Similar matrices and main diagonal summation?

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Homework Statement



True or False? If A is an n × n matrix, P is an n × n invertible matrix, and B = P −1AP, then
a11 + a22 + . . . + ann = b11 + b22 + . . . + bn

Homework Equations


Diagnolization, similar matrixes

The Attempt at a Solution


the question is asking if the summation of the main diagnols of A and B are the same. B is known to be similar to A since B = P^-1 AP. I can't find a counterexample, so I am assuming the summation of the diagnols of both are in fact equal, but this is hardly a proof. I know A and B share the same determinant, rank, and eigenvalues and rank, but I'm not sure how these relate to the main diagnol of the matrices.
 
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Are you familiar with the concept of trace of a matrix?
 
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The key is that A and B have the same characteristic polynomial. The sum of elements on the main diagonal (called the trace of the matrix) is the first-order coefficient of the characteristic polynomial. To prove that you can either use induction or start with the first order Vieta formula and prove that the trace equals the sum of the eigenvalues.
 
The trace of a matrix ##A## is the sum of the diagonal elements, namely ##\textrm{Tr }A = \sum_{i=1}^n A_{ii}##. This means the trace of a product of two matrices will be ##\textrm{Tr }AB = \sum_{i=1}^n \sum_{j=1}^n A_{ij}B_{ji}##.

Since you have a product of three matrices, the the trace reads
$$
\textrm{Tr }ABC = \sum_{i=1}^n \sum_{j=1}^n (AB)_{ij}C_{ji}
$$
Now write ##(AB)_{ij}## in terms of the sum of products between the elements of ##A## and ##B##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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