Jordan normal form

Transform the following matrix into Jordan normal form and determine eigenvectors, and if necessary also generalized eigenvectors.

• Variant

$$\mathbf A=\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 0\\ -1 & 0 & 3 \end{pmatrix}$$

• Hint

The generalized eigenvector $$\mathbf x_i$$ can be obtained from the system$$(\mathbf A-\lambda \mathbf I)\mathbf x_i=\mathbf x_{i-1}$$.

• Resolution

Characteristic polynomial $$p_\mathbf A(t)= \begin{vmatrix} 1-t & 1 & 1\\ 0 & 1-t & 0\\ -1 & 0 & 3-t \end{vmatrix} = (1-t)(2-t)^2$$.

The system $$(\mathbf A-2\mathbf I)\mathbf x^1=\mathbf 0$$ has a solution $$\mathbf x^1=p(1{,}0,1)^T$$.

The eigenvalue $$\lambda=2$$ has geometric multiplicity 1 and algebraic 2, we shall find a generalized eigenvector. In the sequel we choose $$p=1$$, i.e. $$\mathbf x^1=(1, 0, 1)^T$$.

The generalized eigenvector $$\mathbf x^2$$ we get from $$(\mathbf A-2\mathbf I)\mathbf x^2=\mathbf x^1$$. It has a solution $$\mathbf x^2=q(1, 0, 1)^T+(-1, 0, 0)^T$$.

The system $$(\mathbf A-1\mathbf I)\mathbf x=\mathbf 0$$ has solution $$\mathbf x^3=r(2, -1, 1)^T$$.

By a suitable choice of parameters (to have $$\mathbf x^1, \mathbf x^2$$ linearly independent) $$q=1$$ and $$r=1$$ we get the desired matrix $$\mathbf R$$. We also calculate its inverse $$\mathbf R^{-1}$$.

$$\mathbf R= \begin{pmatrix} 1 & 0 & 2\\ 0 & 0 & -1\\ 1 & 1 & 1 \end{pmatrix} \qquad \mathbf R^{-1}= \begin{pmatrix} 1 & 2 & 0\\ -1 & -1 & 1\\ 0 & -1 & 0 \end{pmatrix}$$

• Result

The given matrix can be factorized into Jordan normal form as

$$\mathbf A= \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 0\\ -1 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 2\\ 0 & 0 & -1\\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 0\\ -1 & -1 & 1\\ 0 & -1 & 0 \end{pmatrix} = \mathbf R\mathbf J\mathbf R^{-1}$$

• Variant

$$\mathbf A= \begin{pmatrix} 2 & 1 & 0 & 2\\ 0 & 2 & 0 & -2\\ 0 & 0 & 2 & 2\\ 0 & 0 & 0 & 2 \end{pmatrix}$$
• Solution

The matrix has a single eigenvalue $$\lambda = 2$$ of algebraicmultiplicity 4, eigenvectors are nontrivial solutions of a system of equation with matrix $\mathbf A-\lambda \mathbf I = \begin{pmatrix} 0&1&0&2\\ 0&0&0&-2\\ 0&0&0&2\\ 0&0&0&0 \end{pmatrix} \sim \begin{pmatrix} 0&1&0&0\\ 0&0&0&1\\ \end{pmatrix}$ The dimension of $$\ker(\mathbf A-\lambda \mathbf I)$$ is 2, so the matrix $$\mathbf A$$ has two linearly independent eigenvectors, e.g. $$\mathbf x_1=(0{,}0,1{,}0)^T$$ and $$\mathbf x_2=(1{,}0,0{,}0)^T$$

To get a regular matrix $$\mathbf R$$ we need to find two generalized eigenvectors. Instead the system $$(\mathbf A-\lambda \mathbf I)\mathbf x=\mathbf 0$$ we choose the right-hand side the eigenvector (or in the next step the generalized eigenvector), i.e. we solve a system with augmented matrix (for $$\mathbf x_1$$) $\left( \begin{array}{cccc|c} 0&1&0&2&0\\ 0&0&0&-2&0\\ 0&0&0&2&1\\ 0&0&0&0&0 \end{array} \right)$ which has no solution (see the second and third rows). From this we deduce that this vector corresponds to a Jordan cell of size 1.

For the second eigenvector $$\mathbf x_2$$ we thus need two generalized eigenvectors and the other cell will be of size 3. We follow the same approach, the right-hand side is $$\mathbf x_2$$ $\left( \begin{array}{cccc|c} 0&1&0&2&1\\ 0&0&0&-2&0\\ 0&0&0&2&0\\ 0&0&0&0&0 \end{array} \right) \sim \left( \begin{array}{cccc|c}0&1&0&0&1\\ 0&0&0&1&0\\ \end{array} \right)$ The set of solutions is an affine space of dimension 2, more precisely of form $$(p,1,q,0)^T$$. We choose $$p=q=-1$$ so thus $$\mathbf x_3=(-1{,}1,-1{,}0)^T$$.

