Can equation of trace of a matrix be deduced from properties of trace???

sto

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Sep 29, 2010, 10:50:02 AM9/29/10

to

Let A,B be nxn matrices over a field F and define a function tr on the
nxn matrices by the relation tr(A) = sum_i a_ii.

It is simple to show that, for u,v in F,

tr( u A + v B ) = u tr(A) + v tr(B)

and that

tr(AB) = tr(BA)

in which case, if PP' = P'P = I and C = P' A P, then

tr(C) = tr(P' {A P})
= tr({AP} P')
= tr(AI)
= tr(A)

so that similar matrices have the same trace. Furthermore, it is clear that

tr(I) = sum_i (1) = n

However, if I instead define tr as a linear map on the nxn matrices
having the property that tr(AB) = tr(BA) and tr(I) = n, how does that
imply that the trace must satisfy the relation tr(A) = sum_i (a_ii)? In
other words, how do you prove that the trace is the only linear function
having these two properties?

I suspect this has something to do with the linearity of tr and the fact
that a linear function is fully specified by its action on a basis. But
I don't see how the relation tr(I) = n would be enough to solve the
problem that way. Recall also that not all matrices are similar to a
diagonal matrix, so I don't see how similarity can be useful here.

Thanks
-sto

achille

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Sep 29, 2010, 12:05:20 PM9/29/10

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Hint: Let E_ij, i,j = 1..n be the nxn-matrix with a
1 at i-th row, j-th column and 0 at other positions.
For i != j, let P_ij = E_ij + E_ji:

What are the products between E_ij and E_jj ?
What are the products between P_ij and E_ii P_ij ?

Herman Rubin

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Sep 29, 2010, 12:09:25 PM9/29/10

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> and that

> tr(AB) = tr(BA)

> Thanks
> -sto

There is enough information. The matrices similar to a diagonal
matrix form a basis for all matrices, so it enough to
prove it for diagonal matrices. In fact, the matrices
which are 0 except for an element on the main diagonal
and either 0 or another value on the same row or column,
and it is easily seen that the off-diagonal element does
not change the similarity class.

Also, the matrices which are 0 except for one diagonal
element 1 are all similar, so must have the same trace.
Since their sum is I, their trace is 1. The rest follows
easily.

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

quasi

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Sep 29, 2010, 1:18:56 PM9/29/10

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Hints:

Call your new function Tr and the old one tr, say, so as not to
confuse things. You want to prove Tr = tr.

As you noted, Tr is linear on the space of n x n matrices so it
suffices to prove Tr = tr on a basis.

Consider the set S of n x n matrices with exactly one entry equal to 1
and the rest 0.

First look at those elements of S where the entry of 1 is on the
diagonal. Note that they are all similar to each other, hence Tr gives
the same value on all them. Deduce that Tr must yield a value of 1 for
each of them.

Next look at those elements of S where the entry of 1 is off the
diagonal. Pick one, A, say. Find a matrix B such that AB = 0 but BA =
A. Deduce that Tr(A) is zero.

quasi