- #1
regina_lee
- 2
- 0
Let [itex]e_i [/itex] be a unit vector with one 1 in the [itex]i[/itex]-th element. Is the following expression has a recursive presentation?
$$y_N = \sum_{k=0}^N {\frac{{{X^k} e_i}}{\|{{{X^k} e_i}\|}_2}} $$
where [itex]X[/itex] is a [itex]n \times n[/itex] square matrix, and [itex]{\| \cdot \|}_2[/itex] is a vector norm defined as [itex]{\|z\|}_2 = \sqrt{|z_1|^2+|z_2|^2+...+|z_n|^2}[/itex].
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EDIT: I know that if [itex]y_N = \sum_{k=0}^N {{X^k} e_i} [/itex], it is easy to obtain the following recursive formula:
$$y_{k+1} = X y_{k} + e_i, \quad (k=0,1,2,...) \textrm{ with } \ \ \ y_0=e_i$$
However, after we add a normalized factor, is there a similar recursive expression? Thanks.
$$y_N = \sum_{k=0}^N {\frac{{{X^k} e_i}}{\|{{{X^k} e_i}\|}_2}} $$
where [itex]X[/itex] is a [itex]n \times n[/itex] square matrix, and [itex]{\| \cdot \|}_2[/itex] is a vector norm defined as [itex]{\|z\|}_2 = \sqrt{|z_1|^2+|z_2|^2+...+|z_n|^2}[/itex].
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EDIT: I know that if [itex]y_N = \sum_{k=0}^N {{X^k} e_i} [/itex], it is easy to obtain the following recursive formula:
$$y_{k+1} = X y_{k} + e_i, \quad (k=0,1,2,...) \textrm{ with } \ \ \ y_0=e_i$$
However, after we add a normalized factor, is there a similar recursive expression? Thanks.