Solving for $L$, where $P_n \cdots P_1 = I – XLX^T$, with $\Vert x_i \Vert_2=1$, $P_i := I – 2x_i x_i^T$


Problem

Solve for $ L$ , where $ P_n \cdots P_1 = I – XLX^T$

where $ \Vert x_i \Vert_2=1$ , $ P_i := I – 2x_i x_i^T$ , and $ X_{m \times n} = [x_1 |\cdots | x_n]$ .


Try

Note that $ L$ is a lower triangular matrix. Denoting $ (i,j)$ component of $ L$ as $ L_{ij}$ , let us find $ L_{ij}$ ‘s explicitly. We have

$ $ \begin{align} P – I &= (I – 2x_nx_n^T) \cdots (I – 2x_1x_1^T) – I\ &= – L_{11} x_1x_1^T – L_{21}x_2x_1^T – \cdots -L_{nn}x_nx_n^T \end{align} $ $

where we can note that $ -L{ij}$ is the coefficient for $ x_ix_j^T$ in $ (I – 2x_nx_n^T) \cdots (I – 2x_1x_1^T)$ .

Let us observe more carefully. For $ L_{kk}$ ‘s, we have

$ $ L_{kk} = 2 $ $

since these terms only come from $ I \cdots I (-2x_kx_k^T) I \cdots I$ . Next, we note that

$ $ L_{(k+1)k} = -2^2 $ $

and, next,

$ $ -L_{(k+2)k} = 2^2 – 2^3(v_{k+2}^Tv_{k+1})(v_{k+1}^Tv_k) $ $

but I cannot see any simpler rule for $ L_{ij}$ ‘s… Anyone to help me with finding all the $ L_{ij}$ s?

If $\sum c_n$ is convergent where $c_n = a_1b_n + \cdots +a_nb_1$, then $\sum c_n = \sum a_n\sum b_n$

$ \sum a_n$ and $ \sum b_n$ is convergent and $ c_n = a_1b_n + \cdots + a_nb_1$ . Prove that if $ \sum c_n$ is convergent then $ \sum c_n = \sum a_n \sum b_n$ .

This question has been answered using Abel’s Lemma in this post. But I am looking forward to a more basic solution which only uses basic definition and Cauchy’s convergence test.

Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$

Suppose $ a>1,b>0$ are real numbers. Consider the summation of the infinite series: $ $ S=\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$ $ How can I give a tight estimation on the summation? Apparently, one can get the upper bound: $ $ S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b-1}$ $ But it is not tight enough. For example, fix $ a\rightarrow 1$ , and $ b=0.001$ , then $ S=38.969939$ , it seems that $ S=O(\sqrt{1/b})$ . Another example: $ a=1$ , and $ b=0.00001$ ,$ S=395.039235$ .

$\bigcap I_\gamma$ if $I_0 \supset \cdots \supset\I_{\gamma} \cdots$ are unbounded

Let be $ \kappa$ a regular cardinal and $ I_0 \supset \cdots \supset I_{\gamma} \cdots$ are unbounded subsets of $ \kappa$ for $ \gamma < \lambda <\kappa$ where $ \lambda$ is limit. I want to show $ \bigcap I_\gamma$ is unbounded. Is it true? It’s easy show it’s not empty for regularity of $ \kappa$ .

On constructing free action of the cyclic group $\Bbb Z/3\Bbb Z$ on $S^n \times \cdots \times S^n$($n$ is odd)

Can we construct a free action of the cyclic group $ \Bbb Z/3\Bbb Z$ on $ S^n \times \cdots \times S^n$ ($ n$ is odd) without multiplying any coordinate by $ e^{2\pi i/3} $ ? In other words, do I Need to multiply at least one coordinate by a 3rd root of unity to get a free action of $ \Bbb Z/3\Bbb Z$ ? I have tried to construct such examples but could not find any.

Thank you so much in advance.