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Suppose $A = [A_1, A_2, \ldots, A_n]$ is a matrix, and each $A_i$ is a column-wise sub-matrix of $A$. Given a vector $v \in \mathbb{R}^n$, a bound of $\| \sum_{i=1}^n A_i v_i \|_2$ (spectral norm) in terms of $\|A\|_2$ (spectral norm) and $\|v\|_2$ (Euclidean norm) is desired. A naive approach is \begin{align*} \| \sum_{i=1}^n A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i \|_2 |v_i| \leq \|A\|_2 \|v\|_1 \leq n\|A\|_2 \|v\|_2. \end{align*} However, this may be too loose. Can I get rid of $n$ on R.H.S.? That`s to say, \begin{align} \| \sum_{i=1}^n A_i v_i \|_2 \leq \|A\|_2 \|v\|_2, (1) \end{align} Eqn. (1) is trivially true when each $A_i$ is a column of $A$. Then is this also true when each $A_i$ is a column-wise sub-matrix with arbitrary number of columns, i.e., different $A_i$ can have different number of columns? By numerical experiments using random trails, it seems the answer is positive.

Suppose $A = [A_1, A_2, \ldots, A_n]$ is a matrix, and each $A_i$ is a column-wise sub-matrix of $A$. Given a vector $v \in \mathbb{R}^n$, a bound of $\| \sum_{i=1}^n A_i v_i \|_2$ (spectral norm) in terms of $\|A\|_2$ (spectral norm) and $\|v\|_2$ (Euclidean norm) is desired. A naive approach is \begin{align*} \| \sum_{i=1}^n A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i \|_2 |v_i| \leq \|A\|_2 \|v\|_1 \leq n\|A\|_2 \|v\|_2. \end{align*} However, this may be too loose. Can I get rid of $n$ on R.H.S.? That`s to say, \begin{align} \| \sum_{i=1}^n A_i v_i \|_2 \leq \|A\|_2 \|v\|_2, (1) \end{align} Eqn. (1) is trivially true when each $A_i$ is a column of $A$. Then is this also true when each $A_i$ is a column-wise sub-matrix with arbitrary number of columns, i.e., different $A_i$ can have different number of columns? By numerical experiments using random trails, it seems the answer is positive.

Suppose $A = [A_1, A_2, \ldots, A_n]$ is a matrix, and each $A_i$ is a column-wise sub-matrix of $A$. Given a vector $v \in \mathbb{R}^n$, a bound of $\| \sum_{i=1}^n A_i v_i \|_2$ (spectral norm) in terms of $\|A\|_2$ (spectral norm) and $\|v\|_2$ (Euclidean norm) is desired. A naive approach is \begin{align*} \| \sum_{i=1}^n A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i \|_2 |v_i| \leq \|A\|_2 \|v\|_1 \leq n\|A\|_2 \|v\|_2. \end{align*} However, this may be too loose. Can I get rid of $n$ on R.H.S.? That`s to say, \begin{align} \| \sum_{i=1}^n A_i v_i \|_2 \leq \|A\|_2 \|v\|_2, (1) \end{align} Eqn. (1) is trivially true when each $A_i$ is a column of $A$. Then is this also true when each $A_i$ is a column-wise sub-matrix with arbitrary number of columns? By numerical experiments using random trails, it seems the answer is positive.

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user103308
user103308
  • 85
  • 3

tighter bound on the sum of sub-matrices

Suppose $A = [A_1, A_2, \ldots, A_n]$ is a matrix, and each $A_i$ is a column-wise sub-matrix of $A$. Given a vector $v \in \mathbb{R}^n$, a bound of $\| \sum_{i=1}^n A_i v_i \|_2$ (spectral norm) in terms of $\|A\|_2$ (spectral norm) and $\|v\|_2$ (Euclidean norm) is desired. A naive approach is \begin{align*} \| \sum_{i=1}^n A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i v_i \|_2 \leq \sum_{i=1}^n \|A_i \|_2 |v_i| \leq \|A\|_2 \|v\|_1 \leq n\|A\|_2 \|v\|_2. \end{align*} However, this may be too loose. Can I get rid of $n$ on R.H.S.? That`s to say, \begin{align} \| \sum_{i=1}^n A_i v_i \|_2 \leq \|A\|_2 \|v\|_2, (1) \end{align} Eqn. (1) is trivially true when each $A_i$ is a column of $A$. Then is this also true when each $A_i$ is a column-wise sub-matrix with arbitrary number of columns, i.e., different $A_i$ can have different number of columns? By numerical experiments using random trails, it seems the answer is positive.