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Let B be a nilpotent n×n matrix with complex entries. Set A = B- I. Find the determinant of A. Please someone give a hint..

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marked as duplicate by Mike Pierce, Martin Argerami, Namaste, user1551 matrices Aug 21 '16 at 17:59

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Nilpotent matrices have all eigenvalues being equal to zero (see here). So, for any eigenvector $v_i$ of $B$ we have that $Bv_i = \vec 0$. Note that $B = A+I$, so we can write that $Av_i+Iv_i = \vec 0\implies A v_i = -v_i$. This holds for any eigenvector $v_i$, so it follows that each eigenvalue of $A$ is $-1$. The determinant is the product of the eigenvalues, so $$\operatorname{det}A = \prod_{i = 1}^n\lambda_i = (-1)^n$$

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  • $\begingroup$ Thanks a lot. I was thinking to solve it with the expansion of $(B- I)^n$, so got confused. Eigenvalues method is much revealing. $\endgroup$ – user362331 Aug 22 '16 at 5:30

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