# Computing the Jordan Form of a Matrix

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!

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 $\pmatrix{3 && 2 && 0 \\ 1 && 1 && 0 \\ 8 && 5 && 1}$, $PAP^{-1}=J_A$.

• 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. – kingdras Dec 11 '15 at 8:23
• @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 $Me_j = a_{1j}e_1+a_{2j}e_2+...+a_{nj}e_n$. – Abstraction Dec 11 '15 at 8:33
• 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. – kingdras Dec 11 '15 at 8:39
• @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. – Abstraction Dec 11 '15 at 8:46
• 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! – kingdras Dec 11 '15 at 8:57

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.)