# Thread: Trace problem

1. ## Trace problem

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

Proof.

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

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

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

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

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

Thanks!

4. Originally Posted by tttcomrader
Prove that if A and B are similar nxn matrices, then trace of (A) = trace of (B).

Proof.

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

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:
$\displaystyle tr(A) = tr(Q^{-1}BQ) = tr(QQ^{-1}B) = tr(IB) = tr(B)$

-Dan