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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
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The AA Similarity Postulate states that two triangles are similar if they have?
You would use the AA Similarity Postulate to prove that the following two triangles are similar. True or false?
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