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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 3x3) 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:

$$\begin{pmatrix} -7 & 8 & 2\\ -4 & 5 & 1\\ -23 & 21 & 7\\ \end{pmatrix}$$

  1. 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!
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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 $\lambda=1$: $$A-\lambda I = \pmatrix{-8 && 8 && 2 \\ -4 && 4 && 1 \\ -23 && 21 && 6}, \ker(A-\lambda I) = L(\pmatrix{3 \\ 1 \\ 8})$$ 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 $\pmatrix{3 && 2 && 0 \\ 1 && 1 && 0 \\ 8 && 5 && 1}$, $PAP^{-1}=J_A$.

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    $\begingroup$ OK, I have a bunch of stupid questions. First, what does your L stand for when you find (A-$\lambda$I)v? Second, in step 3, you found the matrix in basis, but how did you know it gave us that particular jordan form? Thanks for the help by the way. $\endgroup$ – kingdras Dec 11 '15 at 8:23
  • $\begingroup$ @kingdras $L(v_1, v_2, ..., v_n)$ is linear hull of vectors $v_1, v_2, ..., v_n$ (set of all values of their linear combinations). Also, my bad: for equation we have one vector as a solution (fixed that in the answer). For your second question... let's name my three vectors $v_1, v_2, v_3$. We have $(A-1 I)v_1=0 \sim Av_1=v_1$, $(A-2I)v_2=0 \sim Av_2=2v_2$ and $(A-2I)v_3=v_2 \sim Av_3=v_2+2v_3$. Coefficients $a_{ij}$ of a matrix $M$ in basis $(e_1,e_2, ..., e_n)$ are equivalent to statements $Me_j = a_{1j}e_1+a_{2j}e_2+...+a_{nj}e_n$. $\endgroup$ – Abstraction Dec 11 '15 at 8:33
  • $\begingroup$ OK, I'm getting closer! So we had to find $v_3$ because we needed that other eigenvector, and now understand how we got those numbers after working it out. Sorry if I missed this, but we have $\lambda=1,2$ mult. 2. Why wouldn't my matrix have the diagonal 2 2 1 instead of having 1 2 2? That's what's throwing me off. Is that supposed to pop out of one of our matrices? I know we have those two options, but don't know when to pick which one. $\endgroup$ – kingdras Dec 11 '15 at 8:39
  • $\begingroup$ @kingdras Don't make a mistake, $v_3$ isn't an eigenvector ($Av_3 \neq \lambda v_3$). It's called generalized eigenvector (and we needed it because nullity of $A-2I$ was less then algebraic multiplicity of root $2$). As about the order - it doesn't matter. The only restriction is that generalized eigenvectors must follow specific (generalized) eigenvector they were derived from (that is, $v_3$ must follow $v_2$). In basis $(v_2, v_3, v_1)$ $A$ would have form $J_A'=\pmatrix{2 && 1 && 0 \\ 0 && 2 && 0 \\ 0 && 0 && 1}$ - which is also Jordan form. $\endgroup$ – Abstraction Dec 11 '15 at 8:46
  • $\begingroup$ Oh, I see. So if I picked $v_2$ first, then $v_3$ must follow and then I get 2 2 1. Otherwise, if I picked $v_1$, then $v_2$ is next and $v_3$ follows from that, and so I would get 1 2 2. It's safe to assume then that if I get one independent solution, then I can get 2 generalized eigenvectors that would work the same way, is that correct? This is finally starting to make some sense, what a relief! $\endgroup$ – kingdras Dec 11 '15 at 8:57
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If you are not interested in computing $P$, then the Jordan form can be computed by using this:

  1. The number of Jordan blocks with diagonal entry as $\lambda$ is the geometric multiplicity of $\lambda$.

  2. The number of Jordan blocks of order $k$ with diagonal entry $\lambda$ is given by $rank(A-\lambda I)^{k-1}-2\, rank(A-\lambda I)^k + rank(A-\lambda I)^{k+1}.$

Here, the geometric multiplicities of $\lambda =1,2$ are each $1.$ And $1$ has algebraic multiplicity $1$ where as of $2$ the algebraic multiplicity is $2.$ So, using the condition (1) only, we see that there is a Jordan block of order $1$ with $\lambda=1$ and one Jordan block with $\lambda=2.$. So, the Jordan form is as computed above. (of course, upto a permutation of the Jordan blocks.)

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