# Jordan Canonical Form of matrix

I am having trouble figuring out computing Jordan Canonical Form. Can someone explain how to get there with this example matrix?

$A=\begin{bmatrix}1&1&1\\0&2&0\\0&0&2\end{bmatrix}$

Also, what would the transformation matrix $D$ be, if $D^{-1}AD$ is in Jordan Form.

• What did you try? – egreg Dec 16 '13 at 22:11

We have:

$$A = \begin{bmatrix}1&1&1\\0&2&0 \\ 0&0&2\end{bmatrix}$$

We find the eigenvalues of $|A - \lambda I| = 0$, hence:

$$\lambda_1 = 1, \lambda_{2,3} = 2$$

That is, we have a single root and a double root eigenvalue, algebraic multiplicity.

To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$.

For $\lambda_1 = 1$, we get the eigenvector:

$$v_1 = (1,0,0)$$

For $\lambda_{2,3} = 2$, we get the eigenvectors (normally, we do not get two linearly independent eigenvectors):

$$v_2 = (1,0,1), v_3 = (1,1,0)$$

We now can write $P$ using the eigenvectors as columns. We have,

$$P = [v_1 | v_2 | v_3 ] = \begin{bmatrix}1&1&1\\0&0&1 \\ 0&1&0\end{bmatrix}$$

We can write the Jordan Normal Form (notice that we do not have any Jordan blocks), $J$, using the corresponding eigenvalues:

$$J = P^{-1} A P$$

However, we can also write this straight off from the eigenvalues and knowing we do not need any Jordan blocks.

$$J = \begin{bmatrix}1&0&0\\0&2&0 \\ 0&0&2\end{bmatrix}$$

Lastly, we should verify:

$$A = P J P^{-1}$$

I purposely left things so you can fill in the details of the calculations.

• So, just a quick question. Why did it come out this way so that matrix J, which is the jordan form (right?), has no one's on the super diagonal? – Eddi Dec 16 '13 at 22:25
• Because we were able to write two linearly independent eigenvectors. If we were not able to do that, we would have had a Jordan Block. I would review the link to understand the mathematical reasons why! Please recall to upvote and/or accept answers that are helpful. Regards – Amzoti Dec 16 '13 at 22:30
• Needs another UV! +1 – Namaste Dec 17 '13 at 0:10