Find the jordan canonical form of the following matrix

Question

A = $$ \begin{matrix} 2 & 1 & 1 & 1 \\ 0 & 2 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \\ \end{matrix} $$ I calculated the eigenvalue to be 2 with a algebraic multiplicity of 4 and found the eigenvectors to be $(1,0,0,0)$ and $(0, 1, -1, 0)$. This means the matrix $J$ can either have a 3 by 1 and 1 by 1 block or two 2 by 2 blocks. I'm lost as to how to determine which one it is. I tried to generate generalized eigenvectors by solving the equation $(A-2I)(V_1) = (1, 0, 0, 0)$ and the same for the other eigenvector. However, for the first eigenvector $(1,0,0,0)$, I got one generalized eigenvector $(0,1,0,0)$ and got an inconsistent system when I tried to keep generating more. Can someone help?

Answer 1

Your characteristic poly is: $(x - 2)^4$, with $\lambda=2$.

$$ A-2I=\left(\begin{array}{rrrr} 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right) $$ has eigenvectors of $b_1=(1,0,0,0), b_4=(0,1,-1,0)$. As such, your Jordan block will have $2$ blocks. A good way to see what sort of form your Jordan form will take is the minimal polynomial, which is the smallest polynomial that annihilates $A$. We see that minpoly = $(x-2)^3$. The power 3 tells us the biggest size of the Jordan block. So your Jordan form will be: $$ \left(\begin{array}{rrrr} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{array}\right) $$ You already know that $(0,1,-1,0)$ will give an inconsistent solution, so RREF

$$ \left(\begin{array}{rrrr|r} 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{rrrr|r} 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ to obtain that your solution set is $\{(0,1,0,0)+\lambda(0,-1,1,0)+\mu(1,0,0,0)\}$. The problem is that you are not choosing vectors to solve for in the range of $A-2I$. Choose $(0,1/2, 1/2, 0)$ instead and solve for: $$ \left(\begin{array}{rrrr|r} 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{rrrr|r} 0 & 1 & 1 & 0 & -\frac{1}{2} \\ 0 & 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ You see a vector you can choose is $(0, -1/2, 0, 1/2)$. The transition matrix:

$$ P=\left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & \frac{1}{2} & 0 & -1 \\ 0 & 0 & \frac{1}{2} & 0 \end{array}\right)$$ will give you your Jordan form.

Add on: you can infer the fact that $A-2\lambda$ will have minimal polynomial of degree 3 without calculating it directly. This is because for any matrix of the form: $$ K=\left(\begin{array}{rrrr} 0 & a & b & c \\ 0 & 0 & d & e \\ 0 & 0 & 0 & f \\ 0 & 0 & 0 & 0 \end{array}\right)$$, we have: $$ K^2= \left(\begin{array}{rrrr} 0 & 0 & a d & a e + b f \\ 0 & 0 & 0 & d f \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \ \ \& \ \ K^3=\left(\begin{array}{rrrr} 0 & 0 & 0 & a d f \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) $$ Since $d=0$, your minpoly has degree 3.

Answer 2

If it’s really the case that all the problem is asking for is the Jordan canonical form (JCF) and not the complete decomposition, then there’s no need to compute any generalized eigenvectors at all. Since the only eigenvalue has a geometric multiplicity of $2$, you know that there will be two Jordan blocks so that, up to a permutation of blocks, the JCF must be either $$\begin{bmatrix}2&1&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{bmatrix}$$ or $$\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&2&1\\0&0&0&2\end{bmatrix}.$$ The exponent of $\lambda-2$ in the minimal polynomial gives us the size of the largest block, so we compute $$(A-2I)^2 = \begin{bmatrix}0&0&0&2\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}.$$ This is nonzero, which eliminates the second possibility.

Answer 3

Let $$u=\begin{bmatrix}1\\0\\0\\0\end{bmatrix},v=\begin{bmatrix}0\\1\\-1\\0\end{bmatrix},$$ Let $$B=A-2I=\begin{bmatrix}0&1&1&1\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{bmatrix}$$ The system $$B\begin{bmatrix}a'\\b'\\c'\\d'\end{bmatrix}=v, i.e$$ $$\begin{bmatrix}0&1&1&1\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{bmatrix}\begin{bmatrix}a'\\b'\\c'\\d'\end{bmatrix}=\begin{bmatrix}0\\1\\-1\\0\end{bmatrix} \text { is inconsistent.}$$ Let $u_1=u,u_4=v.$ Let $$u_2=\begin{bmatrix}a'\\b'\\c'\\d'\end{bmatrix},Bu_2=u_1.$$ $$\begin{bmatrix}0&1&1&1\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{bmatrix}\begin{bmatrix}a'\\b'\\c'\\d'\end{bmatrix}=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$$ Thus $d'=0,b'+c'=1.$ Let $$u_3=\begin{bmatrix}a''\\b''\\c''\\d''\end{bmatrix},Bu_3=u_2$$ $$\begin{bmatrix}0&1&1&1\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{bmatrix}\begin{bmatrix}a''\\b''\\c''\\d''\end{bmatrix}=\begin{bmatrix}a'\\b'\\1-b'\\0\end{bmatrix}$$ Then, on comparing 2-nd and 3-rd coordinates $d''=b',d''=1-b'$,so $d''=1/2,b'=1/2,c'=1/2$. On comparing 1-st coordinates, $b''+c''+1/2=a'.$ Thus we can take $a'=1,b''=1/2,c''=0,a''=1.$ We have $$u_2=\begin{bmatrix}1\\1/2\\1/2\\0\end{bmatrix}u_3=\begin{bmatrix}1\\1/2\\0\\1/2\end{bmatrix}$$ Thus $$P=\begin{bmatrix}1&1&1&0\\0&1/2&1/2&1\\0&1/2&0&-1\\0&0&1/2&0\end{bmatrix} $$ $$AP=P\begin{bmatrix}2&1&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{bmatrix}$$ $$P^{-1}AP=Jord=\begin{bmatrix}2&1&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{bmatrix}$$