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Find the Jordan decomposition of $$ A := \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix} \in M_3(\mathbb{F}_5), $$ where $\mathbb{F}_5$ is the field modulo 5.

What I've done so far The characteristic polynomial is \begin{equation} P_A(t) = (4 - t)(1-t)(3-t) - (1-t) = -t^3 + 8t^2-18t+1 \equiv 4t^3 + 3t^2 + 2t + 1 \mod5. \end{equation} Therefore, $\lambda = 1$ is a zero of $P_A$, since $4+3+2+1 = 10 \equiv 0 \mod 5$. By polynomial division one obtains $$ P_A(t) = (t + 4)(4t^2 + 2t + 4) = (t + 4)(t + 4) (4t + 1) \equiv 4 (t + 4)^3 $$ Therefore $\lambda = 1$ is the only eigenvalue of $A$. Calculating the eigen space we calculate the kernel of $A + 4 E_3$ and obtain $$ \text{span}\left( \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right) $$ Since $(A + 4 E_3)^2 = 0$, the kernel of $(A + 4 E_3)^2$ is the whole space. Now, I choose $v := (1,0,0) \in \text{ker}(A + 4 E_3)^2$ such that $v \not\in \text{ker}(A + 4 E_3)$. We calculate $(A + 4E)v = (3,0,1)$ and then $$ (A + 4E) \begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, $$ but the zero vector can't be a basis vector of our Jordan decomposition.

Have I made a mistake in my calculations?

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    $\begingroup$ The Jordan form is given by $J=\begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 0 &1 \end{pmatrix}$. Over $\Bbb C$ it is $diag(1,\lambda,\mu)$, where $\lambda,\mu$ are the roots of $t^2-7t+11$. $\endgroup$ – Dietrich Burde May 20 at 18:36
  • $\begingroup$ Conjugate the matrix with $S=\begin{pmatrix} 3 & 2 & 4\cr 1 & 0 & 0 \cr 1 & 4 & 0\end{pmatrix}$ to obtain $J$. $\endgroup$ – Dietrich Burde May 20 at 18:39
  • $\begingroup$ I really like the minimal polynomial for this. You see $A-I$ has rank one. Fine. Both $(A-I)^3$ and $(A-I)^2$ are a bunch of multiples of 5, meaning both are zero over the given field. Then take any column vector $w$ that is NOT an eigenvector, let $v = (A-I)w$ will be an eigenvector, finally choose column $u$ as an eigenvector that is independent of $v.$ The matrix $P = (u,v,w)$ provides $P^{-1}AP = J.$ Meanwhile, take care to confirm your $P P^{-1} = P^{-1}P = I$ over that field $\endgroup$ – Will Jagy May 20 at 18:50
  • $\begingroup$ By the way, $(0,1,0)^T$ is a much cleaner basis element for the eigenvectors. $\endgroup$ – Thomas Andrews May 20 at 19:36
  • $\begingroup$ @DietrichBurde I don't see how decomposition over $\mathbb{C}$ helps in this case. And what does "conjugate the matrix with $X$" mean? $\endgroup$ – Viktor Glombik May 20 at 22:15
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Your calculations are fine. However, by definition of kernels, an element of the kernel of $(A+4E)^2$ vanishes when applying $A+4E$ to it twice, so you should not be surprised. You just made the wrong conclusion. The eigenvector $(3,0,1)$ together with the generalized eigenvector $(1,0,0)$ form part of a Jordan basis giving you a Jordan block of size $2$. All you need to do is add another eigenvector which is linear independent to $(3,0,1)$, for example $(1,1,2)$.

Then $$ \begin{pmatrix} 3 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 3 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$

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Since $(A-I)(1,0,0)^T=(3,0,1)^T=3(1,0,2)^T$, a desired ordered basis is given by $\{(1,0,2)^T,\frac13(1,0,0)^T,(1,1,2)^T\}=\{(1,0,2)^T,(2,0,0)^T,(1,1,2)^T\}$.

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  • $\begingroup$ Why is $\frac{1}{3} = 2$? $\endgroup$ – Viktor Glombik May 20 at 19:13
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    $\begingroup$ @ViktorGlombik Because $3\times2=1$. $\endgroup$ – user1551 May 20 at 19:21
  • $\begingroup$ Because $1=2\cdot 3=6$. Recall that $5=0$. $\endgroup$ – Dietrich Burde May 20 at 19:21
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You want three independent vectors, $v_1,v_2,v_3$ that have the properties:

$$Av_1=v_1, Av_2=v_2+v_1, Av_3=v_3.\tag{1}$$

For $v_2=(1,0,0)^T$ not in the eigenspace, we get $v_1=Av_2-v_2=(3,0,1)^T$ [*] and then you need a $v_3$ which is in the eigenspace of $A$ but not a multiple of $v_1.$ We'll choose $v_3=(0,1,0)^T.$ Then if $$S=\begin{pmatrix}v_1&v_2&v_3\end{pmatrix}=\begin{pmatrix}3&1&0\\0&0&1\\1&0&0\end{pmatrix}$$ then we can show:

$$J=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}=S^{-1}AS$$

Since $Se_i=v_i$ and the equalities in (1) and and $S^{-1}v_i=e_i,$ so we get $$Je_1=e_1,Je_2=e_1+e_2,Je_3=e_3$$

Since $\det S = 1$ even in the integers, we can use Wolfram Alpha to invert $S$ over the integers and get $$S^{-1}=\begin{pmatrix}0&0&1\\1&0&-3\\0&1&0 \end{pmatrix}$$


[*] Note that since $(A-E_3)^2=0,$ you have $A^2-A=A-E_3$ and hence, when $v_1=Av_2-v_2,$ you have $Av_1=(A^2-A)v_2=(A-E_3)v_2=Av_2-v_2=v_1.$


We could have started with any $v_2$ not in the eigenspace. We'll always get some multiple of our original $v_1$ for $v_1.$ Then $v_3$ can be any of $20$ vectors in the eigenspace not a multiple of $v_1.$

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Here is a 4th answer that takes into account that we are in a particular case yielding a reduction to a $2 \times 2$ matrix.

Indeed, up to a simultaneous permutation $P$ on lines and columns, $A=PBP^{-1}$ is similar to

$$B:= \left(\begin{array}{cc|c} 4 & 1 & 0 \\ 1 & 3 & 0 \\ \hline 0 & 0 & 1 \end{array}\right)$$

On this form, we have reduced the issue to find a Jordan form for $2 \times 2$ upper block $U$.

A quick computation shows that $1$ is a double eigenvalue of $U$.

The Jordan form of $U$ is :

$$\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)$$

because (due to a quick reasoning, that has been developed in the other answers, but with simpler computations) :

$$\underbrace{\left(\begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array}\right)}_{Q}\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)\underbrace{\left(\begin{array}{cc} 0 & 3 \\ 3 & 1 \end{array}\right)}_{Q^{-1}}=\left(\begin{array}{cc} 4 & 1 \\ 1 & 3 \end{array}\right).$$

The Jordan form of $B$, which is the same as the Jordan form of $A$, is thus :

$$J:= \left(\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 1 & 0 \\ \hline 0 & 0 & 1 \end{array}\right).$$

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