# Expansion of the Characteristic polynomial via exterior product

Looking at wiki, I want to

$$p_A (t) = \sum_{k=0}^n t^{n-k} (-1)^k \operatorname{tr}(\Lambda^k A)$$

where $$\operatorname {tr} (\Lambda ^{k}A)={\frac {1}{k!}}{\begin{vmatrix}\operatorname {tr} A&k-1&0&\cdots &\\\operatorname {tr} A^{2}&\operatorname {tr} A&k-2&\cdots &\\\vdots &\vdots &&\ddots &\vdots \\\operatorname {tr} A^{k-1}&\operatorname {tr} A^{k-2}&&\cdots &1\\\operatorname {tr} A^{k}&\operatorname {tr} A^{k-1}&&\cdots &\operatorname {tr} A\end{vmatrix}}~.}$$

Wikipedia states this can be proven via the language of exterior algebra, How?

• What are you looking for? A proof of the formula for $\mathrm{tr}(\Lambda^k A)$ using exterior algebra, or a proof of the formula for $p_A(t)$? – Nick Jan 13 at 6:44
• @Nick. I want to know both, but the purpose of this post is to know the former one. – phy_math Jan 13 at 11:46

Let $$v_1,\dots,v_n$$ be a basis of $$V$$, consisting of eigenvectors with eigenvalues $$\lambda_1,\dots,\lambda_n$$. Then $$\mathrm{tr}(A) = \sum \lambda_i$$.
As a basis for $$\Lambda^k V$$, you can take the collection of all $$v_{i_1} \wedge \cdots \wedge v_{i_k}$$, where $$i_1 < \cdots < i_k$$. Since by definition, $$(\Lambda^k A)(v_{i_1} \wedge \cdots \wedge v_{i_k}) = (Av_{i_1}) \wedge \cdots \wedge (Av_{i_k}) = \lambda_{i_1} \cdots \lambda_{i_k} \, v_{i_1} \wedge \cdots \wedge v_{i_k}$$ you see that this basis also consists of eigenvectors, with corresponding eigenvalues $$\lambda_{i_1} \cdots \lambda_{i_k}$$. So the trace of $$\Lambda^k A$$ is the sum of its eigenvalues, giving $$\mathrm{tr}(\Lambda^k A) = \sum_{i_1 < \cdots < i_k} \lambda_{i_1} \cdots \lambda_{i_k}$$ If you are familiar with symmetric polynomials, this is the "elementary symmetric polynomial" $$e_k(x_1,\dots,x_n)$$, evaluated at $$x_i = \lambda_i$$ for all $$i$$.
On the other hand, $$\mathrm{tr}(A^k) = p_k(\lambda_1,\dots,\lambda_n)$$, where these are the "power sum symmetric polynomials" $$p_k(x_1,\dots,x_n) = x_1^k + x_2^k + \cdots + x_n^k$$.