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.
4
-
$\begingroup$ I don't understand your question. Isn't $\sum A_i v_i=Av$? So eqn (1) simply holdd by definition of the spectral norm. $\endgroup$– user35593Commented Jan 8, 2017 at 9:47
-
$\begingroup$ no, each $A_i$ can be a matrix, not a column vector. $\endgroup$– user103308Commented Jan 8, 2017 at 15:24
-
$\begingroup$ Then I dont now what v is and how you define the 2-norm of v. $\endgroup$– user35593Commented Jan 8, 2017 at 15:27
-
$\begingroup$ each $v_i$ is a scalar, so $\sum_i A_i v_i$ is a weighted sum of matrices $A_1,\ldots, A_n$. $\endgroup$– user103308Commented Jan 8, 2017 at 15:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
$$\left\|\sum A_iv_i\right\|=\sup_{\|x\|=1} \left\|\sum A_iv_ix\right\|\leq \sup_{\|x_i\|=|v_i|} \left\|\sum A_ix_i\right\|=\sup_{\|x_i\|=|v_i|} \left\|A \begin{pmatrix} x_1\\ \vdots\end{pmatrix}\right\|\\\leq \sup_{\|x_i\|=|v_i|} \|A\| \left\|\begin{pmatrix} x_1\\ \vdots\end{pmatrix}\right\|= \|A\| \|v\|.$$