# Finding a Chain Basis and Jordan Canonical form for a 3x3 upper triangular matrix

I have my matrix A, which is

$$A=\begin{bmatrix} 3 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{bmatrix}$$

and I have been instructed to find a chain basis for it and then put it into Jordan-Canonical form.

I can see that it is upper triangular so finding the eigenvalues isn't hard, you just take the determinant of A - $\lambda$I and set it equal to zero. For that I have:

$det(A - \lambda I) = p(\lambda) = (\lambda - 2)(\lambda - 3)^2$

and then by setting $p(\lambda) = 0$ I find that $\lambda \in \{ 2,3 \}$

so now we find the eigenspaces for each corresponding eigenvalue.

When $\lambda = 2$, the eigenspace is

$$E_{\lambda = 2}=\left\{\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=r\begin{bmatrix}-1\\1\\0\end{bmatrix} : r \in \mathbb{R} \right\}$$

and for $\lambda = 3$, we find the eigenspace to be (I think)

$$E_{\lambda = 3}=\left\{\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=s\begin{bmatrix}1\\0\\0\end{bmatrix} : s \in \mathbb{R} \right\}$$

And now this is where I get lost. I know I need a $\lambda$-chain for $\lambda = 2$ and for $\lambda = 3$, but I don't know how to figure out how long the chains are supposed to be and how many $\lambda$-chains I need for $\lambda = 2$ and $\lambda = 3$ so I can't even proceed to put it in JCF because I'm not sure how to find my chain basis.

Thanks in advance for all the help!

• Actually, since the matrix is upper-triangular, it’s eigenvalues already appear in their algebraic multiplicities along the main diagonal. That should tell you which eigenvalues might need $\lambda$-chains. – amd Apr 30 '17 at 20:12
• @Moo, those examples kind of helped a bit, definitely a good resource, but in my linear algebra class we aren't talking about ODEs so it was hard to connect the material to what we're doing in my LA class. Though I did just have a test this past wednesday in ODE involving eigenvectors/values and there was a 2x2 matrix where I had to do some similar work to theorem 2.2 in that link to get to the general solution (then graph its phase portrait). – Matthew Graham Apr 30 '17 at 20:31
• As far as fixing the formatting is concerned, set the entire equation off with dollar instead of picking out individual symbols. – amd Apr 30 '17 at 22:59