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.



tighter bound on the sum of sub-matrices
$$\left\|\sum A\_iv\_i\right\|=\sup\_{\|x\|=1} \left\|\sum A\_iv\_ix\right\|\leq \sup\_{\|x\_i\|=|v\_i|} \left\|\sum A\_ix\_i\right\|=\sup\_{\|x\_i\|=|v\_i|} \left\|A \begin{pmatrix} x\_1\\ \vdots\end{pmatrix}\right\|\\\leq \sup\_{\|x\_i\|=|v\_i|} \|A\| \left\|\begin{pmatrix} x\_1\\ \vdots\end{pmatrix}\right\|= \|A\| \|v\|.$$


