We were talking about how the symmetric matrix
$$A=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
is not diagonalizable in the field consisting of only $0$ and $1$, since the eigenvalues are $0$ and $2$, but in this field $0=2$, and the two eigenvectors are the same
$$\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
and I was able to find the jordan normal form of the matrix being
$$J = P^{-1} A P$$
$J=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,
$P^{-1}=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$,
$A=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$,
$P=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$
since again in the field, $0=2=4=\dots$ and $-1=1=3=\dots$.
However I need to find another symmetric matrix in the same field that is also not diagonalizable, and also find its jordan normal form. I'm having a hard time with looking for this matrix.
The hint also said that
$$
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}
\quad\text{ or }\quad
\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}
$$
wouldnt work, even though they have no eigenvalues in ${\Bbb F}_2$; but they have two distinct eigenvalues in ${\Bbb F}_4$. so they are diagonalizable in ${\Bbb F}_4$.
What is ${\Bbb F}_4$? Is it a finite field that has only $0,1,2,3$?
If that's the case, the eigenvalues for
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}
are not even integers, how are they in ${\Bbb F}_4$?
Thanks in advance!
Best Answer
$F_4$ is quite different from $\mathbb Z/4 \Bbb Z$, which is not a field. In $F_4$ we have elements $0$, $1$, $\alpha$, and $\alpha+1$, where $\alpha$ is a root of $f(t)=t^2+t+1$ (working over $F_2$), which just happens to be the characteristic polynomial of your matrix.