answersLogoWhite

0

The trace of an nxn matrix is usually thought of as the sum of the diagonal entries in the matrix. However, it is also the sum of the eigenvalues. This may help to understand why the proof works. So to answer your question, let's say A and B are matrices and A is similar to B. You want to prove that Trace A=Trace B If A is similar to B, there exists an invertible matrix P such that A=(P^-1 B P) Now we use the fact that Trace (AB)= Trace(BA) for any nxn matrices A and B.This is easy to prove directly from the definition of trace. (ask me if you need to know) So using this we have the following: Trace(A)=Trace(P^-1 B P)=Trace (BPP^-1)=Trace(B) and we are done! Dr. Chuck

User Avatar

Wiki User

16y ago

Still curious? Ask our experts.

Chat with our AI personalities

CoachCoach
Success isn't just about winning—it's about vision, patience, and playing the long game.
Chat with Coach
MaxineMaxine
I respect you enough to keep it real.
Chat with Maxine
TaigaTaiga
Every great hero faces trials, and you—yes, YOU—are no exception!
Chat with Taiga

Add your answer:

Earn +20 pts
Q: Prove that trace of the matrix is invariant under similarity transformation?
Write your answer...
Submit
Still have questions?
magnify glass
imp