# Trace of the Inverse Matrix of a Finite Order Matrix

## Problem 46

Let $A$ be an $n\times n$ matrix such that $A^k=I_n$, where $k\in \N$ and $I_n$ is the $n \times n$ identity matrix.

Show that the trace of $(A^{-1})^{\trans}$ is the conjugate of the trace of $A$. That is, show that $\tr((A^{-1})^{\trans})=\overline{\tr(A)}$.

Contents

## Hint.

- Note that $\tr(B)=\tr(B^{\trans})$ for any square matrix $B$.
- Use the Jordan canonical form of $A$.
- Show that eigenvalues are $k$-th roots of unity.

## Proof.

First note that the trace of a matrix is the same as the trace of its transpose. Thus we only have to show $\tr(A^{-1})=\overline{\tr(A)}$.

There is an invertible matrix $P$ such that $P^{-1}AP$ is the Jordan canonical form. That is, $P^{-1}AP=T$, where $T$ is an upper triangular matrix whose diagonal entries are eigenvalues of $A$.

Since $A$ is invertible (to see this take the determinant of $A^k=I_n$), the matrix $T$ is also invertible and $P^{-1}A^{-1}P=T^{-1}$. Then we have

\begin{align*}

\tr(A)&=\tr(P^{-1}AP)=\tr(T)\\

\tr(A^{-1})&=\tr(P^{-1}A^{-1}P)=\tr(T^{-1})

\end{align*}

Now let $\lambda_1, \lambda_2,\dots, \lambda_n$ be eigenvalues of $A$. Then the upper triangular matrix $T$ and its inverse matrix are

\[ T=\begin{bmatrix}

\lambda_1 & * & * & * \\

0 &\lambda_2 & * & * \\

0 & 0 & \ddots & \vdots \\

0 & 0 & \dots & \lambda_n

\end{bmatrix},

\,\,

T^{-1}=\begin{bmatrix}

\lambda_1^{-1} & * & * & * \\

0 &\lambda_2^{-1} & * & * \\

0 & 0 & \ddots & \vdots \\

0 & 0 & \dots & \lambda_n^{-1}

\end{bmatrix}\]

Thus we have

\begin{align*}

\tr(A)&=\lambda_1+\lambda_2+\cdots + \lambda_n \\

\tr(A^{-1})&=\lambda_1^{-1}+\lambda_2^{-1}+\cdots +\lambda_n^{-1} \end{align*}

Next, we show that $\lambda_i^{-1}=\overline{\lambda_i}$ for $i=1,2,\dots, n$.

This follows from the fact that if $A^k=I_n$ then the eigenvalues are $k$-th roots of unity.

Assuming this, we have $1=|\lambda_i|=\overline{\lambda_i}\lambda_i$, hence $\lambda_i^{-1}=\overline{\lambda_i}$.

To prove the fact, let $\lambda$ be an eigenvalue of $A$ and let $x$ be an eigenvector corresponding to $\lambda$.

Then we have $Ax=\lambda x$. Using this relation successively we have

\begin{align*}

x&=I_nx=A^kx=\lambda A^{k-1}x=\lambda^2 A^{k-2}x =\cdots=\lambda^kx

\end{align*}

Since $x$ is a nonzero vector, we have $\lambda^k=1$, and $\lambda$ is a $k$-th root of unity.

Now we have

\begin{align*}

\tr(A^{-1})&= \lambda_1^{-1}+\lambda_2^{-1}+\cdots +\lambda_n^{-1} \\

&= \overline{\lambda_1}+\overline{\lambda_2}+\cdots +\overline{\lambda_n}\\

&=\overline{ \lambda_1+\lambda_2+\cdots +\lambda_n} \\

&=\overline{\tr(A)}.

\end{align*}

This completes the proof.

### Analogous Problem.

See also Finite order matrix and its trace for a similar problem.

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