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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!

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  • $\begingroup$ 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. $\endgroup$ – amd Apr 30 '17 at 20:12
  • $\begingroup$ @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). $\endgroup$ – Matthew Graham Apr 30 '17 at 20:31
  • $\begingroup$ As far as fixing the formatting is concerned, set the entire equation off with dollar instead of picking out individual symbols. $\endgroup$ – amd Apr 30 '17 at 22:59
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Your question is similar to Example 2 in Wikipedia article "Generalized eigenvectors," on which I base my answer.

For a given eigenvalue, the number of chains equals the number of linearly-independent eigenvectors for that eigenvalue. So for your matrix A, each eigenvalue has one chain.

For eigenvalue 2, because the algebraic multiplicity is one, the chain length is one and consists of the corresponding eigenvector x₁ = [–1, 1, 0]ᵀ.

For eigenvalue 3, the algebraic multiplicity is two, but there is only one corresponding eigenvector, so you need to find one more generalized eigenvector to make chain of length two. To do that, solve the matrix equation (A – 3 I)y₂ = y₁ for y₂ where y₁ is your eigenvector [1, 0, 0]ᵀ. You get y₂ = [0, 1, 1]ᵀ.

Hence, you have one chain for each eigenvector, and a chain basis is {x₁, y₁, y₂}.

With the chain basis in that order, the first Jordan block is a one-by-one block for eigenvalue 2, and the second block is a two-by-two block for eigenvalue 3.

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