# norm of the matrix series

The goal is to obtain an upper bound for the norm of the vector $$\left\|\sum\limits_{k=0}^{\infty}(I−A)^kAw_k\right\|$$ for any symmetric matrix $A\in{\mathbb R}^{n×n}$ which $0\preceq A\preceq I$ ($I$ is identity matrix) and for any vectors $w_k\in{\mathbb R}^n$ such that $\|w_k\|\leq1,\,\,k=0,1,\ldots$ (all norms are euclidean).

It is easy to show this norm is less $\sqrt{n}$, but it seems it should not depends on $n$.

• What do you mean by $0\preceq A\preceq I$? – smyrlis Dec 30 '13 at 9:42
• For symmetric matrix $A\preceq0\iff \lambda(A)\leq0$ or, equivalently, $x^TAx\leq0\,\,\forall x$. Moreover, $A\preceq B\iff A-B\preceq0$. – user115538 Dec 30 '13 at 9:44
• $\lambda(A)$ mean eigenvalues of $A$ – user115538 Dec 30 '13 at 9:52
• Isn't this series easy to sum after you set $B=I-A$? – Federico Poloni Dec 30 '13 at 11:10
• I think, setting $B=I-A$ doesn't help... – user115538 Dec 30 '13 at 12:23

The answer certainly does depend on dimension, because in infinite dimension it can be unbounded. To show this, let's take as $A$ the operator that multiplies by $t$ in $L^2[0,1]$, and let's take $w_k(t) := c_k \cdot (1-t)^k t$, where the normalizing constants $c_k$ should be taken of order $k^{3/2}$, so that $\Vert w_k \Vert$ is of order $1$.

For this choice of vectors,

$$\sum_k (1-t)^k t w_k(t) \sim \sum_k k^{3/2} (1-t)^{2k} t^2 \sim t^{-1/2}, t \to 0$$

which is not in $L^2$.

This example suggests that the growth should be at least logarithmic in the dimension.

This is to complement Alexander Sharmov's answer with some details. The reason $w_k(t)$ is chosen as it is there is that it maximizes $\int_0^1 (1-t)^ktw_k(t)dt$ under the constraint $\int_0^1w_k(t)^2dt=1$. By Cauchy-Schwartz inequality $\big(\int_0^1 (1-t)^ktw_k(t)dt\big)^2\le\int_0^1 (1-t)^{2k}t^2dt\int_0^1w_k(t)^2dt$ achieves maximum, when $w_k(t)=c_k(1-t)^kt$ for some $c_k>0$. So

$$\Big(\frac{1}{c_k}\Big)^2=\Big(\frac{1}{c_k}\Big)^2\int_0^1 w_k(t)^2dt=\int_0^1 (1-t)^{2k}t^2dt=\frac{\Gamma(2k+1)\Gamma(3)}{\Gamma(2k+4)}=\frac{1}{2^2\big(k+\frac{1}{2}\big)\big(k+1\big)\big(k+\frac{3}{2}\big)}.$$