Algebra 441255.There is a more general property, of which the problem is a particular case.
Riesz lemma.IfVis a finite-dimensional vector space with inner product〈·,·〉,then
any linear functionalf:V→Ris of the formf(x)=〈x, z〉for some uniquez∈V.
This result can be generalized to any (complex) Hilbert space, and it is there where
it carries the name of F. Riesz.
We prove it as follows. Iffis identically zero, thenf(x)=〈x, 0 〉. Otherwise, let
Wbe the kernel off, which has codimension 1 inV. There exists a nonzero vector
yorthogonal toW such thatf(y)=1. Setμ=〈y, y〉and definez=μ−^1 y. Then
〈z, z〉=μ−^1. Any vectorx∈Vis of the formx′+λz, withx′∈W. We compute
f(x)=f(x′)+λf (z)=λμ−^1 =λ〈z, z〉=〈x′,z〉+λ〈z, z〉=〈x, z〉.Note thatzis unique, because if〈x, z〉=〈x, z′〉for allx, thenz−z′is orthogonal to all
vectors, hence is the zero vector. There exists a simpler proof, but the one we gave here
can be generalized to infinite-dimensional Hilbert spaces!
For our particular case,V =Mn(R)and the inner product is the famous Hilbert–
Schmidt inner product〈A, B〉=tr(ABt).
For the second part of the problem, the condition from the statement translates to
tr((AB−BA)C)=0 for all matricesAandB. First, let us show that all off-diagonal
entries ofCare zero. Ifcijis an entry ofCwithi =j, letAbe the matrix whose entry
aikis 1 and all others are 0, andBthe matrix whose entrybkjis 1 and all others are 0, for
some numberk. Then tr((AB−BA)C)=cij=0. SoCis diagonal. Moreover, choose
aij=bij=1, withi =j. ThenAB−BAhas two nonzero entries, the(i, i)entry, which
is 1, and the(j, j )entry, which is−1. Therefore, tr((AB−BA)C)=cii−cjj=0. We
deduce that all diagonal entries ofCare equal to some numberλ, and hence
f (A)=tr(AC)=tr(λA)=λtr(A),as desired.
Remark.The conditionf (AB)=f(BA)gives
tr(AC)=f (A)=f (ABB−^1 )=f(B−^1 AB)=tr(B−^1 ABC)=tr(ABCB−^1 );hence by uniqueness ofC, we have shown thatC=BCB−^1 for allB,orBC=CB.
The solution of the problem is essentially a proof that ifCcommutes with all invertible
matricesB, thenC=λInfor some scalarλ.
256.Fixx∈Rnwith‖x‖=1, and lety=U−^1 V−^1 x. BecauseUandVare isometric
transformations,‖y‖=1. Then
‖UVU−^1 V−^1 x−x‖=‖UVy−VUy‖