Trace of the Inverse Matrix of a Finite Order Matrix

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

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 $\tr((A^{-1})^{\trans})=\overline{\tr(A)}$.

 

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Hint.

  1. Note that $\tr(B)=\tr(B^{\trans})$ for any square matrix $B$.
  2. Use the Jordan canonical form of $A$.
  3. 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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