Given the matrix $A = \begin{pmatrix} 7 &1 &2 &2 \\ 1 &4 &-1 &-1 \\ -2 &1 &5 &-1 \\ 1 &1 &2 &8 \end{pmatrix}$. I found its characteristic polynomium, which is $f(x) = (x-6)^4$, and I found this eigenvectors: $(-1,-1,1,0),(-1,-1,0,1)$. My questions are:
- Am I right?
- How can I do to find another matrix $C$ such that $\mathcal{J} = C^{-1}AC$, where $\mathcal{J} = \begin{pmatrix} 6 &1 &0 &0 \\ 0 &6 &0 &0 \\ 0 &0 &6 &1 \\ 0 &0 &0 &6 \end{pmatrix}$
Thanks for your help.
Let $B = A - 6 I$ $$ B=\pmatrix{ 1& 1& 2& 2\\ 1& -2& -1& -1\\ -2& 1& -1& -1\\ 1& 1& 2& 2 }$$ Verify that $B^3=0$ but $B^2 \ne 0$. So it must have a jordon block of size $3$. So you will have one block of size 3 and the other of size 1, i.e of the form $$ \pmatrix{ 6& 1& 0 & 0\\ 0& 6& 1&0\\ 0&0& 6& 0\\ 0& 0& 0& 6 }$$
How to find the transformation matrix?
First find the eigenvectors by solving $$ B e = 0 $$ and you have already done it to get $e_1$ and $e_2$. Now try and solve $$ B e_3 = e_1$$ If this is not possible, try $$ B e_3 = e_2$$ Only one of them is possible.
Next solve $$ B e_4 = e_3 $$ Now the four vectors are the columns of $C$.
Note in all this it goes without saying that $e_i$ should not be zero.
Sorry, I pulled a fast one on this
In general, if you start with $e_1$ and $e_2$ you may not be able to get $e_3$. However, there is one linear combination of $e_1$ and $e_2$ that is in the column space of $B$. That is the one you want to use to get $e_3$. I.e. a vector in the intersection of the space spanned by columns of $B$ and the space spanned by $e_1$ and $e_2$.