In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable.
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 $\alpha \neq 0$. Add to solve later
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 $\alpha=0$.
Each Element in a Finite Field is the Sum of Two Squares
Let $F$ be a finite field.
Prove that each element in the field $F$ is the sum of two squares in $F$.
Proof.
Let $x$ be an element in $F$. We want to show that there exists $a, b\in F$ such that
\[x=a^2+b^2.\]
Since $F$ is a finite field, the characteristic $p$ of the field […]
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As an example, we solve the following problem.
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\[A=\begin{bmatrix}
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\[A=\begin{bmatrix}
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-4 &-6 &-3 \\
3 & 3 & 1
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Let
\[A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}.\]
Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.
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Definitions/Hint.
Recall the relevant definitions.
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