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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.

Mateusz
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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
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    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
  • For the record, this is one more statement that fails for the $0\times 0$ matrix (while valid for all other square matrices). Though that is mainly due to the fact that the conclusion says "singular" where it should have said "nilpotent", and "nilpotent implies singular" fails exactly for the $0\times 0$ matrix. – Marc van Leeuwen May 06 '15 at 07:52

2 Answers2

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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.

Marc van Leeuwen
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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.

Ben Grossmann
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