Similar matrices and main diagonal summation?

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Homework Help Overview

The discussion revolves around the properties of similar matrices, specifically focusing on the relationship between the sums of the main diagonal elements (traces) of two matrices A and B, where B is derived from A through a similarity transformation involving an invertible matrix P.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of the trace of a matrix and its relationship to the diagonal elements. The original poster attempts to understand whether the diagonal sums of matrices A and B are equal and questions how properties like determinant and eigenvalues relate to this summation.

Discussion Status

Some participants have provided insights into the concept of the trace and its connection to the characteristic polynomial, suggesting that A and B share this property. There is ongoing exploration of how to formally prove the relationship between the traces of A and B.

Contextual Notes

The original poster expresses uncertainty about the proof and is seeking clarification on the relationship between the diagonal sums and other matrix properties. There may be constraints related to the level of detail expected in the homework context.

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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?
 
Last edited:
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##.
 

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