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I looked at https://math.stackexchange.com/a/537632/582205 to help understand why/how trace can be both the sum of the diagonal elements in a matrix and the sum of eigenvalues.

In the specific response that Rob Arthan gives - a polynomial of the form $$f(x) = x^n + a_{n-1}x^{n-1} + ... $$ with roots $r_1, r_2, ...,r_n$ has coefficients of the form $$a_{n-1} = -(r_1 + r_2 + ... + r_n)$$

Did he mean that the function $f(x)$ only has one coefficient that isn't equal to 1 meaning $a_i = 1$ for all $i \neq n-1$?

If not, how does this make sense/ how do I prove this? I feel like this would be incredibly powerful since you can just taylor expand a function and then find its roots.

I have some (rather pathetic) start:

We'll evaluate at one of the roots (say $r_1$), so $f(x = r_1) = 0$

$$f(x = r_1) = x^n + a_{n-1}x^{n-1} + ... + a_0 = 0$$ $$a_{n - 1}r_1^{n-1} = -{\sum_{i = 0\\i \neq n-1}^{n}a_ir^i}$$ $$a_{n - 1} = -\frac{1}{r_1^{n_1}}{\sum_{i = 0\\i \neq n-1}^{n}a_ir^i}$$

Where do I go from here? Is there a better way to prove this?

(This kind-of little off topic but why does trace equal both the sum of the diagonals of any matrix and the sum of eigenvalues? I was taught trace was equal to sum of diagonals on a lower/upper triangular matrix(so the diagonal entries are eigenvalues).

FafaDog
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    This is the first of Vieta's formulas: https://en.wikipedia.org/wiki/Vieta%27s_formulas As for the question about trace, this amounts to proving that the trace is invariant under conjugacy and there are several ways to do this. The worst one is an unenlightening calculation. Others require more sophistication or different definitions of trace. – Qiaochu Yuan Oct 14 '20 at 04:28
  • What do you mean by conjugacy? – FafaDog Oct 14 '20 at 04:58
  • I mean the operation $X \mapsto AXA^{-1}$, which you might also know as similarity. So to say that the trace is invariant under conjugacy means $\text{tr}(X) = \text{tr}(AXA^{-1})$. Actually a slightly stronger statement is true: $\text{tr}(XY) = \text{tr}(YX)$. Either of these can be used to prove that the trace is the sum of the eigenvalues. – Qiaochu Yuan Oct 14 '20 at 05:40

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He doesn't mean that all of the other coefficients are $1$. For instance, the constant coefficient is the product of the roots/eigenvalues, and thus the determinant of the matrix (the negative of the determinant when $n$ is odd).

The matrix is, by the Jordan canonical form, similar to an upper triangular matrix with the eigenvalues on the diagonal. Similar matrices have the same characteristic polynomial, trace and determinant.

Elementary symmetric polynomials describe this situation, as well as Vieta's formulas.

Try writing $p(x)=(x-r_1)\dots(x-r_n)$, and expanding it.