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Thread: A Staircase of Unit Matrices

  1. #1
    MHF Contributor

    May 2008

    A Staircase of Unit Matrices

    Let n \geq 2 be a given integer and A be the vector space of all n \times n matrices with, say, real entries. Let \{e_{ij}: \ \ 1 \leq i,j \leq n \} be the standard basis for A and define

    f(x_1,x_2, \cdots , x_{2n-1})=\sum_{\sigma \in S_{2n-1}} \text{sgn}(\sigma) x_{\sigma(1)}x_{\sigma(2)} \cdots x_{\sigma(2n-1)}. Evaluate f(e_{11},e_{12},e_{22},e_{23}, e_{33}, e_{34}, \cdots , e_{nn}).

    Note: If you solve the problem for n = 2, you'll definitely guess the right answer for the general case. But ... can you give a rigorous proof of your correct guess?

    (It's not important but anyway, that Unit Matrices in the title should be Matrix Units! Sorry!)
    Last edited by NonCommAlg; Dec 23rd 2009 at 01:51 AM.
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  2. #2
    Member Black's Avatar
    Nov 2009
    I'll give it a shot.

    The answer is f(e_{11},e_{12},e_{22},e_{23},\dots ,e_{nn})=e_{1n}.

    I claim the following:

    (1) e_{ij}e_{kl} = e_{il} when j=k and zero otherwise.

    (2) If we let s_1=e_{11}, s_2=e_{12}, s_3=e_{22}, s_4=e_{23},\dots, s_{2n-1}=e_{nn}, then

    s_{\sigma(1)}s_{\sigma(2)}\cdots s_{\sigma(2n-1)} = 0,

    where \sigma \in S_{2n-1}- \{\varepsilon\} ( \varepsilon = identity).

    If these two claims hold, then

    f(e_{11},e_{12},e_{22},e_{23},\dots ,e_{nn})=sgn(\varepsilon)e_{11}e_{12}e_{22}e_{23}\  cdots e_{nn}=e_{1n}.

    Proof of (1): The (i,j)-th entry in e_{ij}e_{kl} is

    (e_{ij}e_{kl})_{i,j}=\sum_{r=1}^{n} (e_{ij})_{i,r}(e_{kl})_{r,j}.

    The expression (e_{ij})_{i,r} is 1 only when r=j (zero otherwise) and (e_{kl})_{r,j} is 1 only when r=k and j=l (zero otherwise). Therefore, (1) follows.

    Proof of (2): Let \sigma \in S_{2n-1}-\{ \varepsilon \}. Then \exists r such that \sigma(r+1) \not= \sigma(r) +1.

    Case 1: \sigma(r) is an odd number, \sigma(r+1) is an odd number.

    s_{\sigma(r)}=e_{[(\sigma(r)+1)/2][(\sigma(r)+1)/2]} and s_{\sigma(r+1)}=e_{[(\sigma(r+1)+1)/2][(\sigma(r+1)+1)/2]}, so


    Clearly, (\sigma(r)+1)/2 \not= (\sigma(r+1)+1)/2, so from (1), s_{\sigma(r)}s_{\sigma(r+1)}=0.

    Case 2: \sigma(r) is odd, \sigma(r+1) is even.

    s_{\sigma(r)}=e_{[(\sigma(r)+1)/2][(\sigma(r)+1)/2]} and s_{\sigma(r+1)}=e_{[\sigma(r+1)/2][(\sigma(r+1)/2)+1]}, so


    Since \sigma(r+1) \not= \sigma(r) +1, (\sigma(r)+1)/2 \not= \sigma(r+1)/2, so from (1), s_{\sigma(r)}s_{\sigma(r+1)}=0.

    Case 3: \sigma(r) is even, \sigma(r+1) is odd.


    Since \sigma(r+1) \not= \sigma(r) +1, (\sigma(r)/2)+1 \not= (\sigma(r+1)+1)/2, so s_{\sigma(r)}s_{\sigma(r+1)}=0.

    Case 4: \sigma(r) is even, \sigma(r+1) is even.

    s_{\sigma(r)}s_{\sigma(r+1)}=e_{[\sigma(r)/2][(\sigma(r)/2)+1]}e_{[\sigma(r+1)/2][(\sigma(r+1)/2)+1]} .

    If \sigma(r) > \sigma(r+1), then (\sigma(r)/2)+1 \not= \sigma(r+1)/2 \Longrightarrow s_{\sigma(r)}s_{\sigma(r+1)}=0 (same with \sigma(r) < \sigma(r+1) and \sigma(r+1)-\sigma(r)>2).

    If \sigma(r) < \sigma(r+1) and \sigma(r+1)-\sigma(r)=2, then let \alpha be the odd number between \sigma(r) and \sigma(r+1) and pick a such that \sigma(a)=\alpha. Then \sigma(a + 1) is an odd number or an even number that's not \sigma(r+1), so from Cases 1 & 2, s_{\sigma(a)}s_{\sigma(a+1)}=0.

    This finishes the proof.

    Happy Holidays!
    Last edited by Black; Dec 27th 2009 at 02:52 PM.
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