Trace of outer power Endomorphism eqauls coefficients of characteristic Polynomial

Question

For a Vector Space over the Complexe Numbers and a linear and diagonalizable map $g:V\rightarrow V$ the trace of ${ \wedge }^{ r }g:{ \wedge }^{ r }V\rightarrow { \wedge }^{ r }V$ is given by the Coefficient of the characteristic polynomial ${ x }^{ m }-{ c }_{ 1 }{ x }^{ m-1 }+....+{ (-1) }^{ m }{ c }_{ m }$

In my book the proof is not omitted so i tried my self but im stuck. What I know: Since we have a diagonalizable Representation Matrix we know that V has a basis of Eigenvectors so since a diagonalizable Matrix T always statisfies $T({ v }_{ 1 })\wedge .....\wedge T({ v }_{ m })=det(T){ v }_{ 1 }\wedge ...\wedge { v }_{ m }$ for v1,...vm a Basis of eigenvectors we can conclude that det(T) is equal to the Product of the Eigenvalues What do i have to proceed?

Answer 1

1. First check how $\wedge^k g$ acts on a specific basis of $\wedge^k V$. In particular, use the basis $v_{i_1} \wedge \ldots \wedge v_{i_k}$, where $v_i$ is an eigenbasis of $g$ and $i_1 < i_2 < \ldots < i_k$, so that these are also eigenvectors of $\wedge^k g$.
2. Summing over the eigenvalues of these ordered $k$ gives you the trace of $\wedge^k g$.
3. Next, compute the coefficient of $t^k$ in $(t - \lambda_1) \ldots ( t - \lambda_n)$. Check here if stuck: https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial
4. Compare.

By the way, both sides of this equation, $tr (\wedge^r g)$ and $r$th coefficient of the characteristic polynomial, are continuous functions of the entries of $g$ (actually polynomial functions -- for example, the coefficients of the characteristic polynomial are sums of determinants of certain minors, and the trace of the exterior powers are sums of products of certain elements from $g$)). Since the diagonalizable matrices are dense in the space of matrices, this implies that the formula is true for all matrices. (Over the complex numbers.)