If Every Trace of a Power of a Matrix is Zero, then the Matrix is Nilpotent

Problem 21

Let $A$ be an $n \times n$ matrix such that $\tr(A^n)=0$ for all $n \in \N$.
Then prove that $A$ is a nilpotent matrix. Namely there exist a positive integer $m$ such that $A^m$ is the zero matrix.

Contents

Steps.

1. Use the Jordan canonical form of the matrix $A$.
2. We want to show that all eigenvalues are zero. (Review Nilpotent matrix and eigenvalues of the matrix)
3. Seeking a contradiction, assume some eigenvalues are not zero.
4. Using  $\tr(A^n)=0$, create a system of linear equations.
5. Calculate the determinant of the coefficient matrix of the system using the Vandermonde matrix formula.

Proof.

We first want to prove that all the eigenvalues of $A$ must be zero.
Seeking a contradiction, assume that some of the eigenvalues of $A$ are not zero.
So assume that $\lambda_i$, $i=1, \dots, r$ are distinct nonzero eigenvalues of $A$ and each $m_i \geq 1$ is a multiplicity of $\lambda_i$.

We use the Jordan canonical form of the matrix $A$.
There exists an invertible matrix $S$ such that $S^{-1}AS=T$, where $T$ is an upper triangular matrix.The diagonal entries are eigenvalues of $A$.

Then we have for any positive integer $n$,
\begin{align*}
0 &= \tr(A^n)=\tr((STS^{-1})^n)=\tr(ST^n S^{-1})=\tr(T^{n}) \\
&=m_1 \lambda_1^n +m_2 \lambda_2^n+ \cdots + m_r \lambda_r^n.
\end{align*}

Note that since $T$ is an upper triangular matrix, the nonzero diagonal entries of $T^n$ are $\lambda_i^n$ appearing $m_i$ times.
Changing $n$ from $1$ to $r$, we obtain the system of linear equation. (Think $m_i$ as variables.)
\begin{align*}
m_1 \lambda_1^1 +m_2 \lambda_2^1+ \cdots + m_r \lambda_r^1 &=0 \\
m_1 \lambda_1^2 +m_2 \lambda_2^2+ \cdots + m_r \lambda_r^2 &=0 \\
& \vdots \\
m_1 \lambda_1^r +m_2 \lambda_2^r+ \cdots + m_r \lambda_r^r &=0 \\
\end{align*}
Equivalently, we have the matrix equation
\begin{align*}
\begin{bmatrix}
\lambda_1^1 & \lambda_2^1 & \cdots & \lambda_r^1 \\
\lambda_1^2 & \lambda_2^2& \cdots & \lambda_r^2 \\
\vdots & \vdots & \vdots & \vdots \\
\lambda_1^r & \lambda_2^r&\cdots & \lambda_r^r \\
\end{bmatrix}
\begin{bmatrix}
m_1 \\
m_2 \\
\vdots \\
m_r
\end{bmatrix}
=
\begin{bmatrix}\tag{*}
0 \\
0 \\
\vdots \\
0
\end{bmatrix}.
\end{align*}

Let $B$ denote the matrix above whose entries are powers of eigenvalues $\lambda_i$.
We calculate the determinant of $B$.
\begin{align*}
\det(B)&=\lambda_1 \lambda_2 \cdots \lambda_r
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
\lambda_1 & \lambda_2& \cdots & \lambda_r \\
\vdots & \vdots & \vdots & \vdots \\
\lambda_1^{r-1} & \lambda_2^{r-1}&\cdots & \lambda_r^{r-1} \\
\end{bmatrix} \6pt] &=\lambda_1 \lambda_2 \cdots \lambda_r \prod_{1 \leq i<j \leq n}(\lambda_j-\lambda_i) \neq 0. \end{align*} (Note that the matrix above is a Vandermonde matrix.) Thus the matrix B is invertible. This means that the matrix equation (*) has unique solution \begin{align*} \begin{bmatrix} m_1 \\ m_2 \\ \vdots \\ m_r \end{bmatrix} = \begin{bmatrix}\tag{*} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{align*} But this is a contradiction because each multiplicity m_i is grater than zero. Thus we proved that all eigenvalues of A are zero. Recall that a matrix is nilpotent if and only if its eigenvalues are zero. See the post ↴ Nilpotent Matrix and Eigenvalues of the Matrix for a proof of this fact. Hence it follows from this fact that A is a nilpotent matrix. Sponsored Links More from my site • If Eigenvalues of a Matrix A are Less than 1, then Determinant of I-A is Positive Let A be an n \times n matrix. Suppose that all the eigenvalues \lambda of A are real and satisfy \lambda <1. Then show that the determinant \[ \det(I-A) >0, where $I$ is the $n \times n$ identity matrix. We give two solutions. Solution 1. Let […]
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