Page 230 - Linear Algebra Done Right

Trace
No one knows an exact formula for any of the eigenvalues of this op-
erator. However, we do know that the sum of the eigenvalues equals 0
because the sum of the diagonal entries of the matrix above equals 0. 221
The theorem above also allows us easily to prove some useful prop-
erties about traces of operators by shifting to the language of traces
of matrices, where certain properties have already been proved or are
obvious. We carry out this procedure in the next corollary.
10.12 Corollary: If S, T ∈L(V), then

trace(ST) = trace(TS) and trace(S + T) = trace S + trace T.

Proof: Suppose S, T ∈L(V). Choose any basis of V. Then

trace(ST) = trace M(ST)

= trace M(S)M(T)

= trace M(T)M(S)
= trace M(TS)
= trace(TS),

where the first and last equalities come from 10.11 and the middle
equality comes from 10.9. This completes the proof of the first asser-
tion in the corollary.
To prove the second assertion in the corollary, note that

trace(S + T) = trace M(S + T)

= trace M(S) +M(T)
= trace M(S) + trace M(T)
= trace S + trace T,

where again the first and last equalities come from 10.11; the third
equality is obvious from the definition of the trace of a matrix. This
completes the proof of the second assertion in the corollary.

The techniques we have developed have the following curious corol-
lary. The generalization of this result to infinite-dimensional vector
spaces has important consequences in quantum theory.