplitz matrix $\endgroup$ Commented Feb 18 at 6:54

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