# Finding the Jordan form and basis for a matrix

I have this matrix $$A=\left[ {\begin{array}{ccc} -3&3&-2\\ -7&6&-3\\ 1&-1&2\\ \end{array} } \right]$$

I computed the characteristic polynomial $$C_A(t)=-(t-2)^2(t-1)$$

When I go to try and find the eigenvectors and generalized eigenvectors I compute $$A-2I=\left[ {\begin{array}{ccc} -5&3&-2\\ -7&4&-3\\ 1&-1&0\\ \end{array} } \right]$$

Using row reduction $$\left[ {\begin{array}{ccc} 1&-1&0\\ 0&1&1\\ 0&0&0\\ \end{array} } \right]x=0$$ gives me eigenvectors having the form $$e=r\left[ {\begin{array}{c} 1\\ 1\\ -1\\ \end{array} } \right]$$

$$v=\left[ {\begin{array}{c} 1\\ 1\\ -1\\ \end{array} } \right]$$ Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.

At this point I know the Jordan form is $$J=\left[ {\begin{array}{ccc} 2&1&0\\ 0&2&0\\ 0&0&1\\ \end{array} } \right]$$

Now to find the generalized eigenvector I can take an augmented matrix $$\left(\begin{array}{ccc|c} 5&3&-2&1\\ -7&4&1&1\\ 1&-1&0&-1 \end{array}\right)$$ I row reduced this to the form $$\left(\begin{array}{ccc|c} 1 & -1 & 0&-1\\ 0 & 1 & 1&2\\ 0&0&0&0 \end{array}\right)$$

So I need a vector that satisfies $$x_1-x_2=-1$$ and $$x_2+x_3=2$$ I let $$x_3=1$$

Which gives me a vector $$x=\left(\begin{array}{c} 0\\ 1\\ 1 \end{array}\right)$$ which seems to work, so I have a basis for the generalized eigenspace for $$\lambda=2$$

then I need to find a solution for $$(A-I)x=0$$ for $$\lambda=1$$.

I row reduce $$A-I=\left[\begin{array}{ccc} -4&3&-2\\ -7&5&1\\ 1&-1&1 \end{array}\right]$$ to get $$\left[\begin{array}{ccc} 1&0&-1\\ 0&1&-2\\ 0&0&0 \end{array}\right]$$ and using $$(A-I)x=0$$ found that all eigenvectors for $$\lambda=1$$ have the form $$v=s\left(\begin{array}{c} 1\\ 2\\ 1 \end{array}\right)$$

So I can make a matrix $$Q=\left[\begin{array}{ccc} 1&1&0\\ 2&1&1\\ 1&-1&1 \end{array}\right]$$ s.t $$A=QJQ^{-1}$$ where the columns of $$Q$$ are a jordan basis for $$J$$

• I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal? – A. Pongrácz Dec 8 '18 at 17:38
• Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1. – AColoredReptile Dec 8 '18 at 17:42
• Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.) – A. Pongrácz Dec 8 '18 at 17:48
• I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$. – AColoredReptile Dec 8 '18 at 17:51
• You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2\times 2$ block, which it might not...) – A. Pongrácz Dec 8 '18 at 17:54

## 1 Answer

Because you are looking for a Jordan canonical basis, $$Ax = 2x+v$$, where $$x$$ is the generalized eigenvector you're looking for. Equivalently, $$(A-2I)x = v$$. Applying $$A-2I$$ one more time to each side will give you the following statement: $$(A-2I)^{2}x = 0$$.

So, your generalized eigenvector is in the nullspace of $$(A-2I)^2$$, and it is not in the nullspace of $$(A-2I)$$. Can you finish from there?

• I believe I finished. – AColoredReptile Dec 8 '18 at 18:31