Why Is Nilpotent Singular?

Alexander Ross
Alexander Ross
• • 6 min read

Theorem: Let a Be a Square Matrix. Furthermore, Let a Be Nilpotent, That Is, Ak=0 for Some Natural Number K. Then, a Is Singular, That Is, |A|=0. .. . Hence...

Why Is Nilpotent Singular?

Theorem: Let A be a square matrix. Furthermore, let A be nilpotent, that is, Ak=0 for some natural number k. Then, A is singular, that is, |A|=0. ... Hence, we cannot have an invertible matrix

invertible matrix
A is invertible, that is, A has an inverse, is nonsingular, or is nondegenerate. A is row-equivalent to the n-by-n identity matrix In. A is column-equivalent to the n-by-n identity matrix In. ... In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring.
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which is nilpotent.

How do you prove that a nilpotent matrix is singular?

Proof 1. We use the fact that a matrix is nonsingular if and only if its determinant is nonzero. 0=det(O)=det(Am)=det(A)m. This implies that det(A)=0, and hence the matrix A is singular.

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Are all nilpotent matrices singular?

Every singular matrix can be written as a product of nilpotent matrices. A nilpotent matrix is a special case of a convergent matrix.

Alexander Ross

Alexander Ross

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Alexander Ross has covered the video game industry for a decade, writing deep dives on game design, esports tournaments, VR developments, and gaming culture.