find the Jordan form and $P$ such that $P^{-1}AP = J$.

Question

Consider the matrix $$A = \left(\begin{array}{cccc} -11&0&-9\\32&1&24\\16&0&13 \end{array}\right)$$

I want to find the Jordan form of $A$, with $1$-s at the bottom and the jordan basis, which is $P$ columns such that $P^{-1}AP = J$.

I evaluated the charechteristic polynomial which is $f_A(x) = -(x-1)^3$. So we have one eigenvalue which is $\lambda =1$.

Now,

$$A-I = \left(\begin{array}{cccc} -12&0&-9\\32&0&24\\16&0&12 \end{array}\right)$$ $$\ker (A-I) = \text{span}\{(0,1,0),(1,0,3/4)\}$$ $$\ker (A-I)^2 =\mathbb{R}^3$$

As I understand I need to find a vector, $v_3$ such that $\ker (A-I) \cup v_3 = \ker (A-I)^2 = \mathbb{R}^3$.

Then, $$v_2 = (A-I)v_3 \\ v_1 = (A-I)v_2$$

And we (suppose to) get $P = (v_1,v_2,v_3)$ such that $P^{-1}AP = J$.

I chose $v_3 = (1,0,0)$, but it doesn't work, and for the matter of fact I'm confused by all this.

I'd be glad if you could explain to me how to correct this.

Thanks.

Answer 1

Saying that 1 is an eigenvalue for matrix $A$ means that there exist non-zero vectors, $v$, such that $Av= v$ or $(A- I)v= 0$.

Here, that means $$\begin{bmatrix}-12 & 0 & -9 \\ 32 & 0 & 24 \\ 16 & 0 & 12\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\begin{bmatrix}-12x- 9z \\ 32x+ 24z \\ 16x+ 12z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$

You basically have that. Now note that this gives you the three equations $$-12x- 9z= 0,$$ $$32x+ 24z= 0,$$ and $$16x+ 12z= 0.$$ From the first equation, $9z= -12x$ so $z= (-12/9)x= (-4/3)x.$ Setting $z= (-4/3)x$ in the second and third equations gives $$32x+ 24(-4/3)x= 32x- 32x= 0$$ and $$16x+ 12z= 16x+ 12(-4/3)x= 16x- 16x= 0.$$
That is, for any value of $x, z= (-4/3)x$ will satisfy all those equations. Since $y$ did not enter into this, $y$ can be anything. That is, the eigenvectors must be of the form $$\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix} x \\ y \\ (-4/3)x \end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ -4/3\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}.$$

We can use $\begin{pmatrix}3\\ 0\\ -4\end{pmatrix}=3\cdot \begin{pmatrix}1\\ 0\\ -4/3\end{pmatrix}$ and $\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}$ form a basis for the Eigen space. since that is a two dimensional subspace, the "Jordan Normal Form" will be of the form $$J= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 0 \\0 & 0 & 3\end{bmatrix}$$ and we can form the matrix $$P^{-1}$$ using the vectors $$\begin{pmatrix}3 \\ 0\\ -4\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$$ as columns.

Answer 2

You've got a lot of the right ingredients here. Let's put them together by following the algorithm described by Stefan Friedl here.

The characteristic polynomial of $A$ is $$ \chi_A(t)=-(t-1)^3 $$ so the only eigenvalue of $A$ is $\lambda=1$ with algebraic multiplicity $m=3$. Note that \begin{align*} A-I&=\begin{bmatrix}-12&0&-9\\ 32&0&24\\16&0&12\end{bmatrix}& (A-I)^2&=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} \end{align*} so $\DeclareMathOperator{null}{null}$ \begin{align*} \dim\null(A-I) &= 2 & \dim\null(A-I)^2&=3 \end{align*} We then compute the numbers \begin{align*} d_1 &= \dim\null(A-I) & d_2 &= \dim\null(A-I)^2-\dim\null(A-I) \\ &= 2 & &= 3-2 \\ &&&= 1 \end{align*} so we must fill the boxes $$ \begin{matrix} \Box & \Box \\ \Box \end{matrix} $$ with vectors. Note that $u=\begin{bmatrix}1&0&0\end{bmatrix}^\top$ satisfies $u\in\null(A-I)^2$ but $u\notin\null(A-I)$. Put $$ v=(A-I)u=\begin{bmatrix}-12\\32\\16\end{bmatrix} $$ so the diagram takes the form $$ \begin{matrix} v & \Box \\ u \end{matrix} $$ Since $w=\begin{bmatrix}0&1&0\end{bmatrix}^\top$ is linearly independent from $v$ we complete the Jordan basis $\{u,v,w\}$. Finally, put $$ P=\begin{bmatrix} -12 & 1 & 0 \\ 32 & 0 & 1 \\ 16 & 0 & 0 \end{bmatrix} $$ and note that the Jordan form of $A$ is $$ J=P^{-1}AP=\begin{bmatrix}1&1&0\\ 0&1&0\\ 0&0&1\end{bmatrix} $$ Now, if we want the $1$'s on the superdiagonal to be on the subdiagonal, we need only reverse the order of the columns of $P$. Indeed, putting $$ Q=\begin{bmatrix}0&1&-12\\ 1 & 0 & 32\\ 0&0&16\end{bmatrix} $$ gives $$ Q^{-1}AQ=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&1&1\end{bmatrix} $$

Answer 3

As $\ker(A-I)^2=\mathbf R^3$, and $\dim\ker (A-I)=2$, all you have to do is to take a vector $v_1\in \ker(A-I)^2\smallsetminus\ker (A-I)$. Then $v_2=(A-I)v_1$ is a non-zero vector of the eigenspace ; you can complete this vector in a basis of the eigenspace.

As the equation of the eigenspace is $4x+3y=0$, we can take $e_1={}^{\mathrm t }(-1,0,1)$. Then $e_2=(A-I)e_1={}^{\mathrm t}(3,-8,-4)$. Complete the basis with, say $e_3={}^{\mathrm t}(0,1,0)$. In this basis, by construction the matrix of the endomorphism and the change of basis matrix are: $$P^{-1}AP=\begin{bmatrix}1&0&0\\1&1&0\\0&0&1\end{bmatrix}, \quad P=\begin{bmatrix}-1&3&0\\0&-8&1\\1&-4&0\end{bmatrix}.$$