# Finding the Jordan Basis and the Canonical form corresponding to the Jordan basis

\begin{pmatrix} 4 & 0 &0 \\ 2 &1 &3 \\ 5& 0 &4 \end{pmatrix} I know that the Characteristic polynomial is : $$(t-4)^2(t-1)$$ I started with eigenvalues $λ=1$ and got in the null space: \begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}

$λ=4$ and got :

\begin{pmatrix} 0\\ -1 \\ 1 \end{pmatrix},

in $A- 4I$ and got

\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix},

in $(A- 4I)^2$

Which vector(s) do I choose correctly for the Jordan Basis?

I chose $[0,1,0],[1,0,1]$ and got $b_3$ as $A-4I[b_2] = [0,5,5]$

What procedures should I use for choosing the correct Jordan basis vectors? Secondly am I right?

Lastly, How do I get the Canonical form "corresponding to the basis"?

is it \begin{pmatrix} 1 & 0 &0 \\ 0 &4 &1 \\ 0& 0 &4 \end{pmatrix} or \begin{pmatrix} 4 & 1 &0 \\ 0 &4 &0 \\ 0& 0 &1 \end{pmatrix} How do I set it up correctly?

• Thats only for the canonical form is it not? – GuestCalc Dec 19 '15 at 2:06
• even better wolframalpha.com/input/… to guide oneself – janmarqz Dec 19 '15 at 2:09

You have the right idea, but you made a slight computational error: the $4$-eigenspace of $A$ is spanned by $(0,1,1)$, not $(0,-1,1)$. You correctly picked a vector $(1,0,1) \in \ker (A-4I)^2$ that doesn't belong to $\ker(A-4I)$, which let you construct the Jordan chain $((1,0,1),(A-4I)(1,0,1)) = ((1,0,1),(0,5,5))$. (This now works because $(0,5,5)$ is an eigenvector for the eigenvalue $4$.)

Then you just place the Jordan chains as columns of the change of basis matrix (reading a chain backwards if you want $1$s on the superdiagonal):

$$P = \begin{pmatrix} 0 & 0&1\\ 1&5&0 \\ 0&5&1 \end{pmatrix}.$$

And then we get

$$P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{pmatrix}.$$

$$P = \begin{pmatrix} 0&1&0\\ 5&0&1 \\ 5&1&0 \end{pmatrix},$$
$$P^{-1}AP = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
• for $(A-4I)^2$ both vectors were not in the null space of A-4I, would either work? – GuestCalc Dec 19 '15 at 2:13
• @GuestCalc The other vector $(-1,1,0)$ would also work; you would get the chain $((-1,1,0),(0,-5,-5))$. – Alex Provost Dec 19 '15 at 2:17
• @GuestCalc The ordering of the blocks is arbitrary and depends on the ordering you chose when placing the Jordan chains as columns of $P$. – Alex Provost Dec 19 '15 at 2:17