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I am trying to prove a more simple analog of a determinant property. The main one being that if $A$ is an $n \times n$ square matrix with characteristic polynomial $\Delta (t) = t^n+c_{n-1}t^{n-1}+\cdots+c_0$ than $c_{n-1}=-\operatorname{trace}(A)$.

I want to try to prove the property that $-c_{n-1}=$ sum of the eigenvalues.

However I am just confused on how I can do this, I know I should consider the expansion of the polynomial ie that it can be written as $(t-\lambda_1) \cdots (t-\lambda_n)$ but I am just really confused on how I can prove this. Any help please? Thank you

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Think what happens when you when you open the brackets in $$(t-\lambda_1)\cdots(t-\lambda_n)$$

You get lots of products ($2^n$ to be precise), where each product is made of one term from each bracket.

The only way to get $t^n$ is by picking one $t$ from each bracket, that's why the first term is exactly $t^n$.

Now, to get $t^{n-1}$ you need $n-1$ $t's$ from $n$ brackets. That means picking $t$'s from all but one bracket, and the other term from the remaining bracket.

When you pick the $t's$ from all but first bracket, you get $(-\lambda_1)t^{n-1}$. When you pick the $t's$ from all but second bracket, you get $(-\lambda_2)t^{n-1}$. ...When you pick the $t's$ from all but last bracket, you get $(-\lambda_n)t^{n-1}$.

This means that all the terms with $t^{n-1}$ are $$ (-\lambda_1)t^{n-1}+(-\lambda_2)t^{n-1}+\cdots+(-\lambda_n)t^{n-1} $$

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The eigenvalues are, by definition, the roots of the caracteristic polynnomial. So: $$ \Delta(t)= (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n) $$ where $\lambda_i$ with multiplicity $>1$ are repeated. And if you perform multiplication you see that $-c_1$ is the sum of all eigenvalues, each repated for his multiplicity.

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