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Let $A$ be a square $n \times n$ over a field (say $\mathbb{R}$ or $\mathbb{C}$). As we know, the main diagonal $(a_{1,1},...,a_{n,n})$ is important in linear algebra while the off-diagonal is far less important. The deep question is why?

We try to test it using the trace, which lies heavily on the main diagonal of a matrix.

Recall that the trace is defined as $$ \operatorname{Tr}(A) := \sum_{i=1}^n a_{i,i} \ .$$ We know that it is a good operator, in the sense it is invariant under matrix conjugation (matrix similarity: $B = M A M^{-1} \Rightarrow \operatorname{Tr}(A) = \operatorname{Tr}(B)$).

Let us look on the symmetry group $S_n$ of $n!$ permutations. Let us define a more general trace: let $\sigma \in S_n$ then $$ \operatorname{Tr}_{\sigma}(A) := \sum_{i=1}^n a_{i,\sigma(i)} \ . $$ Note that $\operatorname{Tr}(A) = \operatorname{Tr}_{\operatorname{id}}(A)$ I checked on few examples that $\operatorname{Tr}_{\sigma}(A)$ is not generally invariant under matrix conjugation. The question is why?

My thought is that it is due to the fact $\operatorname{Tr}(A) = \operatorname{Tr}_{\operatorname{id}}(A)$ and that $\operatorname{id} \in S_n$ is the unit element of the group $S_n$ and hence a distinguished element. However, I did not managed to show how this fact implies that the trace is invariant under matrix conjugation while $\sigma$-trace is generally not. What I did managed to show is that $$ \operatorname{Tr}_{\sigma}(AB) \ne \operatorname{Tr}_{\sigma}(BA) $$ unless $\sigma = \operatorname{id}$. I think this implies that the $\sigma$-trace isn't invariant under conjugation (as $\operatorname{Tr}(MAM^{-1}) = \operatorname{Tr}(AM^{-1}M) = \operatorname{Tr}(A)$).

I want to connect between the deep fact the the identity is the unit element of $S_n$ to the fact the only the $\sigma$-trace $\operatorname{Tr}_{\operatorname{id}}(A)$ is invariant under matrix conjugation.

Vincenzo Tibullo
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LinAlgMan
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    Is it possible to write your $\sigma$-trace in a form like tr$\sigma A = $ tr${P\sigma A}$ where $P_{\sigma}$ is an appropriate permutation matrix? If so, that'd give a natural explanation of why it only is cyclic if $\sigma=$id. – Semiclassical Jul 23 '14 at 13:27
  • While this is just an idea, you could maybe think of some properties which are useful for the trace to have, and then see if these are enough to characterize the trace completely up to a scalar multiple (just like how volume forms are the determinant up to a scalar). This probably requires some idea as to what the trace means geometrically though - from Wikipedia it says "Geometrically, the trace can be interpreted as the infinitesimal change in volume (as the derivative of the determinant)", so that might help. – Andrew D Jul 23 '14 at 13:30
  • I think yes (I checked it for $2 \times 2$ and $3 \times 3$ examples, but havn't prove it). This actually show that $Tr(P_\sigma AB ) \ne Tr(P_\sigma BA)$ which I showed explicitly using index notation. However, this explains while $Tr_{id}$ is the only good trace but don't explain why the main diagonal is distiguished. – LinAlgMan Jul 23 '14 at 13:35

1 Answers1

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Let $M$ be an invertible matrix. Let $m_k$ me the $k^{th}$ row of $M$, $n_k$ the $k^{th}$ row of $M^{-1}$ and $a_k$ the $k^{th}$ column of $A$.

Then the $k^{th}$ row of $MA$ is $(m_k\cdot a_1,\dots, m_k\cdot a_n)$. Thus $(MAM^{-1})_{kk}=(m_k\cdot a_1,\dots, m_k\cdot a_n)\cdot n_k$. Hence $$\text{tr}(MAM^{-1})=\sum_{k=1}^n (m_k\cdot a_1,\dots, m_k\cdot a_n)\cdot n_k=\sum_{k=1}^n \sum_{i=1}^n\sum_{j=1}^n m_{ki}a_{ij}n_{jk}=\sum_{i=1}^n\sum_{j=1}^n a_{ij}\sum_{k=1}^n m_{ki}n_{jk}$$ while for each $k$ by definition $\sum_{k=1}^n m_{ki}n_{jk}=\delta_{ij}$. From this we get the result.

But notice that if we apply this argument to $\text{tr}_\sigma(MAM^{-1})$ the same calculation will go through except $\sum_{k=1}^n m_{ki}n_{j\sigma(k)}=\delta_{ij}$ will not hold unless $\sigma(k)=k$ for all $k$. But there's only one element in $S_n$ that does this!

JonHerman
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