# Find Jordan Decomposition of $\begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}$ over $\mathbb{F}_5$

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?

• 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 $t^2-7t+11$. – Dietrich Burde May 20 at 18:36
• Conjugate the matrix with $S=\begin{pmatrix} 3 & 2 & 4\cr 1 & 0 & 0 \cr 1 & 4 & 0\end{pmatrix}$ to obtain $J$. – Dietrich Burde May 20 at 18:39
• 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 – Will Jagy May 20 at 18:50
• By the way, $(0,1,0)^T$ is a much cleaner basis element for the eigenvectors. – Thomas Andrews May 20 at 19:36
• @DietrichBurde I don't see how decomposition over $\mathbb{C}$ helps in this case. And what does "conjugate the matrix with $X$" mean? – Viktor Glombik May 20 at 22:15

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

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

• Why is $\frac{1}{3} = 2$? – Viktor Glombik May 20 at 19:13
• @ViktorGlombik Because $3\times2=1$. – user1551 May 20 at 19:21
• Because $1=2\cdot 3=6$. Recall that $5=0$. – Dietrich Burde May 20 at 19:21

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 $$v_1.$$

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 :

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