Find $P$ such that $P^{-1}AP = J$

Question

Let $$A = \begin{bmatrix} 1 & 1 & 0 & -1 \\[0.3em] 0 & -1 & 1 & 2 \\[0.3em] -1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

I already found that the Jordan form of $A$ is $$J = \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.3em] 1 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Now, I wish to find an invertible $P$ such that $P^{-1}AP = J$. I know there are several ways doing it. One way is finding a basis of $V$ consisting linearly independent eigen-vectors.

So $$V_{\lambda=0} = \text{sp}\{(-1,1,1,0)\} \\ V_{\lambda=1} = \text{sp}\{(-1,1,0,1)\} $$

Now, I am lacking of two vectors to complete a basis of $V=\mathbb{R}^4$.

What should I do in this kind of scenario?

Thanks.

EDIT
$A$ should be

$$ \begin{bmatrix} 1 & 1 & 0 & -1 \\[0.3em] 0 & -1 & 1 & 2 \\[0.3em] -1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Answer 1

The characteristic polynomial (and the minimal polynomial of $A$, as one can check) is $\chi_A(x)=x^4$.

Thus $\ker A^4=\mathbf R^4$. One checks $\DeclareMathOperator{\rk}{rank}\; \rk A=3,\ \rk A^2=2,\ \rk A^3=1 $. Hence the Jordan normal form is: $$J=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}.$$ To find a change of basis matrix $P$ such that $J=P^{-1}AP$, we have to find a basis of generalised eigenvectors.

For that, we take a vector $v$ that is in $\mathbf R^4\smallsetminus \ker A^3$. As $$A^3=\begin{bmatrix}0&0&0&-1\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{bmatrix},$$ we see a choice is $\;v_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$. Then: $$v_3=Av_4=\begin{bmatrix}-1\\2\\0\\0\end{bmatrix},\enspace v_2=Av_3=\begin{bmatrix}1\\-2\\-1\\0\end{bmatrix},\enspace\text{and finally} \enspace v_1=Av_2=\begin{bmatrix}-1\\1\\1\\0\end{bmatrix}$$ The change of basis matrix is:

$$P=\begin{bmatrix}-1&1&-1&0\\1&-2&2&0\\1&-1&0&0\\0&0&0&1\end{bmatrix}.$$

Answer 2

Hint:

The eigenvalue $\lambda =0$ has algebraic multiplicity $4$ and only one eigenvector $v_1\ne 0 =(-1,1,1,0)^T$. So you needs three generalized eigenvectors that you can chose as: $$ v_2=(0,-1,0,0)^T $$ $$ v_3=(-1,1,0,0)^T $$ $$ v_4=(-1,1,0,1)^T $$

Answer 3

You need to look at the Generalized Eigenvectors of this matrix. See what the algebraic and geometric multiplicities and find the generalized eigenvectors for the ones where the g.m < a.m.

That will give you the full set of basis for $ \mathbb R ^4$.

B.t.w the J matrix you posted is not the same returned by MATLAB.

Matlab gives the following:

(1)

A =

1     1     0    -1
0    -1     1     2
-1    -1     0     0
0     0     0     0


jordan(A)
ans =

 0     1     0     0
 0     0     1     0
 0     0     0     1
 0     0     0     0

Now to find the full set of vectors:

You get one vector from : (same as in your question)

[V,D] = eig(sym(A))

V =

-1
1
1
0


D =

[ 0, 0, 0, 0]
[ 0, 0, 0, 0]
[ 0, 0, 0, 0]
[ 0, 0, 0, 0]

and follow the steps in the link below:

Finding generalized Eigenvectors

For theory, refer to the download links for notes present on this link:

Nilpotent Matrices/Jordan canonical form

The idea of JCF comes from theory of Nilpotent matrices. (Annihilators)