Jordan form of the matrix $\left(\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{smallmatrix}\right)$ 
Determine the Jordan form of the matrix $A= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}.$ 

I've calculated the characteristic polynomial $\det(\lambda I-A)= (\lambda-1) \left| \begin{array}{ccc} \lambda - 1 & 0 \\ 1& \lambda - 1 \\ \end{array} \right| - 1 \left| \begin{array}{ccc} 0 & 0 \\ 0& \lambda - 1 \\ \end{array}  \right|= (\lambda-1)^3$. So there are $3$ eigenvalues all equal to $1$. I found the null space corresponding to the $3$ eigenvalues $\text{null}(I-A)= \text{span} \left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right)$, in attempts to get a matrix $P$ such that $J=P^{-1}AP$ where $J$ is the Jordan form of $A$. How do I proceed? I am not sure what to do with the null space, or how to find the eigenvectors
 A: There are various things to do for different matrices.  Way too long to explain it all, so I will just explain the easiest way to do this specific example.


*

*Find the generalised eigenspace.  Since $A$ has only one eigenvalue, there is only one generalised eigenspace and it must be $\Bbb R^3$.

*Find lengths of chain(s) of generalised eigenvectors.  The number of chains will be the number of (independent) eigenvectors, in this case $2$.  The total length of the chains will be the multiplicity of the eigenvalue, in this case $3$.  So there will be a chain of length $2$ and a chain of length $1$.

*Find the actual chains.  The chain of length $2$ must start with a generalised eigenvector which is not an eigenvector.  In this case, any element of $\Bbb R^3$ which is not an eigenvector, for example,
$$\def\v#1{{\bf#1}}\v v_1=\pmatrix{0\cr1\cr0\cr}\ .$$
To continue the chain, multiply by $A-\lambda I$: we have
$$\v v_2=(A-\lambda I)\v v_1=(A-I)\v v_1=\pmatrix{1\cr0\cr1\cr}\ .$$
It is advisable to check that $(A-\lambda I)\v v_2=\v0$.  The chain of length $1$ will be just an eigenvector, and must be independent of the eigenvector we have already.  There are many choices, for example,
$$\v v_3=\pmatrix{1\cr0\cr0\cr}\ .$$

*Put everything together to find $P$ and $J$.  The columns of $P$ will be the vectors $\v v_1,\v v_2,\v v_3$, but it is important to respect the order of these vectors in chains.  The first column must be the end of the chain so in this case $\v v_2$ must come before $\v v_1$.  Since $\v v_3$ is in a separate chain it can come before or after these two.  In $J$, a chain of length $k$ produces a Jordan block of size $k$.  So there are two options:
$$P=(\v v_2\ \v v_1\ \v v_3)=\pmatrix{1&0&1\cr0&1&0\cr1&0&0\cr}\ ,\quad
  J=\pmatrix{1&1&0\cr0&1&0\cr0&0&1\cr}$$
or
$$P=(\v v_3\ \v v_2\ \v v_1)=\pmatrix{1&1&0\cr0&0&1\cr0&1&0\cr}\ ,\quad
  J=\pmatrix{1&0&0\cr0&1&1\cr0&0&1\cr}\rlap{\ .}$$


Note that this is a very early example and you will not learn much until you have done some harder ones.  Good luck!
A: This particular example is much simpler than requires a general approach. You can see from the matrix that both $\begin{bmatrix}1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}0\\0\\1\end{bmatrix}$ are fixed, and the matrix is not the idetity matrix. Therefore the normal form has to have two fixed vectors in two columns, and the third column has to complete a Jordan block: $$\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$$
