Jordan Canonical form in $C_2$

So if you have the matrix $A = \left( \begin{array}[cc] & 1 & 1 \\ 1 & 1 \end{array} \right)$ then its Jordan form in $\mathbb{Z}$ is $\left( \begin{array}[cc] & 2 & 0 \\ 0 & 0 \end{array} \right)$ right but if you regard $A$ instead as a member of the cyclic field $C_2$ then $\left( \begin{array}[cc] & 2 & 0 \\ 0 & 0 \end{array} \right) \cong \left( \begin{array}[cc] & 0 & 0 \\ 0 & 0 \end{array} \right)$ so the Jordan form of $A$ is $\left( \begin{array}[cc] & 0 & 0 \\ 0 & 0 \end{array} \right)$ so $A$ is similar to $\left( \begin{array}[cc] & 0 & 0 \\ 0 & 0 \end{array} \right)$ so $A = \left( \begin{array}[cc] & 0 & 0 \\ 0 & 0 \end{array} \right)$. But that makes no sense, so where did I go wrong, and what actually is the Jordan form of $A$ in $C_2$?

• Probably two eigenvectors that are independent over $\mathbb Z$ become dependent when reduced mod $2$? – Nishant Apr 7 '15 at 20:09
• Well, in $\,C_2\;$ we actually have that $\;2=0\;$, so both matrices are in fact one and the same! – Timbuc Apr 7 '15 at 20:11
• No, $A$ is not conjugated over $\mathbb{Z}$ to the diagonal matrix $diag(2,0)$, only over $\mathbb{Q}$. We have $SAS^{-1}=diag(2,0)$, where $S$ or $S^{-1}$ has half-integral coefficients. – Dietrich Burde Apr 7 '15 at 20:18
• Over $\mathbb{Z}_2$ the matrix $A$ is nilpotent, so its JNF is $\left(\begin{matrix} 0 & 1\\ 0 & 0\end{matrix}\right)$. – user 59363 Apr 7 '15 at 20:25

1 Answer

First $\mathbb Z$ is not a field so it's not true that every matrix can be put into Jordan normal form. If we think of $A$ as being defined over $\mathbb Q$ then as you've calculated the Jordan normal form is $$\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}$$ The important thing to notice is that the conjugating matrix $$B = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$$ has determinant $-2$ so while it's invertible in $\mathbb Q$, it's not invertible in $\mathbb Z$ and it's not invertible in $C_2$. So the Jordan normal form of this matrix over $\mathbb Q$ doesn't need to agree with the Jordan normal form over $\mathbb Z$ or over $C_2$ because it's not possible to just take the above equation and consider it as an equation over $\mathbb Z$ or $C_2$.

So let's look at $\mathbb Z$ and $C_2$. If a Jordan normal form over $\mathbb Z$ existed then it would be the same as the Jordan normal form over $\mathbb Q$ (cause all the matrix elements would be valid entries in $\mathbb Q$). So does $$B\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}B^{-1} = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}$$ have a solution over $\mathbb Z$? If you move the $B^{-1}$ to the other side (to get rid of the inverse) this is just a system of equations and you get that $B$ must have the form $$B = \begin{bmatrix} a & a \\ b & -b \end{bmatrix}$$ but then $\det B = -2ab$ and if $a, b \in \mathbb Z$ then $-2ab$ is not $\pm1$ so $B$ is not invertible in $\mathbb Z$. Thus $A$ cannot be put into Jordan normal form over $\mathbb Z$.

Now what about $C_2$? This is a field so $A$ definitely has a Jordan normal form. You can check that over $C_2$ we have $A^2 = 0$ so all eigenvalues are zero. This means the Jordan normal form is either $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \qquad \text{or} \qquad \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ As you noted, having zero as the Jordan normal form would imply $A = 0$, and that's not the case, so the Jordan normal form of $A$ over $C_2$ is $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ I suggest you do the following to help you see what's going on here: Find the matrix $B$ over $C_2$ which conjugates $A$ to it's Jordan normal form. Then take a look at what happens when you think of the elements of $B$ as integers. Does $B^{-1}$ have integer entries (there is a way to choose the integers such that this is the case)? If so calculate $BAB^{-1}$. You should find that $BAB^{-1}$ is not in Jordan normal form but as soon as you reduce the coefficients modulo $2$ it is!