Over \(\mathbb F_3=\mathbb Z/3\mathbb Z\), find the eigenvalues, eigenvectors, and Jordan canonical form of \[ A= \begin{pmatrix} 2&1&1\\ 1&2&1\\ 1&1&2 \end{pmatrix}. \]
Solution
Let \[ N=A-I = \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{pmatrix}. \]
1. One has\[ N^2=0 \]in \(M_3(\FF_3)\).
Every entry of \(N^2\) is \[ 1+1+1=3=0 \] in \(\FF_3\). Hence \(N^2\) is the zero matrix.
2. The only eigenvalue of \(A\) is\[ \boxed{1}. \]
Suppose \(Av=\lambda v\) for a nonzero \(v\). Since \(A=I+N\), \[ Nv=(\lambda-1)v. \] Applying \(N\) and using step 1 gives \[ 0 = N^2v = (\lambda-1)Nv = (\lambda-1)^2v. \] Because \(v\neq0\) and \(\FF_3\) is a field, \[ (\lambda-1)^2=0, \] so \(\lambda=1\).
3. The eigenspace for the eigenvalue \(1\) is\[ \boxed{ E_1 = \left\{ \begin{pmatrix}x\\y\\z\end{pmatrix} \in\FF_3^3: x+y+z=0 \right\} }. \]
The equation \(Av=v\) is equivalent to \(Nv=0\). For \[ v= \begin{pmatrix}x\\y\\z\end{pmatrix}, \] one has \[ Nv = \begin{pmatrix} x+y+z\\ x+y+z\\ x+y+z \end{pmatrix}. \] Thus \(Nv=0\) exactly when \(x+y+z=0\). This is one nontrivial linear equation in three variables, so \(E_1\) has dimension \(2\).
4. The nilpotent matrix \(N=A-I\) has rank \(1\).
All three rows of \(N\) are equal and nonzero, so its row space is one-dimensional.
5. The Jordan canonical form of \(A\) consists of one block of size\(2\) and one block of size \(1\), both for eigenvalue \(1\).
Step 1 shows that every Jordan block of \(A\) for eigenvalue \(1\) has size at most \(2\). Step 4 shows that \(N=A-I\) is nonzero, so \(A\) is not diagonalizable and at least one block has size \(2\). Since the total dimension is \(3\), the only possibility is the partition \[ 3=2+1. \]
6. A Jordan basis is given by\[ w= \begin{pmatrix}1\\1\\1\end{pmatrix}, \qquad v= \begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad u= \begin{pmatrix}1\\2\\0\end{pmatrix}. \]
One has \[ Nw=0, \qquad Nv=w, \qquad Nu=0. \] Thus \[ Aw=w, \qquad Av=v+w, \qquad Au=u. \] Moreover, \[ \det \begin{pmatrix} 1&1&1\\ 1&0&2\\ 1&0&0 \end{pmatrix} =2\neq0 \] in \(\FF_3\), so \((w,v,u)\) is a basis.
7. Relative to the basis in step 6, the Jordan form is\[ \boxed{ \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix} }. \]
The coordinate identities in step 6 give the three columns of the displayed matrix.
8. Q.E.D.
Steps 2, 3, and 7 give respectively the eigenvalue, eigenvectors, and Jordan canonical form requested.