A Staircase of Unit Matrices

• Dec 23rd 2009, 12:34 AM
NonCommAlg
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!)
• Dec 25th 2009, 08:25 PM
Black
I'll give it a shot.

Spoiler:
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

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

Clearly, $(\sigma(r)+1)/2 \not= (\sigma(r+1)+1)/2$, 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

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

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

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

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

Since $\sigma(r+1) \not= \sigma(r) +1$, $(\sigma(r)/2)+1 \not= (\sigma(r+1)+1)/2$, 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!