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Shilov's Linear Algebra determinant notation.

  1. Aug 9, 2009 #1
    I just started reading the first chapter of Georgi Shilov's "Linear Algebra" and I have a question about his notation for determinants. His notation, (7), for the determinant of an n x n matrix seems to be [tex]\det ||a_{ij}||.[/tex]

    (4) suggests Shilov would write the 1 x 1 matrix with the single element x as [tex]||x||.[/tex] So in (7), does Shilov mean for [tex]||a_{ij}||[/tex] to be interpreted as a 1 x 1 matrix or am I missing something?

    (4) and (7) can be found by googling for "Shilov determinant."
    Last edited: Aug 9, 2009
  2. jcsd
  3. Aug 9, 2009 #2
    I think your problem is very easily solved. His notation is lousy.

    By ||a_ij|| he just means the whole matrix consisting of the components a_ij, where a_ij is the single component on the ith row and jth column. This is confusing, but also conevenient if you do not want to spell out the entire matrix.

    So strictly speaking, yes you're right it is a 1x1 matrix. But what he abuses the notation slightly for convenience.
  4. Aug 9, 2009 #3
    That sounds like an interpretation I can live with. Thanks for taking the time to clarify his notation!
  5. Nov 28, 2010 #4
    I believe that "det||a_ij||" is referring to the determinant of the matrix, ||a_ij||, where i and j are variables representing the index of any row and column, respectively, and 'a' represents any element in the matrix at that row/column location. So, while a_ij can only take the value of one element at a time, this is not a matrix of just one element--he is just using one variable that can represent any element in the matrix.

    This is similar to describing a set:
    {x such that x is even} (for x in the set of integers)

    Even though I used one variable, x, that doesn't mean my set consists of just one element, since x represents any even integer here, and thus my set contains all even integers.
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