# Finding Jordan Basis of a nilpotent matrix

I have to find the basis of the nilpotent matrix $A$ of size $3\times3$. $$\left(\begin{matrix} 1&1&1\\-1&-1&-1\\1&1&0\end{matrix}\right)$$

I found that

• $A^3=0$
• $ker A^0={0}$
• $ker A^1=Span \left(\begin{matrix} -1\\1\\0\end{matrix}\right)$
• $ker A^2=Span \left(\begin{matrix} 0\\0\\1\end{matrix},\begin{matrix} -1\\1\\0\end{matrix} \right)$

I don't know how to work from here, can you please show me how to use the algorithm for finding the Jordan basis of a nilpotent matrix?

I like this order: you know the minimal polynomial is $\lambda^3.$ So, first you choose (it makes no genuine difference) your favorite column vector $w$ such that $$A^2 w \neq 0.$$ Once that choice is made, everything else is forced. We calculate $$v = Aw,$$ $$u = a v$$ Then the change of basis matrix $R$ that goes on the right will have columns, in order, $(u,v,w)$ I like $$w = \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array} \right)$$ then we get

$$R = \left( \begin{array}{ccc} 1&1&1 \\ -1&-1&0 \\ 0&1&0 \\ \end{array} \right)$$ and $$R^{-1} = \left( \begin{array}{ccc} 0&-1&-1 \\ 0&0&1 \\ 1&1&0 \\ \end{array} \right)$$ and

$$\left( \begin{array}{ccc} 0&-1&-1 \\ 0&0&1 \\ 1&1&0 \\ \end{array} \right) \left( \begin{array}{ccc} 1&1&1 \\ -1&-1&-1 \\ 1&1&0 \\ \end{array} \right) \left( \begin{array}{ccc} 1&1&1 \\ -1&-1&0 \\ 0&1&0 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1&0 \\ 0&0&1 \\ 0&0&0 \\ \end{array} \right)$$

and the order we actually use for calculating things

$$\left( \begin{array}{ccc} 1&1&1 \\ -1&-1&0 \\ 0&1&0 \\ \end{array} \right) \left( \begin{array}{ccc} 0&1&0 \\ 0&0&1 \\ 0&0&0 \\ \end{array} \right) \left( \begin{array}{ccc} 0&-1&-1 \\ 0&0&1 \\ 1&1&0 \\ \end{array} \right) = \left( \begin{array}{ccc} 1&1&1 \\ -1&-1&-1 \\ 1&1&0 \\ \end{array} \right)$$

There is going to be one chain of vectors belonging to the Eigenvalue 0. For the first vector pick anything from $$Ker^{1}$$ and then try to find a vector in Im A that is projected on your first chosen vector and then repeat the process.

HINT

• find the eigenvalues and corresponding eigenvectors
• determine the Jordan canonical form $J$ for $A$
• from the condition $AP=PJ$ find the Jordan basis(i.e. column vectors of P)

You have a true eigenvector to eigenvalue $0$: $v_1 = [1,-1,0]^T$.

Which vector maps onto it? $A^{-1}v_1$ does maybe you found it $v_2=[0,0,1]^{T}$

Now which maps on that one? (Hint: will be same procedure)