# How to compute the change of basis matrix that conjugate a matrix to its Jordan canonical form

Let's make an example: $$A=\left(\begin{array}{rrr} 1&1&1\\-1&-1&-1\\1&1&1\end{array}\right).$$ I calculated that the Jordan canonical form is $J=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$. How can I find a matrix $P$ such that $P^{-1}AP=J$. Could you tell me also how can I find such a matrix in general? In the sense, if I have a matrix $A$ and its Jordan canonical form is $J$, then how can I find $P$ such that $P^{-1}AP=J$?

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Check out this and this (especially starting from pp. 135 & Example 7.2.3). – user2468 Jan 6 '12 at 1:41
Have you found the eigenvalues and eigenvectors of $A$? In this case, that is all you need. In general, you will need to consider the null spaces of $A - \lambda I$ for the different eigenvalues $\lambda$ of $A$. – Michael Joyce Jan 6 '12 at 1:41

Here, because the Jordan canonical form is diagonal, you want to find two linearly independent vectors that are mapped to $0$ (i.e., a basis for the nullspace); here, it can be done by inspection: $(1,-1,0)^t$ and $(1,0,-1)^t$; and a vector that is mapped to itself. Normally you would look for the nullspace of $A-I$, but here you can again do it by inspection: $(1,-1,1)^t$.
Your matrix $P$ will have these vectors as its columns: $$P = \left(\begin{array}{rrr}1&1&1\\ -1 & 0 & -1\\ 0 & -1 & 1 \end{array}\right).$$ And $P^{-1}$ is the inverse of this matrix.
In the more general case that you have several blocks corresponding to the same eigenvalue, of different sizes, you look for the nullspaces of $A-\lambda I$, $(A-\lambda I)^2$, $(A-\lambda I)^3$, and so on. To get a Jordan block of size 3, you start with a vector $\mathbf{v}$ in the nullspace of $(A-\lambda I)^3$ but not in the nullspace of $(A-\lambda I)^2$, and your basis will consist of $(A-\lambda I)^2\mathbf{v}$, $(A-\lambda I)\mathbf{v}$, and $\mathbf{v}$.