I apologize if this has already been answered, but I've seen multiple examples of how to compute Jordan Canonical Forms of a matrix, and I still don't really get it. Could someone help me out with this?
What I know for certain is that I must start off by finding my eigenvalues, and corresponding eigenvectors. OR, (how it was taught in class from my understanding), I can simply plug in the eigenvalues into my original matrix and find the rank. I have no clue what to do from there though… I also know that my Jordan Normal Forms should look like these:
$$\begin{pmatrix}
\lambda_1 & 0 & 0\\
0 & \lambda_2 & 0\\
0 & 0 & \lambda_3\\
\end{pmatrix}$$
or
$$\begin{pmatrix}
\lambda_1 & 1 & 0\\
0 & \lambda_1 & 0\\
0 & 0 & \lambda_2\\
\end{pmatrix}$$
And if we switch 1 and 2, then the 1 will be on the other side of the top. Lastly,
$$\begin{pmatrix}
\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda\\
\end{pmatrix}$$
I've seen from many sources that if given a matrix J (specifically 3×3) that is our Jordan normal form, and we have our matrix A, then there is some P such that $PAP^{-1}=J$.
Here's an example matrix if I could possibly get an explanation on how this works through an example:
$$\begin{pmatrix}
-7 & 8 & 2\\
-4 & 5 & 1\\
-23 & 21 & 7\\
\end{pmatrix}$$
- I don't know how to fill the information in the middle. For instance, what do I do after I find the rank of my matrix or what do I do once I find my rank? Sorry if I made mistakes, very tired, and please try to make this as coherent as possible, because I'm so confused. This is an Advanced Linear Algebra course. Any help is greatly appreciated!
Best Answer
Step 1: find eigenvalues. $\chi_A(\lambda) = \det(A-\lambda I) = -\lambda^3+5\lambda^2-8\lambda+4 = -(\lambda-1)(\lambda-2)^2$. We are lucky, all eigenvalues are real.
Step 2: for each eigenvalue $\lambda_\imath$, find rank of $A-\lambda_\imath I$ (or, rather, nullity, $\dim(\ker(A-\lambda_\imath I))$) and kernel itself. For $\lambda=1$: $$A-\lambda I = \pmatrix{-8 && 8 && 2 \\ -4 && 4 && 1 \\ -23 && 21 && 6}, \ker(A-\lambda I) = L(\pmatrix{3 \\ 1 \\ 8})$$ ($L(v_1, v_2, ..., v_n)$ denotes the linear hull of vectors, the set of all their linear combinations.) Algebraic multiplicity of the root is 1, geometric multiplicity is 1, we're done here. For $\lambda=2$: $$A-\lambda I = \pmatrix{-9 && 8 && 2 \\ -4 && 3 && 1 \\ -23 && 21 && 5}, \ker(A-\lambda I) = L(\pmatrix{2 \\ 1 \\ 5})$$ Algebraic multiplicity of the root is 2, geometric multiplicity is 1. We're unlucky, now we have to solve $$(A-\lambda I)v=\pmatrix{2 \\ 1 \\ 5} \sim v = \pmatrix{0 \\ 0 \\ 1}$$ Step 3: our matrix in basis $(\pmatrix{3 \\ 1 \\ 8},\pmatrix{2 \\ 1 \\ 5},\pmatrix{0 \\ 0 \\ 1})$ has form $J_A = \pmatrix{1 && 0 && 0 \\ 0 && 2 && 1 \\ 0 && 0 && 2}$. Matrix $P$ corresponding to this basis change is $\pmatrix{3 && 2 && 0 \\ 1 && 1 && 0 \\ 8 && 5 && 1}$, i.e. $P^{-1}AP=J_A$.
Note: If you have a root of algebraic multiplicity 3, but there's only one eigenvector $v_1$, then you seek $v_2:(A−λI)v_2=v_1$ and then $v_3:(A−λI)v_3=v_2$ (note the index!). But when the nullity of $A−λI$ is greater than 1 (and less than algebraic multiplicity), things get a bit tricky. You have to find maximal $k : (A−λI)^k≠0$, then find vector(s) $v_k:(A−λI)^k v_k=0,\,(A−λI)^{k−1}v_k≠0$ (chain generators) and then proceed $v_{k−1}=(A−λI)v_k,v_{k−2}=(A−λI)v_{k−1},...$ up to an eigenvector $v_1$ (see "Jordan chains").