In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable.

Problems and solutions in Linear Algebra

Problem 91

Show that the matrix $A=\begin{bmatrix}
1 & \alpha\\
0& 1
\end{bmatrix}$, where $\alpha$ is an element of a field $F$ of characteristic $p>0$ satisfies $A^p=I$ and the matrix is not diagonalizable over $F$ if $\alpha \neq 0$.
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Comment.

Remark that if $A$ is a square matrix over $\C$ with $A^k=I$, then $A$ is diagonalizable.
For a proof of this fact, see If a power of a matrix is the identity, then the matrix is diagonalizable

Thus, over a field of characteristic $p>0$ the condition $A^p=1$ dose not always imply that $A$ is diagonalizable.

Proof.

By induction, it is straightforward to see that
\[A^m=\begin{bmatrix}
1 & m\alpha\\
0& 1
\end{bmatrix}\] for any positive integer $m$.
Thus
\[A^p=\begin{bmatrix}
1 & p\alpha\\
0& 1
\end{bmatrix}=I\] since $p\alpha=0$ in the field $F$.

Since the eigenvalues of $A$ is $1$, if $A$ is diagonalizable, then there exists an invertible matrix $P$ such that $P^{-1}AP=I$, or $AP=P$.
Let $P=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$. Then we have
\[AP=\begin{bmatrix}
a+\alpha c & b+\alpha d\\
c& d
\end{bmatrix}\] and this is equal to $P$, hence
\begin{align*}
a+\alpha c=a \text{ and } b+\alpha d=b \\
\end{align*}
Thus
\begin{align*}
\alpha c=0 \text{ and } \alpha d=0 \\
\end{align*}
If $\alpha \neq 0$, then $c=d=0$ but this implies that the matrix $P$ is non-invertible, a contradiction.
Therefore we must have $\alpha=0$.


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