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Recently, I needed to estimate the operator norm of the tridiagonal operator, but I am sure answers much more refined than my simple observations must be known.

Let $T$ be the linear operator that maps a square matrix to its tridiagonal part. Thus, the action of $T$ on a matrix $X$ can be defined by the Hadamard product $M \circ X$, where $m_{ij}=1$ if $|i-j| \le 1$, and $m_{ij}=0$ otherwise.

What is the operator 2-norm of $T$? (or what is a good approximation thereof?)

The observation $\|M \circ X\| \le \|M\| \|X\|$ shows that $\|T\| \le 3$. A more refined estimate follows from Theorem 5.5.3 of Horn and Johnson's Topics in Matrix Analysis, which says that $\|M\circ X\| \le r_1(M)c_1(X)$, where $r_1$ is maximum row-length (Euclidean norm) and $c_1$ is max column length. This result then implies that $\|T\| \le \sqrt{3}$.

I am sure that significantly more refined estimates of $\|T\|$ are available, and will be thankful if you can provide me a reference, or maybe a short proof itself.

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A good answer is given by R. Bhatia: Pinching, trimming, truncating, and averaging of matrices. Amer. Math. Monthly 107 (2000), no. 7, 602–608.

If you consider the operator $T_r$ that retains the diagonals defined by $|i-j|\le r$ (yours is $T_1$), its norm is accurately bounded by $$L_r=\frac{1}{2\pi}\int_{-\pi}^\pi |D_r(\theta)|d\theta,$$ where $D_r$ is the Dirichlet kernel.

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  • $\begingroup$ ah! i should have known about that paper! thanks Denis. $\endgroup$
    – Suvrit
    Commented Jun 8, 2011 at 15:04
  • $\begingroup$ I like it! I teach the case $r=0$ (diagonal extraction) here. $\endgroup$ Commented Jun 8, 2011 at 15:35

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