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).
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.