7) Consider the field \( F = \mathbb{Z} / 7 \mathbb{Z} \) of order \( 7.\) Consider the following matrix \( A \in \text{Mat}_7(F): \)
\[
A :
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0
\end{pmatrix}.
\]
Find the Jordan canonical form \( J \) for \( A. \) Prove that your answer is correct.
Proof. Multiplying any other \( 7 \times 7 \) matrix \( B \) on the left by \( A \) has the effect of moving down all
the rows of \( B \) by one row, bringing the bottom row of \( B \) to the top. Therefore \( A^7 = 1 \) and \( m_A(x) \) is
a multiple of \( {(x^7 - 1)} = {(x - 1)^7}. \) To see this equality, expand \( (x - 1)^7 \) by the binomial theorem, noting
that all coefficients except the first and last are multiples of 7, but \( 7 = 0 \) in \( \mathbb{Z} / 7 \mathbb{Z}. \)
Because of the row shifting effect, no lower power of \( x - 1 \) is \( 0 \), so \( m_A(x) = (x - 1)^7. \)
Therefore the size of the largest Jordan block for eigenvalue \( \lambda = 1 \) (the only eigenvalue) is \( 7 \). Since \( 7 \)
is the size of the Jordan matrix for \( A \), the Jordan canonical form for \( A \) consists of a single Jordan block of size
\( 7 \) for \( \lambda = 1: \)
\[
\text{JCF}(A) =
\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}.
\]
QED.