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Let $A∈M_{5×6}(R)$ and $B∈M_{6×5}(R)$ such that $(AB)^{100}=0$. Which of the following statements is not necessarily true?

(a) $(AB)^{4}=0$

(b) $(BA)^{6}=0$

(c) $\det(BA)=0$

(d) $\operatorname{tr}(AB)=0$

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:what are you trying ? –  Maisam Hedyelloo Feb 16 '13 at 17:34
    
we just saw this one exactly –  Ross Millikan Feb 16 '13 at 17:38
    
@usermath, please stop flooding the front page with minor edits of old questions. Do three or four a day, not 20 in an hour, please. –  Gerry Myerson Jun 12 at 7:16
    
@GerryMyerson Extremely sorry..I will keep in mind.Thnx for the advice.Sorry again. –  usermath Jun 12 at 7:25

1 Answer 1

up vote 7 down vote accepted

(a) need not be true since $$A=\left( \begin{matrix} 0& 1& 0 & 0 & 0 & 0 \\ 0& 0& 1 & 0 & 0 & 0 \\ 0& 0& 0& 1 & 0 & 0 \\ 0 & 0 & 0& 0& 1 &0 \\ 0 & 0& 0& 0& 0& 0 \\ \end{matrix}\right) \mathrm{\qquad and \qquad} B=\left( \begin{matrix} 1& 0& 0 & 0 & 0 \\ 0& 1& 0 & 0 & 0 \\ 0& 0& 1& 0 & 0 \\ 0 & 0 & 0& 1& 0 \\ 0 & 0& 0& 0& 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right)$$ satisfy $(AB)^4\ne 0 $ but $(AB)^5= 0 $.

Now the characteristic polynomial of $AB$ $\ $is of degree $\le 5$ and divides $X^{100} $ so it must equal $X^5$. (b) is true since $(BA)^6=B(AB)^5A=0$ by Cayley-Hamilton. (c) is true since the determinant of $AB$ is $\pm$ the constant term of the characteristic polynomial of $AB$, and (d) is true since tr$(AB)$ is $-({\mathrm{ coefficient\ of \ }}X^4)$.

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