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Let $A$ be a $n\times n$ matrix with complex entries. Suppose that $\operatorname{tr}(A)=\operatorname{tr}(A^{2})=\cdots \ =\operatorname{tr}(A^{n-1})=0 $, and $A^{n}\neq 0 $. Then $A$ is diagonalisable.

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Is it a homework problem? What did you try? – Davide Giraudo Dec 27 '11 at 19:16
Hint: What is the characteristic polynomial of $A$? – David Speyer Dec 27 '11 at 19:33
@Davide This is not a homework; I try to do some exercises – WLOG Dec 27 '11 at 22:44
up vote 5 down vote accepted

We have $\mathrm{Tr}(A^n)\neq 0$, otherwise $\operatorname{Tr}(A^k)=0$ for all $1\leq k\leq n$ and $A$ would be nilpotent. Let $p:=\operatorname{card}\operatorname{Sp}(A)$. We will show that $p=n$. Let $\lambda_i$, $1\leq i\leq p$, the eigenvalues of $A$, $m_i\geq 1$ their multiplicities. Using the fact that $\operatorname{Tr}(A^k)=0$ for all $1\leq k\leq p$, we have $$\pmatrix{1&1&\cdots& 1&1\\\ \lambda_1&\lambda_2&\cdots&\lambda_{p-1}&\lambda_p\\\ \vdots&\vdots&&\vdots&\vdots\\\ \lambda_1^{p-1}&\lambda_2^{p-1}&\cdots&\lambda_{p-1}^{p-1}&\lambda_p^{p-1}}\pmatrix{m_1\lambda_1\\\ m_2\lambda_2\\\ \vdots \\\ m_p\lambda_p }=\pmatrix{0\\\ 0\\\ \vdots\\\ 0},$$ but since $\lambda_i\neq\lambda_j$ if $i\neq j$, the determinant of this matrix is $\neq 0$, so $m_j\lambda_j=0$ for all $j$. It implies $p=1$ and $\lambda_1=0$. But we would have $\operatorname{Tr}(A^n)=0$. So $p=n$ and we have $n$ distinct eigenvalues: $A$ is diagonalisable.

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what does cardSp mean? is this standard notation? – yohBS Jan 3 '12 at 22:41
I don't know if it's a standard notation, but it's just the cardinality of the spectrum. – Davide Giraudo Jan 3 '12 at 22:49
very nice answer . – GA316 Jul 28 '14 at 10:14

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