Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 11 down vote accepted

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 $$ f(e_{jj})=f(e_{jj})+0=f(e_{jj})+f([e_{ij},e_{ji}]=f(e_{jj}+(e_{ii}-e_{jj}))=f(e_{ii}). $$
share|cite|improve this answer

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,




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

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

share|cite|improve this answer

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 $$ f(A) = f(\sum_{i,j = 1}^n A_{ij} e_{ij}) = \sum_{i,j = 1}^n A_{ij} f(e_{ij}) = \sum_{i=1}^n A_{ii} f(e_{ii}) = \sum_{i=1}^n A_{ii} f(e_{11}) = f(e_{11}) \sum_{i=1}^n A_{ii} = f(e_{11}) \mbox{trace} A $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.