## Jordan normal form

### Task number: 2665

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 $$\mathbf x_1$$) $\begin{pmatrix} 0&1&0&2&0\\ 0&0&0&-2&0\\ 0&0&0&2&1\\ 0&0&0&0&0 \end{pmatrix}$ 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 $$\mathbf x_2$$ $\begin{pmatrix} 0&1&0&2&1\\ 0&0&0&-2&0\\ 0&0&0&2&0\\ 0&0&0&0&0 \end{pmatrix} \sim \begin{pmatrix} 0&1&0&0&1\\ 0&0&0&1&0\\ \end{pmatrix}$ 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 $\begin{pmatrix} 0&1&0&2&-1\\ 0&0&0&-2&1\\ 0&0&0&2&-1\\ 0&0&0&0&0 \end{pmatrix} \sim \begin{pmatrix} 0&1&0&0&0\\ 0&0&0&2&-1\\ \end{pmatrix}$ 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 $$\mathbf R^{-1}$$

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

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