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I found the following proof that the trace of a square matrix is equal to the sum of its eigenvalues (see the second half of page 3):

https://www.adelaide.edu.au/mathslearning/play/seminars/evalue-magic-tricks-handout.pdf

The first half of page 3 contains a proof that show that the determinant of a square matrix is equal to the product of its eigenvalues; I understand all of this.

I am having severe difficulty understanding the second half of page 3, which is the proof that the trace of a square matrix is equal to the sum of its eigenvalues.

My immediate difficulty is in understanding the following section:

In order to get the $\lambda^{n−1}$ term, the $\lambda$ must be chosen from $n − 1$ of the factors, and the constant from the other. Hence, the $\lambda^{n−1}$ term will be $-\lambda_1\lambda^{n - 1} - ... - \lambda \lambda^{n - 1} = -(\lambda_1 + ... + \lambda_n)\lambda^{n−1}$. Thus $c_{n−1} = −(\lambda_1 +···+ \lambda_n)$.

I would greatly appreciate it if people could please clarify what this is saying. What is meant by "In order to get the $\lambda^{n−1}$ term, the $\lambda$ must be chosen from $n − 1$ of the factors, and the constant from the other."? What is the entire section trying to say?

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    Hint: if you expand $(x-a)(x-b)(x-c)$, what is the coefficient of $x^2$?2017-02-24
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    @Aretino $(a + b + c)x^2$. I understand now! Thank you for much for the assistance. :)2017-02-24

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