Then we choose $$\mathbf x_3$$ as the right side of amny times solved system and continue $\left( \begin{array}{cccc|c} 0&1&0&2&-1\\ 0&0&0&-2&1\\ 0&0&0&2&-1\\ 0&0&0&0&0 \end{array} \right)\sim \left( \begin{array}{cccc|c} 0&1&0&0&0\\ 0&0&0&2&-1\\ \end{array} \right)$ with (for sequel calculations nice) solution $$\mathbf x_4=(0{,}0,0,-\frac{1}{2})^T$$

Now we may assemble matrices $$\mathbf J$$ a $$\mathbf R$$ and then calculate $$\mathbf R^{-1}$$

The above calculation does not give a general recipy for finding the Jordan decomposition, since one needs to ensure linear independence of the vectors from different chains. Here, this was guaranteed by ensuring that one of the cells was of size 1. The general correct procedure is shown in the following variant.

$$\mathbf J = \begin{pmatrix} 2&0&0&0\\ 0&2&1&0\\ 0&0&2&1\\ 0&0&0&2 \end{pmatrix} , \mathbf R= \begin{pmatrix} 0&1&-1&0\\ 0&0&1&0\\ 1&0&-1&0\\ 0&0&0&-\frac{1}{2} \end{pmatrix} , \mathbf R^{-1}= \begin{pmatrix} 0&1&1&0\\ 1&0&0&0\\ 0&1&1&0\\ 0&0&0&-2 \end{pmatrix}$$
• Variant

$$\mathbf A= \begin{pmatrix} 0&1&0&0&-1\\ 0&0&0&0&0\\ 0&0&0&1&-1\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{pmatrix}$$
• Solution

It has a single eigenvalue $$\lambda=0$$ of algebraic multiplicity 5.

The rank of $$\mathbf A$$ is 3, so $$\dim(\ker A)=2$$ and we have to linearly independent eigenvectors.

When we would like for the eigenvector $$\mathbf x_1$$ calculate a generalized eigenvector $$\mathbf x_2$$, it would mean to solve $$(\mathbf A-\lambda \mathbf I)\mathbf x_2=\mathbf x_1$$, what is equivalent to solving $$(\mathbf A-\lambda \mathbf I)^2\mathbf x_2 = (\mathbf A-\lambda \mathbf I)\mathbf x_1 = \mathbf 0$$. In other words, $$\mathbf x_2 \in \ker((\mathbf A-\lambda \mathbf I)^2) \setminus \ker(\mathbf A-\lambda \mathbf I)$$ etc. for the other generalized eigenvectors in the chain.

In our case with $$\lambda=0$$ we calculate $\mathbf A^2= \begin{pmatrix} 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{pmatrix}$ a $\mathbf A^3= \begin{pmatrix} 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{pmatrix}$

Clearly $$\dim(\ker \mathbf A^2)=4$$ and $$\dim(\ker \mathbf A^3)=5$$, so there is only one linearly independent vector in $$\ker \mathbf A^3 \setminus \ker \mathbf A^2$$, e.g. $$x_3=(0{,}0,0{,}0,1)^T$$. This shall satisfy $$\mathbf A\mathbf x_3=\mathbf x_2$$, so we claculate $$\mathbf x_2=(-1{,}0,-1{,}1,0)^T$$ and $$\mathbf x_1=\mathbf A\mathbf x_2=(0{,}0,1{,}0,0)^T$$ (this is an eigenvector of the matrix $$\mathbf A$$). This yields a single Jordan cell of size 3.

The other cell will be of size 2 and again we will start at the end of the chain, i.e. by the generalized eigenvector $$\mathbf y_2$$. Tis must satisfy $$\mathbf y_2 \in \ker \mathbf A^2 \setminus \ker \mathbf A$$ and is linearly independent on $$\mathbf x_2$$. We choose e.g. $$\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\mathbf y_1=\mathbf A\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\mathbf A$$).

Now we may assemble matrices $$\mathbf R$$ (columns are $$\mathbf y_1,\mathbf y_2,\mathbf x_1,\mathbf x_2,\mathbf x_3$$) and $$\mathbf J$$ and claculate the inverse matrix $$\mathbf R^{-1}$$

$$\mathbf R= \begin{pmatrix} 1&0&0&-1&0\\ 0&1&0&0&0\\ 0&0&1&-1&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{pmatrix}$$, $$\mathbf J= \begin{pmatrix} 0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{pmatrix}$$, $$\mathbf R^{-1}=\begin{pmatrix} 1&0&0&1&0\\ 0&1&0&0&0\\ 0&0&1&1&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{pmatrix}$$