Trace of the exterior power as a determinant

Question

Let $A$ be a matrix. According to Wikipedia, $$tr(\wedge^k A) = \frac{1}{k!} \det \begin{pmatrix} tr (A) & k-1 & 0 & \cdots \\ tr (A^2) & tr (A) & k-2 & \cdots \\ \cdots & \cdots & \cdots & \cdots \\ tr(A^{k-1}) & tr(A^{k-2}) & \cdots & 1 \\ tr(A^k) & tr(A^{k-1}) & \cdots & tr A \end{pmatrix}$$ Where can I find a proof of this fact?

Answer 1

This should follow from Newton's identities and the fact that the trace of a matrix is the sum of its eigenvalues.

The sum of the (generalized) eigenvalues of $\Lambda^k A$ is the sum all products of $k$ eigenvalues of $A$, i.e. the elementary symmetric polynomials in the eigenvalues.

The trace of $A^n$ is the sum of $n$th powers of eigenvalues of $A$. So the determinant of the thing on the right is the sum of products of the power sum polynomials in the eigenvalues.

Then, try to show that the determinant on the RHS is what you get from applying Newton's identities recursively: http://en.wikipedia.org/wiki/Newton%27s_identities

(Disclaimer: I haven't worked out the details, but this is how I think about these things so feel free to yell at me if I am wrong)