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Thread: A Staircase of Unit Matrices

  1. #1
    MHF Contributor

    May 2008

    A Staircase of Unit Matrices

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

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

    Note: If you solve the problem for $\displaystyle n = 2$, 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 12:51 AM.
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  2. #2
    Member Black's Avatar
    Nov 2009
    I'll give it a shot.

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

    I claim the following:

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

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

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

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

    If these two claims hold, then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    This finishes the proof.

    Happy Holidays!
    Last edited by Black; Dec 27th 2009 at 01:52 PM.
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