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My question is: "If determinant of $A^2$ is $0$ can we also say that determinant of $A$ is $0$?"

I have tried to argue that it is by saying that if $A^2 = 0$ then $A$ must be $0$ so determinant is also $0$.

user3209698
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anon
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    Your argument is wrong: a matrix may be nilpotent without the matrix being $0$ (example: $\begin{pmatrix}0&1\0&0\end{pmatrix}$. – Bernard Apr 10 '16 at 14:08

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Your first sentence is correct, since you can make use of the property of $\mbox{det}(AB)=\mbox{det}(A)\mbox{det}(B)$, finding $$0=\mbox{det}(A^2)=\mbox{det}(A)^2\to \mbox{det}(A)=0$$ However, your last reasoning is uncomplete, since there are matrices (different from zero) which can have determinants equal to zero. For example $$\left(\begin{array}{} 1&1\\1&1\end{array}\right)$$

user3209698
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    Thank you very much! I understand I was wrong with my reasoning but thank god that property saved me :) – anon Apr 10 '16 at 14:17