Page 228 - Linear Algebra Done Right

Trace

trace T = trace M T, (v 1 ,...,v n ) , where (v 1 ,...,v n ) is an arbitrary
basis of V. We already know this is true if (v 1 ,...,v n ) is a basis with
respect to which T has an upper-triangular matrix (if V is complex) or 219
an appropriate block upper-triangular matrix (if V is real). We will need
the following proposition to prove our trace formula for an arbitrary
basis.
10.9 Proposition: If A and B are square matrices of the same size,
then
trace(AB) = trace(BA).
Proof: Suppose
   
a 1,1 ... a 1,n b 1,1 ... b 1,n
 . .   . . 
A =  . . . .  , B =  . . . .  .
   
a n,1 ... a n,n b n,1 ... b n,n
The j th term on the diagonal of AB equals
n

a j,k b k,j .
k=1
Thus
n n

trace(AB) = a j,k b k,j
j=1 k=1
n n

= b k,j a j,k
k=1 j=1
n

th
= k term on the diagonal of BA
k=1
= trace(BA),
as desired.

Now we can prove that the sum of the diagonal entries of the matrix
of an operator is independent of the basis with respect to which the
matrix is computed.

10.10 Corollary: Suppose T ∈L(V).If (u 1 ,...,u n ) and (v 1 ,...,v n )
are bases of V, then

trace M T, (u 1 ,...,u n ) = trace M T, (v 1 ,...,v n ) .