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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?

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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

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

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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.

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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)
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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)

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