Can find the determinant of a matrix A of size $n$ in terms of the traces of $A^m$

Question

We can find the determinant of a matrix A of size $n$ in terms of the traces of $A^m$, for $m=1,…,n$ ?

It's det of a matrix with term are traces, but i saw but i can't remember

see here: http://math.stackexchange.com/questions/668374/determinant-of-a-finite-dimensional-matrix-in-terms-of-trace — martini, Apr 09 '15 at 13:52

score 1 · Accepted Answer · answered Apr 09 '15 at 13:58

Hint If the eigenvalues of $A$ (counting algebraic multiplicity) are $\lambda_1, \ldots, \lambda_n$, then:

$\det A = \lambda_1 \cdots \lambda_n$;
$\text{tr }A = \lambda_1 + \cdots + \lambda_n$;
the eigenvalues of $A^k$ are $\lambda_1^k, \ldots, \lambda_n^k$, and in particular, $\text{tr}(A^k) = \lambda_1^k + \cdots + \lambda_n^k$.

With these facts in hand, apply Newton's Identities.

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