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Powers of a superdiagonal matrix

  1. Aug 22, 2011 #1

    while dealing with non-homogeneous equations with constant coefficients I met a following problem. I need an easy way to calculate powers of a superdiagonal matrix (every power up to n-1):

    [tex]\mathbb N^{n}_{n} \ni \mathbb M_{n}:=\begin{bmatrix} 0&n-1&0&0&...&0&0&0&0\\0&0&n-2&0&...&0&0&0&0\\0&0&0&n-3&...&0&0&0&0\\...&...&...&...&...&...&...&...&...\\0&0&0&0&...&0&3&0&0\\0&0&0&0&...&0&0&2&0\\0&0&0&0&...&0&0&0&1\\0&0&0&0&...&0&0&0&0 \end{bmatrix}[/tex]

    (zeros outside the superdiagonal, an arithmetic progression on the superdiagonal).

    Thanks in advance.
  2. jcsd
  3. Aug 22, 2011 #2
    I think this is quite simple. You would simply raise each matrix element on the diagonal to the order of the power you are computing.

    If you are computing M^2 then the element in a_{11} position would be (n-1)^2, a_{12} = (n-2)^2 and so on down the diagonal.

    The then the (n-1)th power would return (in the a_{11} position) (n-1)^(n-1) and so on down the diagonal.

    I think this is the solution unless I have totally misunderstood your matrix structure definition.
  4. Aug 22, 2011 #3
    The diagonal elements are zeros here, so it won't work. It's not a diagonal matrix.
  5. Aug 22, 2011 #4

    Whoops! Thought it was too easy. didn't read your laTex code correctly. Nor did I fully understand the terminology of "Super"diagonal.

    I'll have another think. My intial thought here is that you would be dealing with matrices in Jordan Normal form. Are you familiar? If not they are upper triangular matrices of which there is a plethora of material on them. Your matrix above is a special type of Jordan Normal form where the upper triangular block is a itself a lower triangular matrix.

    Some thought required here though.
    Last edited: Aug 22, 2011
  6. Aug 22, 2011 #5


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    I think it might be easiest to express in basis format. Define the matrix:
    [itex]e^i\otimes e_j = \Lambda^i_j[/itex]
    To be the matrix which is is zero everywhere except a 1 in the i-th row and j-th column.
    Then your matrix is:
    [tex]\mathbb{M}_n = (n-1)\Lambda^1_2 +(n-2)\Lambda^2_3 + \cdots + 1\Lambda^{(n-1)}_n=\sum_{k=1}^{n-1}(n-k)\Lambda^k_{k+1}[/tex]

    In component form multiplication becomes [itex]\Lambda^i_j \Lambda^k_n = \delta^k_j \Lambda^i_n[/itex] where [itex]\delta[/itex] is the Kronecker delta.

    Expanding the first power gives:
    [tex](\mathbb{M}_n)^2 = \sum_{k=1}^{n-2}(n-k)(n-k-1)\Lambda^k_{k+2}[/tex]
    I think you can see the pattern emerge.
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