Linear Algebra

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

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

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.

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Steps.

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

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.

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