Nilpotent Matrices

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# Nilpotent Matrices

Submitted by finles on Thu, 04/10/2003 - 12:41

Forums:

Im just new to linear algebra, and am having trouble proving that a nilpotent matrix must be singular. Can anyone give me a hand in the right direction slash provide the proof?

Cheers

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new question: Lorenz system by David Bankom

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new correction: examples and OEIS sequences by fizzie

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new correction: Define Galois correspondence by porton

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new correction: Closure properties on languages: DCFL not closed under reversal by babou

new correction: DCFLs are not closed under reversal by petey

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new correction: Many corrections by Smarandache

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new question: how to contest an entry? by zorba

new question: simple question by parag

## Re: Nilpotent Matrices

Let A be a nilpotent matrix. Then by definition for some natural n, A^n = 0.

Let d = |A| (the determinant). |0| = 0 = |A^n| = |A|^n = d^n. Hence, d = 0 and A is singular.

## Re: Nilpotent Matrices

Another way to think about this is to suppose instead that we had an invertible matrix A such that A^n=0. A product of invertible matrices is invertible, so A^n=0 is invertible, but 0 is not invertible.

## Re: Nilpotent Matrices

Yet another approach. Consider N nilpotent (but non-zero) with n the smallest integer (>1) such that N^n = 0. Assume N is non-singular with inverse M, then N^(n-1) = M N^n = M 0 = 0, but n was suppose to be the smallest integer...proof by contradiction.