The trace of a square matrix \(A\) is the sum of the diagonal entries in A and is denoted by \({\mathop{\rm tr}\nolimits} A\). It can be verified that \({\mathop{\rm tr}\nolimits} \left( {FG} \right) = {\mathop{\rm tr}\nolimits} \left( {GF} \right)\) for any \(n \times n\) matrices Fand G. Show that if A and B are similar, then \({\mathop{\rm tr}\nolimits} A = {\mathop{\rm tr}\nolimits} B\).

The Diagonalization Theoremstates that a \(n \times n\) matrix \(A\) isdiagonalizable such that if \(A\) contains \(n\) linearly independent eigenvectors.

In addition, \(A = PD{P^{ - 1}}\), where \(D\) is a diagonal matrix, such that if the columns of \(P\) are \(n\)linearly independent eigenvectorsof \(A\) . Accordingly, the eigenvalues of \(A\)are the diagonal entries of \(D\) which correspond to the eigenvectors in \(P\).

When \(A = PB{P^{ - 1}}\) then

\(\begin{aligned}{}{\mathop{\rm tr}\nolimits} \left( A \right) &= {\mathop{\rm tr}\nolimits} \left( {\left( {PB} \right){P^{ - 1}}} \right)\\ &= {\mathop{\rm tr}\nolimits} \left( {{P^{ - 1}}\left( {PB} \right)} \right)\left( {{\mathop{\rm From}\nolimits} \,\,\,{\mathop{\rm the}\nolimits} \,\,{\mathop{\rm trace}\nolimits} \,\,{\mathop{\rm property}\nolimits} } \right)\\ &= {\mathop{\rm tr}\nolimits} \left( {{P^{ - 1}}PB} \right)\\ &= {\mathop{\rm tr}\nolimits} \left( {IB} \right)\\ &= {\mathop{\rm tr}\nolimits} \left( B \right)\end{aligned}\)

If \(B\) is diagonal, then the diagonal entries of \(B\) should be eigenvalues of A, according to the Diagonalization Theorem.

Therefore, \({\mathop{\rm tr}\nolimits} A = {\mathop{\rm tr}\nolimits} B = \left( {{\mathop{\rm sum}\nolimits} \,\,{\mathop{\rm of}\nolimits} \,\,{\mathop{\rm the}\nolimits} \,\,{\mathop{\rm eigenvalues}\nolimits} \,\,{\mathop{\rm of}\nolimits} \,\,A} \right)\) .

Thus, it is proved that \({\mathop{\rm tr}\nolimits} A = {\mathop{\rm tr}\nolimits} B\).