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I know that may be trivial, but I don't know how to prove that if $$w(\lambda)=(\lambda-\lambda_1)^{r1}\cdot \ldots \cdot (\lambda-\lambda_{k})^{r_k}$$ is a characteristic polynomial of matrix $A\in M_{n \times n}(K)$ where $K=\mathbb{R} \ \vee \ K=\mathbb{C}$ then tr$(A)=r_{1}\lambda_1+_{\cdot \cdot \cdot}+r_k\lambda_k$ and that if $$(\forall \ k\in \left \{ 1,2,... \right \}) \ \text{tr}(A^k)=0$$ then $A$ is singular.

Any hints appreciated.

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This is wrong over, say, the field $\mathbb{F}_p$. Take $A=I_p$, the identity matrix of size $p$. Clearly $tr(A^k)=p=0$ in $\mathbb{F}_p$, but $A$ is not singular. –  Dietrich Burde May 13 '14 at 14:03
For the first question use that the characteristic polynomial is $\omega(\lambda)=\lambda^n+\cdots+\mathrm{det(A)}.$ –  mfl May 13 '14 at 14:08
See "properties" of the characteristic polynomial. If $tr(A^k)=0$ for all $k\le n$ then the characteristic polynomial of $A$ is $t^n$, hence $A$ satisfies $A^n=0$ and is singular. –  Dietrich Burde May 13 '14 at 14:11
@DietrichBurde: But $tr(A)^k$ where $A=I_p$ always equals to $p$ and $p$ is greater than $0$. –  Mateusz May 13 '14 at 14:13
@DietrichBurde: Sorry for not including it in the question earlier, but the question is about matrices over $\mathbb{C}$ and $\mathbb{R}$. –  Mateusz May 13 '14 at 14:19

2 Answers 2

up vote 3 down vote accepted

From the definition of the characteristic polynomial as $\det(I_nX-A)$, it is clear that its coefficient of $X^{n-1}$ is $\def\tr\{\operatorname{tr}}-\tr A$: to get a contribution of degree $n-1$ in $X$ in the Leibniz formula for the determinant you need to pick up those $n-1$ factors $X$ on the diagonal, and the remaining factor will be a $-A_{i,i}$ on the diagonal in the same entry as the one factor$~X$ that was not chosen (this is summed over all$~i$). Also it is clear that the product $(X-\lambda_1)^{r_1}\ldots(X-\lambda_k)^{r_k}$ has a term $(-r_1\lambda_1-\cdots-r_k\lambda_k)X^{n-1}$.

As for the part about traces of $A^k$, these give the sums over the $k$-th powers of the eigenvalues $\lambda_i$, each $k$-th power $\lambda_i^k$ being taken with the multiplicity $r_i$ of its eigenvalue; call this the $k$-th power sum $p_k$ of the eigenvalues. Now if the power sums $p_1,\ldots,p_n$ are all zero, then it follows in characteristic$~0$ from Newton's identities that the elementary symmetric polynomials $e_1,\ldots,e_n$ of the (multiset of) eigenvalues are also all$~0$, and since these are up to a sign the non-leading coefficients of the characteristic polynomial, that polynomial is $X^n$. In particular the constant term of the characteristic polynomial, which is $\det(-A)$, is zero (if $n>0$), so $A$ is singular. In fact one can say more, namely that $A$ is nilpotent.

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A common approach is as follows:

First, note that tr$(SAS^{-1}) =$tr$(A)$. From there, select an $S$ that puts $A$ into some upper triangular form (e.g. find the Schur triangularization of $A$). It is clear how the trace of an upper-triangular matrix relates to its eigenvalues.

As for the second part: you would need Newton's identities.

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