logo

Orthogonal Triangular Matrices are Nilpotent 📂Matrix Algebra

Orthogonal Triangular Matrices are Nilpotent

Theorem1

n×nn \times n Upper triangular matrix AA is a nilpotent matrix.

Explanation

The converse is not true. A simple counterexample is when A=[1111]A = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix},

A2=[1111][1111]=[0000] A^{2} = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

The method of proof is identical, so only the case for upper triangular matrices is discussed.

Proof

We prove this by mathematical induction.

  • When n=1n=1, it holds.

    Let AA be a 1×11 \times 1 upper triangular matrix. A=[0] A = \begin{bmatrix} 0 \end{bmatrix} It is trivially nilpotent.

  • Assume it holds for n=kn=k, then for n=k+1n=k+1, it holds.

    Let AA be a (k+1)×(k+1)(k+1) \times (k+1) upper triangular matrix. Then, for the k×kk \times k upper triangular matrix BB, AA is represented as the following block matrix. A=[B[a1k+1a2k+1akk+1][000][0]]=[BCO1k[0]] A = \begin{bmatrix} B & \begin{bmatrix} a_{1k+1} \\ a_{2k+1} \\ \vdots \\ a_{kk+1} \end{bmatrix} \\ \begin{bmatrix} 0 & 0 & \cdots & 0 \end{bmatrix} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} B & C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} Then, calculating the power of AA yields the following. A2=[BCO1k[0]][BCO1k[0]]=[B2BCO1k[0]]A3=[B2BCO1k[0]][BCO1k[0]]=[B3B2CO1k[0]]Ap+1=[Bp+1BpCO1k[0]] \begin{align*} A^{2} &= \begin{bmatrix} B & C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} \begin{bmatrix} B & C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} B^{2} & BC \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} \\ A^{3} &= \begin{bmatrix} B^{2} & BC \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} \begin{bmatrix} B & C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} B^{3} & B^{2}C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} \\ \vdots & \\ A^{p+1} &= \begin{bmatrix} B^{p+1} & B^{p}C \\ O_{1k} & \begin{bmatrix} 0 \end{bmatrix} \end{bmatrix} \end{align*} Let Bp=OkkB^{p} = O_{kk}. Then since Ap+1=Ok+1k+1A^{p+1} = O_{k+1k+1}, the n=k+1n=k+1 upper triangular matrix is nilpotent.


  1. Stephen H. Friedberg, Linear Algebra (4th Edition, 2002), p512 ↩︎