# Math Help - Trace problem

1. ## Trace problem

Prove that if A and B are similar nxn matrices, then trace of (A) = trace of (B).

Proof.

Write $A = Q^{-1}BQ$

Then we have QA = BQ, tr(QA) = tr(BQ), since trace is commutative, I have tr(AQ) = tr(BQ).

Now, can I say that tr(A) = tr(B)?

2. Remember that the determinant of a matrix is equal to the sum of its eigenvalues.

So looking at

$A-sI=Q^{-1}BQ-sQ^{-1}Q=Q^{-1}(B-sI)Q$

it's evident that $A$ and $B$ have the same eigenvalues, hence equal trace.

3. our class haven't got to engenvalue yet, is there another way to do it?

Thanks!

Prove that if A and B are similar nxn matrices, then trace of (A) = trace of (B).

Proof.

Write $A = Q^{-1}BQ$

Then we have QA = BQ, tr(QA) = tr(BQ), since trace is commutative, I have tr(AQ) = tr(BQ).

Now, can I say that tr(A) = tr(B)?
Remember that the trace is cyclic:
$tr(A) = tr(Q^{-1}BQ) = tr(QQ^{-1}B) = tr(IB) = tr(B)$

-Dan