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Homework Help: Linear Algebra - number of entries

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    a) How many entries can be chosen independently, in a symmetric matrix of order n?

    b) How many entries can be chosen independently, in a skew-symmetric matrix of order n?

    2. Relevant equations

    3. The attempt at a solution

    All I know are the definitions of a symmetric and a skew symmetric matrix.
    I don't understand the problem, what does chosen independently mean? Is this a sum of some sort? Any hints are greatly appreciated!
  2. jcsd
  3. Dec 12, 2008 #2
    a) I think that all the entries in the first row of a symmetric matrix may be chosen independently. And I also think that the "last" entry is independent. (the bottom right corner entry).

    b) I would think that all entries are independent in a skew-symmetric matrix..

    anyone? Thank you.
  4. Dec 12, 2008 #3


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    Science Advisor

    It would have helped if you had given some reason why you "think" those things!

    When you don't understand something, look at simple examples. If A is a 2 by 2 matrix, it can be written as
    [tex]\left[\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right][/tex]

    If no other conditions on A are given, you would be free to choose the four numbers to be anything you want- you have "four independent choices".

    Now suppose A is symmetric- that means [itex]a_{12}= a_{22}[/itex]. You can still choose [itex]a_{11}[/itex] and [itex]a_{22}[/itex] to be anything you want and you can choose either [itex]a_{12}[/itex] or [itex]a_{21}[/itex] to be anything you want but as soon as you choose one of them the other must be the same- you have three independent choices.

    Suppose A is skew-symmetric. Then [itex]a_{ij}= -a_{ji}[/itex]. No, you are NOT free to choose all entries independently! For one thing, on the diagonal you must have [itex]a_{ii}= -a_{ii}[/itex] which means each diagonal element must be 0! In our 2 by 2 case, [itex]a_{11}= a_{22}= 0[/itex]. We can choose either [itex]a_{12}[/itex] or [itex]a_{21}[/itex] but once you have chosen one of them, the other is the negative of that. You have only one independent choice.

    Some of the things you will have to figure out to do this: in an n by n matrix, how many entries are there on the main diagonal? How many entries are there above the main diagonal? How many entries are the below the main diagonal? Think about the definitions of "symmetric" and "skew-symmetric" to see why you need to know those numbers and the answers to these questions should be easy.
  5. Jul 29, 2009 #4
    As it is never good to leave things unresolved, I will give this problem a try several months after posting it.

    There are n entries on the main diagonal.
    There are [tex]\frac{(n-1)n}{2}[/tex] entries above/below the main diagonal.

    I think of this as a triangle with width (n-1) and height n with area [tex]\frac{(n-1)n}{2}[/tex].

    As mister HallsOfIvy writes, a skew-symmetric matrix has [tex]a_{ij} = -a_{ji}[/tex], which means that only the entries above or below the main diagonal can be chosen independently.
    And thus the answer is:


    Thank you.
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