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I have a linear operator with its matrix in certain coordinates to be

$$ \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 & \frac{1}{3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & \cdots & \frac{1}{n} \end{pmatrix} $$

What is this linear operator? How could I construct it without referring to coordinates?

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linear operator it is kernel yes? –  dato datuashvili May 9 '12 at 19:02
    
i think it is null space ,or all set of vectors x,for which $A*x=0$ –  dato datuashvili May 9 '12 at 19:04
    
@dato: this matrix is full rank –  Alex R. May 9 '12 at 19:07
    
yes i see determinant is not zero,but how it is related to linear operator? –  dato datuashvili May 9 '12 at 19:09
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@dato: The main idea of comments under the question is to clarify the question or contribute to an answer without giving a full answer; it's not to ask questions of your own. If you don't know how a matrix is related to a linear operator, you may want to read up on those concepts at Wikipedia. If you don't understand the connection after having learned about them, you can ask your own question about those concepts here. But please don't clutter other people's questions with your own. –  joriki May 9 '12 at 19:13
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1 Answer 1

up vote 4 down vote accepted

Of course it could be any number of things, but one operator with this matrix is the one that assigns to every polynomial $p(x)$ of degree less than $n$ the polynomial $\frac1x\int_0^xp(t)\,\mathrm dt$.

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i did not know it @joriki,what is in this case polynomial?matrix or? –  dato datuashvili May 9 '12 at 19:07
    
@dato: It's nothing special in this case, just a plain vanilla polynomial in one variable, the kind you're likely to meet in the street on your way to the bus stop. –  joriki May 9 '12 at 19:08
    
That's exactly how I came to this operator in connection with math.stackexchange.com/questions/142941/… I just hoped that there is some other interpretation I could rely on. –  Yrogirg May 9 '12 at 19:09
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