Find Jordan basis of a given matrix

Question

$A=\begin{bmatrix} -1 & 4 & -4 \\ 0 & 1 & 0 \\ 1 & -2 & 3 \end{bmatrix}$

The eigenvalues of the matrix are all $1$. The dimension of it's eigenspace is 2 so the Jordan normal form of the matrix is \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

this is all confirmed by WolframAlpha.

Now, an eigenvector for $1$ is $(0,1,1)$ but when I try t solve $AP=PJ$ where

$P=\begin{bmatrix} 0 & a & d \\ 1 & b & e \\ 1 & c & f \end{bmatrix}$

I get $1+b=b$ for the middle element. Why is $(0,1,1)$ a bad choice for the first column here, and how do I find the Jordan basis in this case?

Answer 1

The characteristic polynomial is given by

$$\begin{align} p (\lambda) &= (-1-\lambda)(1 - \lambda)(3 - \lambda) - (-4)(1- \lambda)(1) \\&= (1 - \lambda)[-(-1-\lambda)(3-\lambda) - 4]\\&= -(\lambda - 1)^3\end{align}$$

Which gives you eingenvalue $\lambda = 1$ of multiplicity $3$.

To find the eigenvectors, solve

$$(A - I)\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} = 0.$$

We get $u = \begin{pmatrix}2u_2 - 2u_3\\u_2\\u_3\end{pmatrix}$.

Take two linearly independent eigenvectors, e.g. $u = \begin{pmatrix}0\\1\\1\end{pmatrix}$ (your same idea) and $u' = \begin{pmatrix}2\\0\\-1\end{pmatrix}$. Now to find a generalized vector we make

$$(A - I) v = u'$$

Then $v = \begin{pmatrix}-1 + 2v_2 - 2v_3\\v_2\\v_3\end{pmatrix}$, we may take $v = \begin{pmatrix}-1\\0\\0\end{pmatrix}$. A basis would then be

$$\Bigg\{\begin{pmatrix}0\\1\\1\end{pmatrix},\begin{pmatrix}2\\0\\-1\end{pmatrix},\begin{pmatrix}-1\\0\\0\end{pmatrix}\Bigg\}$$

As we have two linearly independent eigenvectors there is going to be two Jordan blocks, a $1 \times 1$ block corresponding to $u$ and $2\times 2$ block corresponding to $u'$ and $v$. That gives us

$$J =\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$$

Answer 2

You have to find a basis of its eigenspace $\{e_1,e_2\}$ first, then a vector $e_3$ such that $(A-I)e_3=e_2$, which means $Ae_3=e_3+e_2$, hence wrt the basis $\{e_1,e_2,e_3\}$ the matrix has the form:$$\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}.$$

In practice: $(A-I)^2=0$, so $\,\operatorname{im}(A-I)\subset \ker (A-I)$. Take any vector not in the eigenspace as $e_3$, say $\,^{\mathrm t}(1,0,0)$, and $\,e_2=(A-I)e_3= {}^{\mathrm t}(-2,0,1)$; it is an eigenvector.

Complete with an eigenvector $e_1$ independent from $e_2$; since the equation of the eigenspace is $\,x=2(y-z)$, we can take $e_1={} ^{\mathrm t}(2,1,0)$.

In the basis $(e_1,e_2,e_3)$, we have $Ae_3=e_2+e_3$ (by construction), $Ae_2=e_2$, $Ae_1=e_1$, so the matrix of $A$ in that base has the aforementioned form.

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