# Finding the determinant of A nilpotent matrix [duplicate]

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..

## marked as duplicate by Mike Pierce, Martin Argerami, Namaste, user1551 matrices StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 21 '16 at 17:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

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 $$\operatorname{det}A = \prod_{i = 1}^n\lambda_i = (-1)^n$$

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