# Characterization of the trace function

We know that the trace of a matrix is a linear map for all square matrices and that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ when the multiplication makes sense.

On the Wikipedia page for trace, under properties, it says that these properties characterize the trace completely in the following sense: If $f$ is a linear function on the space of square matrices satisfying $f(xy)=f(yx)$, then $f$ and $\operatorname{tr}$ are proportional. A note on the bottom of the page gives the justification, but I do not understand the logic of it. Thanks

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The word linear means, here, $f(A+B) = f(A) + f(B),$ and, for a constant $c,$ also $f(cA) = c f(A).$ – Will Jagy Feb 2 '12 at 5:47
In the note of the bottom of the wiki page, $e_{ij}$ means the matrix with $(i,j)$-entry being equal to $1$, and all the other entries are zero. – Paul Feb 2 '12 at 5:50
@Paul, thanks for your response, but I am having trouble understanding the proof in general, not just what the symbols stand for. – Edison Feb 2 '12 at 5:59

The proof in that footnote depends on the following facts that were left out.

1. If $i\neq j$, then $[e_{ii},e_{ij}]=e_{ii}e_{ij}-e_{ij}e_{ii}=e_{ij}-0=e_{ij}$. So we can write $e_{ij}$ as a commutator $e_{ij}=xy-yx$ and thus $f(e_{ij})=f(xy)-f(yx)=0.$
2. If $i\neq j$, then similarly $[e_{ij},e_{ji}]=e_{ii}-e_{jj}$, and we get that $$f(e_{jj})=f(e_{jj})+0=f(e_{jj})+f([e_{ij},e_{ji}]=f(e_{jj}+(e_{ii}-e_{jj}))=f(e_{ii}).$$
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Let $f$ be such a map and $E_{ij}$ be the matrix with a one in $(i,j)$-entry and zero elsewhere. We have

$$E_{ij}E_{kl}=\left\{\begin{array}{cc} 0, &\mbox{ if } \; j \neq k \\E_{il}, & \mbox{ if } \; j=k\end{array}\right.$$

Therefore, if $i\not=j, \; E_{ij}=E_{i1}E_{1j}.$ Then by hypothesis,

$$f(E_{ij})=f(E_{i1}E_{1j})=f(E_{1j}E_{i1})=f(o)=o,$$

and

$$f(E_{ii})=f(E_{i1}E_{1i})=f(E_{1i}E_{i1})=f(E_{11})$$

for each $1 \leq i \leq n.$

Now, let $C=\sum_{1\leq i,j \leq n}a_{i,j}E_{ij}$ be a vector in the above basis, then

$$f(C)=\sum_{1\leq i,j \leq n}a_{ij}f(E_{ij})=\sum_{i=1}^n a_{ii}f(E_{11})=f(E_{11})\sum_{i=1}^n a_{ii}=f(E_{11})tr(C).$$

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Yeah, we are given a function on square matrices of a fixed size, call it $f,$ with three properties, square matrices $A,B$ and constant $c.$ So: $$f(A + B ) = f(A) + f(B),$$ $$f(AB) = f(BA),$$ $$f(cA) = c f(A).$$

As Paul pointed out, the notation $e_{ij}$ means the matrix with a 1 at position $ij$ and 0 everywhere else.

There is some value for $f(e_{11}).$ E do not know what that is.

First,for some $i \neq 1,$ define $$S_i = e_{i1} + e_{1i}$$ The main thing is that $$S_i e_{11} S_i = e_{ii}$$ and $$S_i^2 = I.$$ So $$f(e_{ii}) = f(S_i (e_{11} S_i)) = f( (e_{11} S_i) S_i) = f( e_{11} S_i^2) = f(e_{11}).$$

Next, with $i \neq j,$ we use $$e_{ii} e_{ij} = e_{ij}$$ while $$e_{ij} e_{ii} = 0,$$ the matrix of all 0's.

Begin with any $B,$ $$f(0) = f(0B) = 0 f(B) = 0.$$

Now, for any $i \neq j,$ $$f(e_{ij}) = f(e_{ii} e_{ij}) = f(e_{ij} e_{ii}) = f(0) = 0.$$

Finally, if the entries of $A$ are $A_{ij},$ we have $$A = \sum_{i,j = 1}^n A_{ij} e_{ij},$$ so $$f(A) = f(\sum_{i,j = 1}^n A_{ij} e_{ij}) = \sum_{i,j = 1}^n A_{ij} f(e_{ij}) = \sum_{i=1}^n A_{ii} f(e_{ii}) = \sum_{i=1}^n A_{ii} f(e_{11}) = f(e_{11}) \sum_{i=1}^n A_{ii} = f(e_{11}) \mbox{trace} A$$

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