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I am working on a square matrix consisting of zero entries, except the entries right above its diagonal are 1's. For example, matrix of 3 by 3 has 2 entries of 1's slanted parallel to its diagonal:

$$\begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0\\ \end{bmatrix},$$ ect. And here are my two questions:
(1) Is there any name for that type of matrix? Any online reference to it?
(2) I am trying to prove, using only elementary means, that if $X_{k \times k}$ is such matrix, then its $k$ power equals to zero, while its $k-1$ power does not. Any link, hints, guidance or help will be very much appreciated.

Thank you for your time and effort.


PS: Getting good hints from Dirk below, I think it's should be called Shift Matrix, see the link here. The link even says that

Clearly all shift matrices are nilpotent; an n by n shift matrix S becomes the null matrix when raised to the power of its dimension n...

Is there any proof relating to its being nilpotent, using elementary means? Thanks again to Dirk.

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    $\begingroup$ One helpful word is "nilpotent". $\endgroup$
    – Dirk
    Oct 8 '16 at 13:31
  • $\begingroup$ I think 2) would fail on a zero matrix with only 1 in top right corner $\endgroup$
    – Pieter21
    Oct 8 '16 at 13:31
  • $\begingroup$ For $\lambda = 0$, this is in Jordan normal form. $\endgroup$ Oct 8 '16 at 13:31
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    $\begingroup$ One interesting fact about the given matrix, call it $A$: for any $t\in\mathbb R$ we have $$e^{tA} = \begin{bmatrix}1 & t & \frac12 t^2\\ 0 & 1 & t\\ 0 & 0 & 1\end{bmatrix} $$ $\endgroup$
    – Math1000
    Oct 8 '16 at 13:34
  • $\begingroup$ Thanks to all. Getting hint from @Drk about Nilpotent, do you think it is what is called Shift Matrix? en.wikipedia.org/wiki/Shift_matrix $\endgroup$
    – A.Magnus
    Oct 8 '16 at 13:36
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It's certainly a matrix that represents a "shift" operation on a vector that represents a list, in the sense that it moves each entry "up" one step (and puts a $0$ in the last entry).

So a good inductive proof would work with "After $k\le n$ iterations, $A^k v$ has zero as its last $k$ entries, no matter what the vector $v$."

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