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Nilpotent Matrix 📂Matrix Algebra

Nilpotent Matrix

Definition1

$n \times n$ For a matrix $A$, if there exists a positive integer $k$ that satisfies $A^{k} = O$, then we call $A$ a nilpotent matrix. In this case, $O$ is the zero matrix of $n \times n$.

Explanation

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“Nil” means ‘zero’ or ’none.’ “Potent” means ‘powerful,’ and is the root of the word “potential.” Therefore, the term “nilpotent” can be understood as ‘having the potential/power to become $0$.’ In mathematics, ‘冪’ denotes exponentiation and ‘零’ denotes the number $0$. Thus, the word “nilpotent” literally means ‘becoming zero upon exponentiation.’

Theorems

Proof

[1]

Shown by mathematical induction.

[2]

$(\implies)$ 2

Since $A$ is a square matrix, a Schur decomposition exists, and it can be represented as $A = Q T Q^{\ast}$ using some unitary matrix $Q$ and upper triangular matrix $T$. Given that all eigenvalues of $A$ are $0$, $T$ is an upper triangular matrix with all diagonal elements equal to $0$. Since an upper triangular matrix is nilpotent, $T$ is nilpotent for some $k \in \mathbb{N}$ such that $T^{k} = O$. Consequently, $A$ is also nilpotent since $A^{k} = Q T^{k} Q^{*} = O$ holds for at least $k$.

$(\impliedby)$ 3

Let us represent the zero vector as $\mathbf{0}$. $$ \exists k \in \mathbb{N} : A^{k} = O $$ For some eigenvalue $\lambda$ of the nilpotent matrix $A$ and its corresponding eigenvector $\mathbf{v}$, we can set it equal to $\lambda \mathbf{v} = A \mathbf{v}$. By repeatedly multiplying both sides by $A$, the following holds for $k$. $$ \lambda^{k} \mathbf{v} = A^{k} \mathbf{v} = O \mathbf{v} = \mathbf{0} $$ Since this must hold for all eigenvectors $\mathbf{v} \ne \mathbf{0}$, all eigenvalues of $A$ must be $\lambda = 0$.

[3]

Proof omitted4.

See Also