Extremely difficult 2nd year linear algebra proof: Let V=M_mn denote the space of all nxn matrices over lR...?

Question

Let V=M_mn denote the space of all nxn matrices over lR. Let T : M_mn → lR denote the trace, that is T(A) = tr(A) for all matrices A∈M_mn. We know that T is a linear transformation with the property that T(AB)=T(BA) for all A, B∈M_mn. If S : M_mn → lR satisfies S(AB)=S(BA) for all A,B∈M_mn, show that S=aT for some a. (here aT is defined by (aT)(A)=aT(A) for all A∈M_mn). 

[Hint: Let E_ij∈M_mn have (i,j)-entry 1 and all other entries 0. you may use the fact that these matrices multiply according to the formula E_ijE_kl={E_il if j=k, 0 if j≠k}. Show that S(E_ij)=0 if i≠j, and S(E_ii)=E_11 for all i.]

I have no idea how to do this proof. Could you please help?

greentunic · Accepted Answer

I'll just write Eij instead of E_ij.  I guess there's a typo in the last part of your hint: since S maps into R, then S(Eii) can't equal E11, which is a matrix.  Instead, I think it should say S(Eii) = S(E11).

First, let's show the two parts in the hint, using the given multiplication formula together with the facts that S(AB) = S(BA) and S is linear.

S(Eij) = S(EikEkj) = S(EkjEik) = S(0) = 0, when i≠j
S(Eii) = S(Ei1E1i) = S(E1iEi1) = S(E11)

Let a = S(E11), which is some real number.

Notice that {Eij : 1 <= i,j <= n} is a basis for what you called "M_mn" (should it say "M_nn"?).  Indeed, if B = (bij) is an arbitrary nxn matrix, then B = ΣbijEij (where the summation runs over all values of i and j from 1 to n).  Notice also that T(B) = tr(B) = Σbii (where i runs through 1,...,n).  Then, using the linearity of S together with the two facts we already showed, we get:

S(B) = S(ΣbijEij) = ΣbijS(Eij) [since S is linear]
= ΣbiiS(Eii) [since S(Eij) = 0 when i≠j]
= ΣbiiS(E11) = Σbii*a = a(Σbii)
= a*tr(B) = aT(B)

which shows that S = aT.

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Extremely difficult 2nd year linear algebra proof: Let V=M_mn denote the space of all nxn matrices over lR...?

2 Answers